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forward
ptproj-
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audioerror
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FcVolume
default
update
buttonUp
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V, #>
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audioerror
playing
FcVolume
default
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default
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audioerror
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FcVolume
BPS-3-
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forward
ptView-10
paused
ptView-
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audioerror
playing
FcVolume
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forward
repeat
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EV&TS-3-
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IntL&P-2-
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Dvpmt(Pr)-2-
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forward
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default
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forward
IntP&S-2-
IntP&S-2-15
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audioerror
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repeat
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EV&TS-1-
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playing
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.&, "
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forward_silent
update
LinVis-3-
default
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repeat_silent
fwd_slnt
update
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.&, "
.&, "
.&, "
LinVis-4-
forward_silent
update
default
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repeat_silent
fwd_slnt
update
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.&, "
.&, "
forward_silent
update
IntS&S-2-
default
buttonUp
repeat_silent
fwd_slnt
update
false
Dvpmt(Pr)-2-2
Dvpmt(Pr)-2-4
audioerror
Dvpmt(Pr)-2-3
enterPage
Dvpmt(Pr)-2-2
Dvpmt(Pr)-2-4
audioerror
Dvpmt(Pr)-2-3
leavePage
IntS&S-4-3
false
IntS&S-4-2
IntS&S-4-4
audioerror
IntS&S-4-5
IntS&S-4-6
enterPage
IntS&S-4-2
IntS&S-4-4
IntS&S-4-3
audioerror
IntS&S-4-5
IntS&S-4-6
leavePage
IntS&S-3-4
false
IntS&S-3-5
IntS&S-3-3
audioerror
IntS&S-3-2
enterPage
IntS&S-3-5
IntS&S-3-3
IntS&S-3-4
audioerror
IntS&S-3-2
leavePage
IntP&S-2-14
IntP&S-2-9
IntP&S-2-2
IntP&S-2-4
false
IntP&S-2-11
IntP&S-2-10
IntP&S-2-15
IntP&S-2-8
IntP&S-2-5
IntP&S-2-7
IntP&S-2-6
IntP&S-2-13
IntP&S-2-3
audioerror
IntP&S-2-12
enterPage
IntP&S-2-14
IntP&S-2-9
IntP&S-2-2
IntP&S-2-4
IntP&S-2-11
IntP&S-2-10
IntP&S-2-15
IntP&S-2-8
IntP&S-2-5
IntP&S-2-7
IntP&S-2-6
IntP&S-2-13
IntP&S-2-3
audioerror
IntP&S-2-12
leavePage
IntP&P(v)-2-7
IntP&P(v)-2-6
IntP&P(v)-2-3
false
IntP&P(v)-2-12
IntP&P(v)-2-9
IntP&P(v)-2-2
IntP&P(v)-2-4
IntP&P(v)-2-11
audioerror
IntP&P(v)-2-10
IntP&P(v)-2-8
IntP&P(v)-2-5
enterPage
IntP&P(v)-2-6
IntP&P(v)-2-3
IntP&P(v)-2-12
IntP&P(v)-2-9
IntP&P(v)-2-2
IntP&P(v)-2-4
IntP&P(v)-2-11
audioerror
IntP&P(v)-2-10
IntP&P(v)-2-8
IntP&P(v)-2-5
IntP&P(v)-2-7
leavePage
IntP&P(v)-1-10
IntP&P(v)-1-8
IntP&P(v)-1-5
false
IntP&P(v)-1-7
IntP&P(v)-1-6
IntP&P(v)-1-3
audioerror
IntP&P(v)-1-9
IntP&P(v)-1-2
IntP&P(v)-1-4
IntP&P(v)-1-11
enterPage
IntP&P(v)-1-10
IntP&P(v)-1-8
IntP&P(v)-1-5
IntP&P(v)-1-7
IntP&P(v)-1-6
IntP&P(v)-1-3
audioerror
IntP&P(v)-1-9
IntP&P(v)-1-2
IntP&P(v)-1-4
IntP&P(v)-1-11
leavePage
DiAng-8
DiAng-5
DiAng-7
DiAng-6
false
DiAng-3
DiAng-9
DiAng-2
audioerror
DiAng-4
DiAng-11
DiAng-10
enterPage
DiAng-8
DiAng-5
DiAng-7
DiAng-6
DiAng-3
DiAng-9
DiAng-2
audioerror
DiAng-4
DiAng-11
DiAng-10
leavePage
false
ptView-9
ptView-2
ptView-4
ptView-10
ptView-8
ptView-5
audioerror
ptView-7
ptView-6
ptView-3
enterPage
ptView-9
ptView-2
ptView-4
ptView-10
ptView-8
ptView-5
audioerror
ptView-7
ptView-6
ptView-3
leavePage
LinVis-2
audioerror
leavePage
false
LinVis-2
audioenable
audioerror
enterPage
showVolume
false
pause
clearVolume
repeat
wavplayable
audioenable
showVolume
false
pause
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repeat
wavplayable
audiodisable
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update
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IntP&P(r)-1-
paused
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IntP&P(r)-1-8
audioerror
playing
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default
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forward
repeat
update
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}gyieldApp
10BasGeom-1
paused
audioerror
playing
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default
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10BasGeom-1
p6-105
paused
audioerror
playing
leavePage
false
audioenable
10BasGeom-1
p6-105
audioerror
wavPlayable
enterPage
showVolume
false
pause
clearVolume
repeat
Geometric1
Geometric1_silent
wavplayable
audioenable
showVolume
false
pause
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repeat
Geometric1
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wavplayable
audiodisable
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p8-105
paused
audioerror
10BasGeom-3
playing
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default
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10BasGeom-4
paused
audioerror
playing
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default
p9-105
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p10-105
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paused
10BasGeom-5
audioerror
playing
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default
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p11-105
}gyieldApp
paused
10BasGeom-6
audioerror
playing
FcVolume
default
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paused
default
10BasGeom-7
audioerror
playing
FcVolume
p12-105
buttonUp
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10BasGeom-9
}gyieldApp
paused
audioerror
playing
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default
p14-105
buttonUp
clearVolume
audioerror
audioenable
wavPlayable
p28-105
enterPage
paused
audioerror
playing
p28-105
leavepage
showVolume
false
pause
clearVolume
repeat
wavPlayable
audioenable
audioenable
audiodisable
program
repeat
forward
fwd_slnt
bwd_slnt
repeat_silent
smpause
smpausedis
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currentPage
focusWindow
scrollPosition
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0,4320
-- These scripts accept commands
the navigation bar, which
a separate
opened
8"e28revad.exe"
Table
4interuptBack
8"dghelp.vwr" --
default
Contents
-- allows user
Previous
(pg-1)
-- Allows users
(pg+1)
HelpBack
4modifiedBack
modBack
-- To
repeated use
HelpSearch
mmstatus
clip
paused
mmstop
mmrewind
? wait
"Press the
arrow
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("IntS&S-3-"&(
playing
picture (fwd)
- 1)
= - 1)
allow students
click through
steps
)their own pace
-1)) --
volume
cVolume
-- control
mmplay
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yieldApp()
update
default
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
b wait
C"ptproj-8"
playing
picture "2"
allow students
click through the steps
)their
-- own pace
"&(2))
volume
cVolume
-- control
mmplay
"&(2))
"&(fwd+1)) --
yieldApp()
audioerror
update
default
normalgraphic
= icon "repeat"
syserrornumber = 0
mmstatus
clip
paused
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mmrewind
[ wait
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playing
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picture (
* - 1)
_ - 1)
allow students
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volume
cVolume
-- control
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yieldApp()
audioerror
update
default
syserrornumber = 0
mmstatus
clip
paused
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mmrewind
[ wait
C"LinVis-4-6"
playing
picture "2"
allow students
click through the steps
)their
-- own pace
"&(2))
volume
cVolume
-- control
mmplay
"&(2))
"&(fwd+1)) --
yieldApp()
audioerror
update
default
normalgraphic
= icon "repeat"
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
[ wait
C"BPS-2-7"
playing
picture "2"
allow students
click through the steps
)their
-- own pace
"&(2))
volume
cVolume
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mmplay
"&(2))
"&(fwd+1)) --
yieldApp()
audioerror
update
default
normalgraphic
= icon "repeat"
syserrornumber = 0
mmstatus
clip
paused
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[ wait
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volume
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yieldApp()
audioerror
update
default
syserrornumber = 0
mmstatus
clip
paused
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mmrewind
[ wait
D"BPS-3-15"
playing
picture "2"
allow students
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)their
-- own pace
"&(2))
volume
cVolume
-- control
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"&(2))
"&(fwd+1)) --
yieldApp()
audioerror
update
default
normalgraphic
= icon "repeat"
syserrornumber = 0
mmstatus
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paused
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mmrewind
[ wait
C"EV&TS-1-14"
playing
`"EV&TS-1-14"
t"EV&TS-1-14"
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allow students
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syserrornumber = 0
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audioerror
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syserrornumber = 0
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normalgraphic
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audioerror
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default
normalgraphic
= icon "repeat"
syserrornumber = 0
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audioerror
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default
normalgraphic
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syserrornumber = 0
mmstatus
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paused
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[ wait
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allow students
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volume
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audioerror
update
default
normalgraphic
= icon "repeat"
syserrornumber = 0
mmstatus
clip
paused
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playing
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allow students
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-- own pace
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volume
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+1)) --
"&(fwd+1)) --
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audioerror
update
B"Forward"
default
normalgraphic
= icon "repeat"
syserrornumber = 0
mmstatus
clip
paused
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[ wait
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playing
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allow students
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default
normalgraphic
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syserrornumber = 0
mmstatus
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volume
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audioerror
update
default
syserrornumber = 0
mmstatus
clip
paused
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[ wait
B"Dvpmt(Con)-1-7"
playing
picture "2"
allow students
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-- own pace
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volume
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-- control
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audioerror
update
default
normalgraphic
= icon "repeat"
syserrornumber = 0
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audioerror
update
default
syserrornumber = 0
mmstatus
clip
paused
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-- own pace
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audioerror
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default
normalgraphic
= icon "repeat"
syserrornumber = 0
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volume
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audioerror
update
default
syserrornumber = 0
mmstatus
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[ wait
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allow students
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volume
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mmplay
"&(2))
"&(fwd+1)) --
yieldApp()
(fwd-1)
audioerror
update
default
normalgraphic
= icon "repeat"
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
[ wait
"Press the
arrow
Bbelow."
("Dvpmt(Con)-3-"&(
playing
4fwd))
picture (
* - 1)
_ - 1)
allow students
click through
steps
)their own pace
fwd-1)) --
volume
cVolume
-- control
mmplay
yieldApp()
audioerror
update
default
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
[ wait
B"Profile-1-5"
playing
picture "2"
allow students
click through the steps
)their
-- own pace
"&(2))
volume
cVolume
-- control
mmplay
"&(2))
"&(fwd+1)) --
yieldApp()
(fwd-1)
audioerror
update
default
normalgraphic
= icon "repeat"
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
[ wait
B"Cut&Fill-1-4"
playing
picture "2"
allow students
click through the steps
)their
-- own pace
"&(2))
volume
cVolume
-- control
mmplay
"&(2))
"&(fwd+1)) --
yieldApp()
(fwd-1)
audioerror
update
default
normalgraphic
= icon "repeat"
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
[ wait
B"Cut&Fill-2-5"
playing
picture "2"
allow students
click through the steps
)their
-- own pace
"&(2))
volume
cVolume
-- control
mmplay
"&(2))
"&(fwd+1)) --
yieldApp()
(fwd-1)
audioerror
update
default
normalgraphic
= icon "repeat"
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
[ wait
"Press the
arrow
Bbelow."
("Cut&Fill-2-"&(
playing
2fwd))
picture (
* - 1)
_ - 1)
allow students
click through
steps
)their own pace
fwd-1)) --
volume
cVolume
-- control
mmplay
yieldApp()
audioerror
update
B"Forward"
default
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
[ wait
B"Shad-1-4"
playing
picture "2"
allow students
click through the steps
)their
-- own pace
"&(2))
volume
cVolume
-- control
mmplay
"&(2))
"&(fwd+1)) --
yieldApp()
(fwd-1)
audioerror
update
default
normalgraphic
= icon "repeat"
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
[ wait
"Press the
arrow
Bbelow."
("Shad-1-"&(
playing
.fwd))
picture (
* - 1)
_ - 1)
allow students
click through
steps
)their own pace
fwd-1)) --
volume
cVolume
-- control
mmplay
yieldApp()
audioerror
update
default
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
[ wait
B"Shad-2-3"
playing
picture "2"
allow students
click through the steps
)their
-- own pace
"&(2))
volume
cVolume
-- control
mmplay
"&(2))
+1)) --
"&(fwd+1)) --
yieldApp()
audioerror
update
default
normalgraphic
= icon "repeat"
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
[ wait
"Press the
arrow
Bbelow."
("Shad-2-"&(
playing
.fwd))
picture (
* - 1)
_ - 1)
allow students
click through
steps
)their own pace
fwd-1)) --
volume
cVolume
-- control
mmplay
yieldApp()
audioerror
update
default
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
[ wait
A"Shad-3-7"
playing
picture "2"
allow students
click through the steps
)their
-- own pace
"&(2))
volume
cVolume
-- control
mmplay
"&(2))
"&(fwd+1)) --
yieldApp()
(fwd-1)
audioerror
update
default
normalgraphic
= icon "repeat"
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
[ wait
"Press the
arrow
Bbelow."
("Shad-3-"&(
playing
.fwd))
picture (
* - 1)
_ - 1)
allow students
click through
steps
)their own pace
fwd-1)) --
volume
cVolume
-- control
mmplay
yieldApp()
audioerror
update
default
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
i wait
"Press the
arrow
Bbelow."
("LinVis-3-"&(
playing
(fwd)
picture (
- 1)
P - 1)
allow students
click through
steps
)their own pace
fwd-1)) --
volume
cVolume
-- control
mmplay
yieldApp()
audioerror
update
default
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
i wait
C"LinVis-3-6"
playing
picture "2"
allow students
click through the steps
)their
-- own pace
"&(2))
volume
cVolume
-- control
mmplay
"&(2))
+1)) --
+1))
"&(fwd+1)) --
yieldApp()
audioerror
update
default
normalgraphic
= icon "repeat"
picture "2"
("CutPlnM-2-"&(
^- 1)
default
update
B"forward_silent"
normalgraphic
= icon "repeat_silent"
fwd_slnt"
picture "2"
("IntP&P(v)-1-"&(
default
update
B"forward_silent"
normalgraphic
= icon "repeat_silent"
fwd_slnt"
picture "2"
("IntP&P(v)-2-"&(
default
update
B"forward_silent"
normalgraphic
= icon "repeat_silent"
fwd_slnt"
picture "2"
("IntP&S-1-"&(
default
update
B"forward_silent"
normalgraphic
= icon "repeat_silent"
fwd_slnt"
picture "2"
("IntP&S-2-"&(
default
update
B"forward_silent"
normalgraphic
= icon "repeat_silent"
fwd_slnt"
picture "2"
("IntS&S-1-"&(
- 1)
default
update
B"forward_silent"
= 10
normalgraphic
= icon "repeat_silent"
fwd_slnt"
picture "2"
("Dvpmt(Pr)-1-"&(
default
update
B"forward_silent"
normalgraphic
= icon "repeat_silent"
fwd_slnt"
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "Cut&Fill-2-2"
mmopen
audioerror
mmclose
picture "2"
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "IntS&S-4-2"
mmopen
audioerror
mmclose
picture "2"
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "IntP&P(r)-2-2"
mmopen
audioerror
mmclose
picture "2"
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "IntP&P(r)-1-2"
mmopen
audioerror
mmclose
picture "2"
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "CutPlnM-2-4"
mmopen
audioerror
mmclose
picture "1"
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "CutPlnM-1-2"
mmopen
audioerror
mmclose
picture "2"
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "IntL&P-2-2"
mmopen
audioerror
mmclose
picture "2"
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "DistPM-4-2"
mmopen
audioerror
mmclose
picture "2"
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "EV&TS-3-2"
mmopen
audioerror
mmclose
picture "2"
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "DistLM-1-2"
mmopen
audioerror
mmclose
picture "2"
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "ptView-2"
mmopen
audioerror
mmclose
picture "2"
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "BPS-2-2"
mmopen
audioerror
mmclose
picture "2"
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "LinVis-3-2"
mmopen
audioerror
mmclose
picture "2"
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "LinVis-2"
mmclose
^<> 0
audioerror
"one"
"two"
audioenable -- doesn't matter
/disable; the fn checks
mmopen
4wavplayable
showVolume
--
B"repeat"
clearVolume
--
audiodisable
--
--
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
[ wait
"Press the
arrow
Bbelow."
("IntP&P(r)-1-"&(
playing
3fwd))
picture (
* - 1)
_ - 1)
allow students
click through
steps
)their own pace
fwd-1)) --
volume
cVolume
-- control
mmplay
yieldApp()
U"l6"
U"l7"
U"l6"
audioerror
update
default
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
[ wait
B"IntP&P(r)-1-8"
playing
picture "2"
U"l6"
U"l7"
allow students
click through the steps
)their
-- own pace
"&(2))
volume
cVolume
-- control
mmplay
"&(2))
"&(fwd+1)) --
yieldApp()
U"l6"
U"l7"
U"l6"
audioerror
update
default
normalgraphic
= icon "repeat"
syserrornumber = 0
mmstatus
clip
paused
mmstop
mmrewind
[ wait
"Press the
arrow
Bbelow."
("IntS&S-1-"&(
playing
0fwd))
picture (
* - 1)
_ - 1)
allow students
click through
steps
)their own pace
fwd-1)) --
volume
cVolume
-- control
mmplay
YieldApp()
yieldApp()
audioerror
update
default
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "10BasGeom-3"
mmclose
mmstatus
3p8-105"
"paused"
Lp8-105"
"playing"
mmstop
kp8-105" wait
audioerror
picture "2"
4fwd, nam
audioenable
mmopen
W = "p8-105"
wavplayable
mmplay
showVolume
--
B"repeat"
B"Geometric3"
B"Geometric3_silent"
clearVolume
--
audiodisable
--
--
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "10BasGeom-5"
mmclose
mmstatus
3p10-105"
"paused"
= "playing"
mmstop
1wait
audioerror
picture "2"
4fwd, nam
audioenable
mmopen
W = "
wavplayable
mmplay
showVolume
--
B"repeat"
B"Geometric5"
B"Geometric5_silent"
clearVolume
--
audiodisable
--
--
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "10BasGeom-7"
mmclose
mmstatus
3p12-105"
"paused"
= "playing"
mmstop
1wait
audioerror
picture "2"
4fwd, nam
audioenable
mmopen
W = "
wavplayable
mmplay
showVolume
--
B"repeat"
B"Geometric7"
B"Geometric7_silent"
clearVolume
--
audiodisable
--
--
4wavPlayable
syserrornumber = 0
mmIsOpen
clip "10BasGeom-9"
mmclose
mmstatus
3p14-105"
"paused"
= "playing"
mmstop
1wait
audioerror
picture "2"
4fwd, nam
audioenable
mmopen
W = "
wavplayable
mmplay
showVolume
--
B"repeat"
B"Geometric9"
B"Geometric9_silent"
clearVolume
--
audiodisable
--
--
J"Page Title"
clearVolume
4wavPlayable, nam
audioenable
"p28-105"
syserrornumber = 0
wavplayable
mmplay clip
D<> 0
audioerror
4thisanim, ref
mmstatus
= "playing"
= "paused"
mmstop
wait
showVolume
B"repeat"
audiodisable
J"Page Title"
clearVolume
4wavPlayable, nam
audioenable
"p32-105"
syserrornumber = 0
wavplayable
mmplay clip
D<> 0
audioerror
4thisanim, ref
mmstatus
= "playing"
= "paused"
mmstop
wait
showVolume
B"repeat"
audiodisable
.&+ +E
.&, "
.&, #>
.&, #>
V, #>
V, #>
V, #>
V, #>
.&, "
}gyieldApp
forward
BPS-2-
BPS-2-7
paused
update
audioerror
playing
FcVolume
default
buttonUp
forward
repeat
update
.&+ +E
.&, "
.&, "
V, #>
V, #>
V, #>
V, #>
.&, "
}gyieldApp
forward
BPS-2-
paused
Press the forward arrow button below.
audioerror
playing
FcVolume
default
update
buttonUp
.&+ +E
.&, "
.&, #>
.&, #>
V, #>
V, #>
V, #>
V, #>
}gyieldApp
forward
paused
default
update
audioerror
playing
FcVolume
BPS-3-15
BPS-3-
buttonUp
forward
repeat
update
.&+ +E
.&, "
.&, #>
.&, #>
V, #>
V, #>
V, #>
V, #>
}gyieldApp
Forward
DistLM-1-11
paused
DistLM-1-
update
audioerror
playing
FcVolume
default
buttonUp
forward
repeat
update
.&+ +E
.&, "
.&, #>
.&, #>
V, #>
V, #>
V, #>
V, #>
}gyieldApp
Forward
paused
DistLM-2-
update
DistLM-2-15
audioerror
playing
FcVolume
default
buttonUp
forward
repeat
update
.&+ +E
.&, "
.&, "
V, #>
V, #>
V, #>
V, #>
}gyieldApp
forward
DistLM-2-
paused
Press the forward arrow button below.
audioerror
playing
FcVolume
default
update
buttonUp
.&+ +E
.&, "
.&, "
V, #>
V, #>
V, #>
V, #>
}gyieldApp
EV&TS-2-
forward
paused
Press the forward arrow button below.
audioerror
playing
FcVolume
default
update
buttonUp
.&+ +E
.&, "
.&, "
V, #>
V, #>
V, #>
V, #>
}gyieldApp
Forward
paused
Press the forward arrow button below.
audioerror
DistPM-1-
playing
FcVolume
default
update
buttonUp
.&+ +E
.&, "
.&, "
V, #>
V, #>
V, #>
V, #>
}gyieldApp
forward
paused
FcVolume
Press the forward arrow button below.
audioerror
playing
IntL&P-2-
default
update
buttonUp
.&+ +E
.&, "
.&, #>
.&, #>
V, #>
V, #>
V, #>
V, #>
}gyieldApp
forward
CutPlnM-1-
CutPlnM-1-7
paused
update
audioerror
playing
FcVolume
default
buttonUp
forward
repeat
update
.&+ +E
.&, "
.&, "
.&, #>
.&, #>
V, #>
V, #>
V, #>
V, #>
.&, "
.&, "
CutPlnM-2-
}gyieldApp
forward
paused
CutPlnM-2-9
playing
FcVolume
audioerror
default
update
buttonUp
forward
repeat
update
.&+ +E
.&, "
.&, #>
.&, #>
V, #>
V, #>
V, #>
V, #>
.&, "
.&, "
}gyieldApp
forward
IntP&P(r)-2-10
IntP&P(r)-2-
paused
update
audioerror
playing
FcVolume
default
buttonUp
forward
repeat
update
.&+ +E
.&, "
.&, "
V, #>
V, #>
V, #>
V, #>
.&, "
.&, "
.&, "
.&, "
}gyieldApp
forward
IntP&S-2-
paused
Press the forward arrow button below.
audioerror
playing
FcVolume
default
update
buttonUp
.&+ +E
.&, "
.&, "
V, #>
V, #>
V, #>
V, #>
}gyieldApp
forward
paused
Dvpmt(Con)-1-
Press the forward arrow button below.
audioerror
playing
FcVolume
default
update
buttonUp
.&+ +E
.&, "
.&, "
V, #>
V, #>
V, #>
V, #>
}gyieldApp
forward
Dvpmt(Con)-2-
paused
Press the forward arrow button below.
audioerror
playing
FcVolume
default
update
buttonUp
.&+ +E
.&, "
.&, #>
.&, #>
V, #>
V, #>
V, #>
V, #>
Profile-1-
Profile-1-5
forward
paused
update
audioerror
playing
FcVolume
}gyieldApp
default
buttonUp
forward
repeat
update
.&+ +E
.&, "
.&, #>
.&, #>
V, #>
V, #>
V, #>
V, #>
}gyieldApp
forward
Cut&Fill-1-4
paused
update
Cut&Fill-1-
audioerror
playing
FcVolume
default
buttonUp
forward
repeat
update
.&+ +E
.&, "
.&, "
V, #>
V, #>
V, #>
V, #>
}gyieldApp
forward
paused
Press the forward arrow button below.
Cut&Fill-1-
audioerror
playing
FcVolume
default
update
buttonUp
.&+ +E
.&, "
.&, #>
.&, #>
V, #>
V, #>
V, #>
V, #>
}gyieldApp
forward
Cut&Fill-2-
Cut&Fill-2-5
paused
update
audioerror
playing
FcVolume
default
buttonUp
forward
repeat
update
.&+ +E
.&, "
.&, "
V, #>
V, #>
V, #>
V, #>
}gyieldApp
Forward
Cut&Fill-2-
paused
Press the forward arrow button below.
audioerror
playing
FcVolume
default
update
buttonUp
.&+ +E
.&, "
.&, #>
.&, #>
V, #>
V, #>
V, #>
V, #>
}gyieldApp
forward
paused
update
audioerror
Shad-1-4
playing
FcVolume
default
Shad-1-
buttonUp
forward
repeat
update
.&+ +E
.&, "
.&, "
V, #>
V, #>
V, #>
V, #>
}gyieldApp
forward
paused
Press the forward arrow button below.
audioerror
playing
FcVolume
default
update
Shad-1-
buttonUp
.&+ +E
.&, "
.&, "
V, #>
V, #>
V, #>
V, #>
.&, "
.&, "
}gyieldApp
forward
paused
Press the forward arrow button below.
LinVis-3-
audioerror
playing
FcVolume
default
update
buttonUp
.&+ +E
.&, "
.&, #>
.&, #>
V, #>
V, #>
V, #>
V, #>
.&, "
.&, "
}gyieldApp
forward
paused
update
LinVis-3-
audioerror
LinVis-3-6
playing
FcVolume
default
buttonUp
forward
repeat
update
.&+ +E
.&, "
.&, "
.&, "
CutPlnM-2-
forward_silent
default
update
buttonUp
repeat_silent
fwd_slnt
update
.&+ +E
.&, "
.&, "
.&, "
.&, "
IntP&P(v)-1-
forward_silent
update
default
buttonUp
repeat_silent
fwd_slnt
update
false
IntS&S-2-2
IntS&S-2-4
IntS&S-2-8
audioerror
IntS&S-2-5
IntS&S-2-7
IntS&S-2-6
IntS&S-2-3
enterPage
IntS&S-2-2
IntS&S-2-4
IntS&S-2-8
audioerror
IntS&S-2-5
IntS&S-2-7
IntS&S-2-6
IntS&S-2-3
leavePage
IntP&S-1-14
false
IntP&S-1-12
IntP&S-1-9
IntP&S-1-2
IntP&S-1-4
IntP&S-1-11
IntP&S-1-10
IntP&S-1-15
IntP&S-1-8
IntP&S-1-5
audioerror
IntP&S-1-7
IntP&S-1-6
IntP&S-1-13
IntP&S-1-3
enterPage
IntP&S-1-14
IntP&S-1-12
IntP&S-1-9
IntP&S-1-2
IntP&S-1-4
IntP&S-1-11
IntP&S-1-10
IntP&S-1-15
IntP&S-1-8
IntP&S-1-5
audioerror
IntP&S-1-7
IntP&S-1-6
IntP&S-1-13
IntP&S-1-3
leavePage
IntP&P(r)-2-9
IntP&P(r)-2-2
IntP&P(r)-2-4
false
IntP&P(r)-2-10
IntP&P(r)-2-8
IntP&P(r)-2-5
IntP&P(r)-2-7
IntP&P(r)-2-6
IntP&P(r)-2-3
audioerror
enterPage
IntP&P(r)-2-2
IntP&P(r)-2-4
IntP&P(r)-2-10
IntP&P(r)-2-8
IntP&P(r)-2-5
IntP&P(r)-2-7
IntP&P(r)-2-6
IntP&P(r)-2-3
audioerror
IntP&P(r)-2-9
leavePage
false
IntP&P(r)-1-2
IntP&P(r)-1-4
IntP&P(r)-1-8
IntP&P(r)-1-5
audioerror
IntP&P(r)-1-7
IntP&P(r)-1-6
IntP&P(r)-1-3
enterPage
IntP&P(r)-1-2
IntP&P(r)-1-4
IntP&P(r)-1-8
IntP&P(r)-1-5
audioerror
IntP&P(r)-1-7
IntP&P(r)-1-6
IntP&P(r)-1-3
leavePage
CutPlnM-2-8
CutPlnM-2-5
false
CutPlnM-2-6
CutPlnM-2-9
CutPlnM-2-7
CutPlnM-2-4
audioerror
enterPage
CutPlnM-2-5
CutPlnM-2-7
CutPlnM-2-6
CutPlnM-2-9
audioerror
CutPlnM-2-4
CutPlnM-2-8
leavePage
CutPlnM-1-4
false
CutPlnM-1-5
CutPlnM-1-7
CutPlnM-1-6
CutPlnM-1-3
audioerror
CutPlnM-1-2
enterPage
CutPlnM-1-4
CutPlnM-1-5
CutPlnM-1-7
CutPlnM-1-6
CutPlnM-1-3
audioerror
CutPlnM-1-2
leavePage
false
DistLM-1-2
DistLM-1-4
DistLM-1-11
DistLM-1-10
DistLM-1-9
DistLM-1-8
DistLM-1-5
DistLM-1-7
DistLM-1-6
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DistLM-1-3
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DistLM-1-9
DistLM-1-2
DistLM-1-4
DistLM-1-11
DistLM-1-10
DistLM-1-8
DistLM-1-5
DistLM-1-7
DistLM-1-6
audioerror
DistLM-1-3
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BPS-2-4
false
BPS-2-5
BPS-2-7
BPS-2-6
BPS-2-3
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BPS-2-4
BPS-2-5
BPS-2-7
BPS-2-6
BPS-2-3
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BPS-2-2
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LinVis-4-5
false
LinVis-4-6
LinVis-4-3
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p7-105
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Geometric2
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Shad-3-4
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Shad-3-5
Shad-3-7
Shad-3-6
Shad-3-3
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enterPage
Shad-3-2
Shad-3-4
Shad-3-5
Shad-3-7
Shad-3-6
Shad-3-3
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leavePage
Cut&Fill-2-2
Cut&Fill-2-4
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Cut&Fill-2-5
Cut&Fill-2-3
audioerror
enterPage
Cut&Fill-2-2
Cut&Fill-2-4
Cut&Fill-2-5
Cut&Fill-2-3
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leavePage
Dvpmt(Con)-3-6
Dvpmt(Con)-3-3
false
Dvpmt(Con)-3-2
Dvpmt(Con)-3-4
audioerror
Dvpmt(Con)-3-5
enterPage
Dvpmt(Con)-3-6
Dvpmt(Con)-3-3
Dvpmt(Con)-3-2
Dvpmt(Con)-3-4
audioerror
Dvpmt(Con)-3-5
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Dvpmt(Con)-2-4
false
Dvpmt(Con)-2-5
Dvpmt(Con)-2-6
Dvpmt(Con)-2-3
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Dvpmt(Con)-2-2
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Dvpmt(Con)-2-4
Dvpmt(Con)-2-5
Dvpmt(Con)-2-6
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Dvpmt(Con)-2-2
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Dvpmt(Con)-1-2
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Dvpmt(Con)-1-4
Dvpmt(Con)-1-5
Dvpmt(Con)-1-7
Dvpmt(Con)-1-6
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Dvpmt(Con)-1-2
Dvpmt(Con)-1-4
Dvpmt(Con)-1-5
Dvpmt(Con)-1-7
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Dvpmt(Cyl)-2-5
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Dvpmt(Cyl)-2-7
Dvpmt(Cyl)-2-6
Dvpmt(Cyl)-2-3
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Dvpmt(Cyl)-2-5
Dvpmt(Cyl)-2-7
Dvpmt(Cyl)-2-6
Dvpmt(Cyl)-2-3
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Dvpmt(Cyl)-2-2
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DistPM-4-15
DistPM-4-17
DistPM-4-16
false
DistPM-4-13
DistPM-4-9
DistPM-4-2
DistPM-4-4
DistPM-4-8
DistPM-4-12
DistPM-4-5
DistPM-4-14
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DistPM-4-7
DistPM-4-6
DistPM-4-11
DistPM-4-3
DistPM-4-10
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DistPM-4-15
DistPM-4-17
DistPM-4-16
DistPM-4-13
DistPM-4-9
DistPM-4-2
DistPM-4-4
DistPM-4-8
DistPM-4-12
DistPM-4-5
DistPM-4-14
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DistPM-4-7
DistPM-4-6
DistPM-4-11
DistPM-4-3
DistPM-4-10
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DistPM-2-16
DistPM-2-13
false
DistPM-2-9
DistPM-2-2
DistPM-2-4
DistPM-2-8
DistPM-2-12
DistPM-2-5
DistPM-2-14
audioerror
DistPM-2-7
DistPM-2-6
DistPM-2-11
DistPM-2-3
DistPM-2-10
DistPM-2-15
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DistPM-2-16
DistPM-2-13
DistPM-2-9
DistPM-2-2
DistPM-2-4
DistPM-2-8
DistPM-2-12
DistPM-2-5
DistPM-2-14
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DistPM-2-7
DistPM-2-6
DistPM-2-11
DistPM-2-3
DistPM-2-10
DistPM-2-15
leavePage
EV&TS-3-7
EV&TS-3-6
EV&TS-3-3
false
EV&TS-3-9
EV&TS-3-2
EV&TS-3-4
EV&TS-3-11
audioerror
EV&TS-3-10
EV&TS-3-8
EV&TS-3-5
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EV&TS-3-6
EV&TS-3-3
EV&TS-3-11
EV&TS-3-9
EV&TS-3-2
EV&TS-3-4
EV&TS-3-7
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EV&TS-3-10
EV&TS-3-8
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EV&TS-2-4
EV&TS-2-11
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EV&TS-2-5
EV&TS-2-7
EV&TS-2-6
EV&TS-2-3
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EV&TS-2-9
EV&TS-2-2
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EV&TS-2-4
EV&TS-2-11
EV&TS-2-10
EV&TS-2-8
EV&TS-2-5
EV&TS-2-7
EV&TS-2-6
EV&TS-2-3
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EV&TS-1-12
EV&TS-1-9
EV&TS-1-2
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EV&TS-1-4
EV&TS-1-11
EV&TS-1-10
EV&TS-1-14
EV&TS-1-8
EV&TS-1-5
EV&TS-1-7
EV&TS-1-6
EV&TS-1-13
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EV&TS-1-12
EV&TS-1-9
EV&TS-1-2
EV&TS-1-4
EV&TS-1-11
EV&TS-1-10
EV&TS-1-14
EV&TS-1-8
EV&TS-1-5
EV&TS-1-7
EV&TS-1-6
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DistLM-2-4
DistLM-2-11
false
DistLM-2-10
DistLM-2-8
DistLM-2-5
DistLM-2-14
DistLM-2-7
DistLM-2-6
DistLM-2-3
DistLM-2-15
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DistLM-2-13
DistLM-2-9
DistLM-2-2
enterPage
DistLM-2-4
DistLM-2-11
DistLM-2-10
DistLM-2-8
DistLM-2-5
DistLM-2-14
DistLM-2-7
DistLM-2-6
DistLM-2-3
DistLM-2-15
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DistLM-2-13
DistLM-2-9
DistLM-2-2
leavePage
BPS-3-14
BPS-3-7
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BPS-3-13
BPS-3-3
false
BPS-3-12
BPS-3-9
BPS-3-2
BPS-3-4
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BPS-3-11
BPS-3-10
BPS-3-15
BPS-3-8
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BPS-3-7
BPS-3-6
BPS-3-13
BPS-3-3
BPS-3-12
BPS-3-14
BPS-3-9
BPS-3-2
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projection2-1
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Show example
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p3-105
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Show example
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Show example
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p70-105
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p57-105
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Show example
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"step3txt"
"highlight3"
play3on
play4off
Cut-and-Fills
Descriptive Geometry Problems -- Contours
Cut-and-Fills
When performing landscape work, it is important to know how the land will look after the
work is done. This usually means that the areas that have been dug out and the areas
that have been filled in need to be shown.
A cut and fill drawing shows where the tops of the cuts and the toes of the fills exist, with
the cut portions and the filled portions cross-hatched to emphasize those areas. These
areas are then assumed to be planar with their respective cut or fill ratios.
Cut or fill ratios are the vertical:horizontal (i.e., vertical to horizontal) displacements for land that has been cut out or filled in to create new surfaces not at the original land height.
1:1 Cut
1:2 Fill[
pause
audioOn
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Cut-and-Fills
pointProjection
EV & TS of Plane - 1
Development (Cone) - 3
Intersections of Planes and Sol
$!&!&!
Cut-and-Fill
Descriptive Geometry Problems -- Contours
Cut-and-Fills, cont'd
To create a level road across variable height land, first consider the profile view of the land.
Any land above the road (120 m) will need to be cut, and any land below needs to be filled.
Filling the land below the road (elevation 115 m) creates new elevation contour lines. If the
fill ratio was 1:2, these new lines would appear as parallel to the road, but 10 m away from
it. The reason is that for every 5 m drop in elevation due to fill, the toe of the fill moves out
by 10 m. The filling stops when the elevation of the land matches the fill elevation below the
land. The same procedure is then used for each successive 5 m fill under the road.oad......................................................
Land+
Road 120 mi
2 $ /
115 m
110 m
115 m
110 m
110 m
Road 120 m
115 m
Front View
Top View=
10 mU
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Repeat Animation
Click on the background to return to Real Intersections.
Cut-and-Fill
Descriptive Geometry Problems -- Contours
Cut-and-Fills, cont'd
To create a road at a grade, say a 10% grade, we need to know the length of the run on the
road.
For a 10% grade for every 50 m of run, the road rises 5 m, and the toe of the fill line extends
outward an additional 10 m. Thus the constant elevation contours appear as lines which
extend from the road at an angle of arctan(outward extension/run), or in this case,
arctan(10/50) for the fill. We are using the same cut and fill ratios as for a level road.....................................
L : I
Road 120 m]
120 m
115 m
130 m
125 m
125 m
Roadc
130 m
Front View
Top Viewg
110 m
130 m
125 m
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Click on the background to return to Real Intersections.
Cut & Fill - 1
false
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Descriptive Geometry Problems -- Contours
Cut-and-Fills
Example 1. Show the cut-and-fill for a level road, going from A to B, at an elevation of 120 m on the land profile shown below.rom
A to B.......from A to B........................to B............
backward
forward
To see the step-by-step solution, click the forward arrow below.................
First, the road is drawn from A to B. Because the level of the road is 120 m, the cuts and fills begin there.
The ratio of the cut is 1:1, so the cut lines are drawn 5 m out from the road for every 5 m increase in elevation.
The ratio of the fill is 1:2, so the fill lines are drawn 10 m out for every 5 m decrease in elevation.
Then intersections of the cut and fill lines and the elevation lines are found for each elevation.
Connecting these points gives the outlines of the cuts and fills.
Note that the connections were splined to give a smooth transition between elevations.
Finally, the cuts and fills are cross-hatched.
The fill area is much larger than the cut area because the fill ratio was lower than the cut ratio.
Click the forward arrow to see this example again.
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Cut & Fill - 2
Descriptive Geometry Problems -- Contours
Cut-and-Fills
Example 2. Show the cut-and-fill for a road which has a 10% grade from left to right, and an
To see the step-by-step solution, click the forward arrow below.................
The first step is to mark the point at which the road is level with the ground (120 m) and points 50 m to the left and right.
We chose 50 m because a 10% grade will give a 5 m offset for the cut and 10 m offset for the fill at that distance.
The constant elevation lines appear as lines which extend from the road at an angle of arctan(5/50) for the cut and arctan(10/50) for the fill.
The intersections of the constant elevation contours are found to give the outlines of the cut and fill.
This is only done for one side of the road in this example.
Finally, the cut and fill are cross-hatched.
The solution (for one side of the road) is now complete. To see the example again, click the forward arrow below.M
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Shadows
Descriptive Geometry Problems -- Shadows
Shadows
Architects, and engineers who work with solar heating and cooling and HVAC
(Heating, Ventilation, and Air Conditioning) worry a lot about shadows. Builders
are concerned with the shadows cast upon their structure by other structures, at
different hours of the day and different times of the year, since this can greatly affect
the comfort within their structure and its appearance. Also, builders need to be
concerned with how shadows from their structure affect the function and appearance
of other structures.
Finding shadows is usually a direct application of the problem of the intersection of
a line and a plane. Solutions for four different cases are outlined below:
1. The shadow cast by a single point from a light source travels a straight line from the
source, through the point, until it strikes the surface.
2. For a planar object, the procedure is extended to finding the striking points of several
points, then connecting the intersection points to form the total shadow of the object.
3. If the object is solid, the simplest approach is to solve for the front plane shadow, then
add the back plane shadow, then connect the corresponding edges of the solid to
account for its depth.
4. For a curved object, a series of points may be taken along the curved surface.
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Shadows
LinVis - 3
Int (Pln & Pln) Virt - 1
Cut/Fill - 2
Cut & Fill - 1
Shadow - 1
false
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Descriptive Geometry Problems -- Shadows
Shadows
Example 1. Find the shadow formed by the point from the light source shown. source shown.
backward
forward
To see the step-by-step solution, click the forward arrow below.................
A ray from the light source travels through the point to the ground. Here, the intersection between the ray and the ground give the shadow of the point.
A projection line is created that travels from the shadow in the frontal view into the horizontal view.
The shadow is found in the horizontal view as the intersection of the projection line and the ray from the light source passing through the point.
The solution is now complete.
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Shadow - 2
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Descriptive Geometry Problems -- Shadows
Shadows
Example 2. Find the shadow cast by the rectangular plane from
the sun.
colors
cgmimw
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backward
forward
To see the step-by-step solution, click the forward arrow below.................
Create lines parallel to the ray in the frontal view and project the shadow points into the horizontal view.
Similarly, create lines parallel to the ray in the horizontal view. The intersections of these rays with the corresponding projections from the frontal view gives the vertices of the shadow of the triangle.
Connecting these points gives the shadow of the triangle in the horizontal view. The final solution of a shadow problem should always be cross-hatched.
The solution is now complete.
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Shadow - 3
Descriptive Geometry Problems -- Shadows
Shadows
Example 3. Find the shadow cast upon the ground by Etcheverry Hall. Etcheverry Hall.
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The first step is to create projection lines parallel to the rays and use the intersection points to (separately) find the shadows of the front and back planes.
The projection points are connected to find the shadow of the first plane, which gives the shadow of the front of the building.
The second plane is parallel to the first, but gives the shadow of the back of the buliding.
The upper and lower planes are connected.3
The portion of the shadow on the ground is cross hatched.
The shadow also falls on a portion of the building itself, so that part must also be cross-hatched.
The solution is now complete.
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To see the step-by-step solution, click the forward arrow below.................
Etcheverry
clearVolume
enterPage
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audiodisable
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The End
You have reached the conclusion of Graphics Interactive.
Please direct comments/suggestions to:
Prof. Dennis K. Lieu
5128 Etcheverry Hall
University of California at Berkeley
Berkeley, CA 94720
dlieu@euler.berkeley.edu
BPS - 2
EV & TS of Plane - 2
Int (Pln & Pln) Real - 1
The edge view and true shape o
The Edge View and True Shape of
BPS - 3
EV & TS of Plane - 3
Dihedral Ang - 1
Distance (PlnMthd) - 1
Int (Sld & Sld) - 2
Shadow - 2
Distance (PlnMthd) - 3
Int (Pln & Sld) - 1
Development (Prsm) - 2
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Int (Sld & Sld) - 1
Descriptive Geometry Problems -- Intersections
Intersection of Two Solids
Example 1. Find the intersection of the triangular prism and the rectangular prism
DEFG.
PostScript
Courier
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backward
forward
To see the step-by-step solution, click the forward arrow below.................
First, the intersection of the prism between lines A and B is found. Lines from line DG in view H are projected into view F..
The intersection points in view F are connected.
The visibility is determined, and the construction lines are removed.
Next, the intersection between lines B and C are found. The virtual intersection point of DG with line C is found.c
Using the virtual intersection point, the intersection is bounded by the edge of the rectangular prism.9
The visibility is determined and the construction lines are removed.
Now the intersection between lines A and C is determined. The virtual intersections of these lines are found in view F.
These construction lines bound the intersection within the area of the rectangular prism.
This part of the intersection is behind the rectangular prism, so the lines are dashed. Note that the portion of the rectangular prism that is visible is shown, but the parts within the intersection are not shown.
The solution is now complete. Click the forward arrow below to see the solution again.
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Int (Sld & Sld) - 2
Descriptive Geometry Problems -- Intersections
Intersection of Two Solids
Example 2. Find the intersection of the cylinder and the rectangular prism DEFG........................................
PostScript
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backward
forward
To see the step-by-step solution, click the forward arrow below.................
First we must draw the end view of the cylinder.
Because this is a circular object and we cannot simply connect the ends of lines to find the intersection, we must find a way of representing the circle with lines.
In order to imagine the circular end of the cylinder as a series of straight lines, points are created around the circle. Imagine that the short lines between these points are straight.
Be careful that you draw the points in the correct positions in adjacent views.
Project each point into views H and F from the end views of the cylinder.
The point of intersection will be where the projection line from view H to view F meets with the projection line from the end view next to view F.
This is done with each point (1 through 7) on the cylinder.
For now, we only consider these points because they are in front in view F.
These points are connected to show the intersection, and the construction lines are removed.
The points are originally connected with straight lines, which are then splined to give a curved appearance. Notice that the larger number of points you use, the more accurate the solution.M
Now projections from the back side of the cylinder are done. These points will be hidden in view F.
The hidden points are connected, then splined to give a smooth curve.
The solution is now complete. In order to see this solution again, click the forward arrow below.
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Int (Sld & Sld) - 3
Descriptive Geometry Problems -- Intersections
Intersection of Two Solids
Example 3. Find the intersection of the two cylinders shown below.....................................
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backward
forward
The first step is to draw the end view of the angled cylinder for views H and F.
Create points around the end of the cylinder. We will only solve for the visible points, that is, points 1 through 7.
Be careful that the points are oriented and labelled correctly in the end views.5
The projection lines are drawn from view H and view F.
The intersections of these projection lines show the intersection points of the angled cylinder with the upright cylinder.]
The intersection points are connected and shown solid (since we only solved for the visible part of the intersection).
The solution is now complete. Click the forward arrow below to see the solution again.
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To see the step-by-step solution, click the forward arrow below.ard arrow below.
Int (Sld & Sld) - 4
Descriptive Geometry Problems -- Intersections
Intersection of Two Solids
Example 4. Find the intersection of the cone and the cylinder shown below...... cylinder shown below.
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backward
forward
To see the step-by-step solution, click the forward arrow below.ard arrow below.
First draw view 1 to get the cylinder in end view.
Create evenly spaced points 1 through 5 in view H, then project them into views F and 1.
We are only using these points because we will only find the visible part of the intersection.
Rays are drawn from the tip of the cone to each of the points we just created.
The intersection of the rays on the surface of the cone and the end view of the cylinder in view 1 give the intersection points.
The intersection points are then projected into view F.
The intersection points are connected to find the intersection line.
The solution is now complete. .lick the forward arrow below to see it again.
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Development of Prismatic Surface
Descriptive Geometry Problems -- Developments
Development of Prismatic Surfaces
A development is the unfolding of a three dimensional object to create a single flat surface. It is a "map" of the surface of the object. Folded correctly, the flat surface will
reform the original 3-D object.
Developments represent a departure from the strict discipline of descriptive geometry.
In a development, all lines are shown in true length, and all planes in true shape, in a
single view.
Note, however, that the development is NOT unique. Many other possibilities exist.
Limitations: Only surfaces that can be generated by straight lines can be developed,
i.e. prismatic solids, cylinders, and cones. Warped surfaces, such as
spheres, cannot be developed.
Process: Obtain true lengths and true shapes, and put them in a single view for assembly. Take advantage of views which contain them already. Use your aptitude with descriptive geometry.........
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Development (Prsm) - 1
!@"^#h$f%2&
Descriptive Geometry Problems -- Developments
Development of Prismatic Surfaces
Example 1. Develope the surface of the triangular prism shown below. below....
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backward
forward
To see the step-by-step solution, click the forward arrow below.................
First, the distances bordering the triangular prism are labelled 1, 2, and 3.
Also note that the vertical lines lengthwise down the prism in view F are in true length because they are in point view in view H.))))
The development is begun by drawing line BC in true length.
We will create the development view next to view F because it is a convenient place to project the true length lines.9
Next, line CR is drawn in the development view.
From view H, we know that the distance between the parallel lines BQ and CR is the same as distance 2 in view H.
Lines BC and QR are connected to somplete this face of the prism.
We will now use this face as the center of the development and draw the other faces around it.
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The next face to be developed is ABPQ. Line AP is drawn in true length and offset from line BQ by a distance equal to line 1 in view H.
AB and PQ are connected and this face is completed.="
Next, face ACPR is developed.
Line AP is drawn again, but this time to the right of the face BCPQ and at a distance equal to line 3 in view H.H.
Lines AC and PR are drawn in.
Now all the side faces of the prism have been developed. Only the ends of the prism remain.
Since the triangular end ABC of the prism has simply the lengths AB, BC, and CA, point A can be located by swinging arcs from points B and C.c%
Lines AB and AC are drawn in, and the construction arcs are removed.
The same is done for end PQR. Arcs are swung from points Q and R to locate point P in the development.
Lines PQ and PR are drawn in.
The solution is now complete. In order to see it again, click the forward arrow below.
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Development (Prsm) - 2
Descriptive Geometry Problems -- Developments
Development of Prismatic Surfaces
Example 2. Develop the surface of the four sided symmetrical hopper shown below...................................
P"E P"y 0"
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The first step is to get one side of the hopper in true shape.
Using simple descriptive geometry, we draw view 1 parallel to line BD in view F, and we get the lengths of the projection lines from view H.
The side of the hopper BDSR is in true shape in view 1.
It is in true shape because view 1 was drawn parallel to BD and SR in view F and lines BS and DR were in point view in view F.
Since the problem states that the hopper is symmetrical, we can simply copy and flip the same side four times in order to get the complete development.
The solution is now complete. In order to see it again, click the forward arrow below.y
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Development of Cones and Cylinde
Descriptive Geometry Problems -- Developments
Development of Cones and Cylinders
In the development of cones and cylindrical surfaces, we use the same basic principles
as with the prismatic surface.
Again, we will use some mathematics to solve problems more easily. We will also use
the method of creating points around circles in order to approximate them as groups of
straight lines, as we did with intersections.
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Devlopment (Cylin) - 1
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Descriptive Geometry Problems -- Developments
Development of Cones and Cylinders
Example 1. Develop the surface of the cylinder below.
1]2\)
1H($2
1991. Al
l rights r
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backward
forward
To see the solution, click the forward arrow below..............................
This problem is very simple to solve. Using geometry, we can easily calculate the circumference of the cylinder. This is will be the length of the developed cylinder, and h will be the height.
The solution is now complete.now complete. complete.. complete.
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Development (Cylin) - 2
Descriptive Geometry Problems -- Developments
Development of Cones and Cylinders
Example 2. Develop the surface of the truncated cylinder below.]
z*H.z*4%
q,H.q,
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U2H.U2
backward
forward
To see the step-by-step solution, click the forward arrow below.................
First, points are created around the cylinder end in view H.
Remember that the larger number of points, the more accurate the solution.
These points are projected into vertical lines around the circumference of cylinder in view F.
Only lines 1 through 7 are visible; the rest of the lines are hidden behind the cylinder.
The development view is begun.
The length of the line, which represents the base of the cylinder, is the circumference. The points are spaced evenly on this line.
The first 3 lines are projected onto the development view as shown.
We also could simply copy the lines from view F, since the vertical lines are shown in true length.;
The rest of the lines are projected.
Noting that the truncated cylinder is symmetrical will save time.
The tops of the lines are connected, then the connection is splined to give a smooth curve.
The solution is now complete. To so it again, click the forward arrow below.A
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Development (Cone) - 1
Descriptive Geometry Problems -- Developments
Development of Cones and Cylinders
Example 3. Develop the surface of the right circular cone below.
PostScript
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S 4/S
&Help
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CREATEP
backward
forward
To see the step-by-step solution, click the forward arrow below.................
Create points around the end view in view H.
Project the points in to view F.
Draw in the lines in view F. These are the lines you will see in the final development.
Find the true length of a line on the side of the cone as shown.
S is the distance between two points on the circle. Remember that the distance S is an approximation. It should actually be the length of an arc rather than a linear distance.
Lines 1 through 12 are drawn in the development view.
Remember that they are in true length and the distance between them at the outside is S.
Spline the curve connecting these points.
The solution is now complete. Click the forward arrow below to see in again.
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Development (Cone) - 2
Descriptive Geometry Problems -- Developments
Development of Cones and Cylinders
Example 4. Develop the surface of the truncated right circular cone below. below.
All Righ
%Q$A&*#
backward
forward
To see the step-by-step solution, click the forward arrow below.................
Create points around the base of the cone, and project the lines into view H.
Find the true length of a line on the untruncated cone and show the development. (This was done in the previous problem.)
Find the true length of each of the lines over the truncated region.
Measure the rise in view F and the run in view H and find the true length either graphically or using the pythagorean theorem.
Using the lengths just measured, draw points on the untruncated development at the appropriate distances from the center point.
Connect all the points and spline the curve to get the final development.
The solution is now complete. Click the forward arrow below to see it again.
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Development (Cone) - 3
Descriptive Geometry Problems -- Developments
Development of Cones and Cylinders
Example 5. Develop the surface of the oblique cone below.A
N!j%d.
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To see the step-by-step solution, click the forward arrow below.................
First, create points evenly spaced around the base of the cone.
Project the points into view F.
Draw in lines from the tip of the cone to the points on the base in both views.y
Find the true length of each line, using the rise from view F and the run from view H.
Also, use a circle to mark the distance along the base between points in view H.
Draw the true length lines in the development view. Make sure that they are appropriately spaced, using the same circle you created in view H.
Draw in the outline, splining the curved edge.
The solution is now complete. Click the forward arrow below to see it again.
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Contours
Descriptive Geometry Problems -- Contours
Contours
A contour map is a map of some constant quality in an area. The mapping may be of
constant temperature, pressure, height, etc.
Most familiar are contour maps of elevation of land. A example is shown below. In this
aerial view of a piece of land, the contours of constant elevation are shown. Thus the land
goes from an elevation of 100 m on the right, to a peak elevation of 145 on the left.
Note that this map is an approximation,
and is only as accurate as the surveyor
could be on the site when it was made.
The areas between the contour lines are
not flat, but must be interpolated to give
a smooth transition.transition.
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Profiles
Descriptive Geometry Problems -- Contours
Profiles
A profile of the land is a section view of the land made by a vertical cutting plane.
A true shape of the section can be contructed by a viewing plane parallel to the vertical
cutting plane. Note that the construction is an approximation, due to the very nature of a
contour map.
Also note that the vertical and horizontal scale may not be the same. Different horizontal
and vertical scales are used to emphasize the vertical displacement of the land....................
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Profiles - 1
Descriptive Geometry Problems -- Contours
Profiles
Example. Generate a profile of the
hillside on a line from point A to point B.................nt B...............
backward
forward
First, draw in the vertical cutting plane between lines A and B.
Mark the points where the contour lines intersect it.
Set up the profile view. In this solution, we will make 1 m = 5 mm in the profile view.u
Project the intersection points into the profile view.
Make sure that each projection line ends at the appropriate elevation.i
Connect the points at the ends of the projection lines to find the final profile and spline the resulting line to get a smooth curve.
The solution is now complete. Click the forward arrow below to see it again.
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Int (Pln & Pln) Real - 2
Descriptive Geometry Problems -- Intersections
Intersection of Two Planes -- Real Intersection
Example 2. Find the intersection of planes ABC and QRS. Show correct visibility.................................
PostScript
Courier
ourie
backward
forward
First, determine which lines (plane edges) are involved in the intersection.
From view F, it is obvious that AB and QS are not in the intersection. Their visbilities are determined and shown in view H and F.
Four lines remain to be determined: AC, BC, QR, and RS.
Using the visibility test, it is determined that lines BC and RS are not involved in the intersection and their visibility is shown accordingly.
The cutting plane method will now be used to determine the point of intersection of edge QR.
Projection lines from the apparent intersection of QR and plane ABC in view H are drawn into view F.
The cutting plane is drawn across plane ABC in view F, and the intersection between the cutting plane and line QR shows the intersection point.
This point is projected into view H and is labelled I.
The visibility is determined in views H and F and line QR is drawn in.
The projection and construction lines have been erased.
The cutting plane method is now done for line AC.
First, projection lines from the cutting plane in view H are drawn into view F.A
The cutting plane is drawn in view F and the intersection of line AC and the cutting plane in view F gives the intersection point.
The intersection point is labelled J and is projected into view H.}
The visibility of line AC is determined on both sides of the intersection point J in views H and F.
The last step is to draw in the intersection line, which is a straight line between the points I and J. Try to picture these planes in three dimensions.
The solution is now complete. Click the forward arrow below to see this solution again.
To see the step-by-step solution, click the forward arrow below.................
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Intersection of Two Planes (Rea
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Intersection of Two Planes (Real Intersection) Quiz - 1/3
Quiz 9.20
Given two intersecting planes, if one edge of a plane is in front of the other plane in the
horizontal view, then that edge is completely visible in the front view.....................false.
Go to the next page after you have tried answering the question.
Quiz 8.20. Given two intersecting planes, if one edge of a plane
is in front of the other plane in the horizontal view, then
that edge is completely visible in the front view..........
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Intersection of Two Planes (Real Intersection) Quiz - 2/3
Quiz 9.21
Given two intersecting planes, if one edge of a plane is in back of the other plane in the
horizontal view, then that edge is completely hidden in the front view.....................r false.
Go to the next page after you have tried answering the question.
Quiz 8.21. Given two intersecting planes, if one edge of a plane
is in back of the other plane in the horizontal view, then
that edge is completely hidden in the front view.
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answer
Inconclusive. Only the portion of the edge inside the intersection area is hidden; the remaining part is visible.
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Intersection of Two Planes (Real Intersection) Quiz - 3/3
Quiz 9.22
The intersection of two bounded planes is a line.nt is
false, or "Inconclusive" (if applicable) if insufficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.22. The intersection of two bounded planes is a line..........
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answer
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Intersection of Two Planes -- Vi
Descriptive Geometry Problems -- Intersections1
Intersection of Two Planes -- Virtual Intersection
In the previous method, we considered the lines of both planes, intersecting the
respective "other planes" to establish the intersection. In this alternate method, we
consider only the intersections of the lines of one plane on the other.
To do this, we imagine that one plane is infinitely large. The intersection of the
other plane must then be a line between two sides. The procedure is to find the
unbounded intersection, then bound it.
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Click on the background to return to Virtual Intersections.
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Int (Pln & Pln) Virt - 1
: f!2"h#
Descriptive Geometry Problems -- Intersections
Intersection of Two Planes -- Virtual Intersection
Example 1. Determine the intersection of planes ABC and XYZ. Show correct visibility...............................
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forward
To see the step-by-step solution, click the forward arrow below.................
Lines AB, AC, XZ, and ZY can be seen in their entirety outside the instersection area, therefore they do not intersect other planes.
Their visbility is easily esablished using the visbility test.Q
In the solution for a real intersection, we considered the lines of both planes. In this solution, however, we will only consider the lines of plane XYZ and imagine that plane ABC is infinite.
For line XY, the cutting plane method is still used and projection lines are drawn into view F.
The cutting plane is projected into view F and the intersection point of line XY with plane ABC, called 2, is found.
Visibility is established for line XY. The procedure has been, up to this point, identical to the real intersection method.
Now we will consider line YZ. Even though we have already established its visibility, it can be used to find the other real intersection point....
Projection lines from ZY's apparent interesection with plane ABC in view H are drawn into view F.
The line representing the cutting plane in view F is extended to intersect with YZ.
This point, labelled point 3, is projected back into view H.
This is a virtual intersection point (where line YZ would have intersected plane ABC if plane ABC were unbounded).
The intersection of XYZ on the unbounded plane ABC would have been a straight line between points 2 and 3.
In reality, plane ABC is bounded at edge BC, thus the intersection line actually stops there, at point 1.
Point 1 is found at the intersection between the line between 2 and 3 and the edge BC. This is also the intersection point of BC with plane XYZ.
Point 1 is projected into view F.
The visibility of line BC is found using the visibility test.
The intersection of planes ABC and XYZ is shown as a straight line between points 1 and 2.
The solution is now complete. Click the forward arrow below to see this solution again.e#
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Int (Pln & Pln) Virt - 2
Descriptive Geometry Problems -- Intersections
Intersection of Two Planes -- Virtual Intersection
Example 2. Determine the intersection of planes ABC and QRS. Show correct visibility..........................................................................
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As in the previous problems, first we determine which lines (plane edges) are involved in the intersection.
From view F, it is obvious that AB and QS are not in the intersection. Their visibilities are determined and shown in views H and F.
Four lines remain to be determined: AC, BC, QR, and RS.
Using the visibility test, it is determined that lines BC and RS are not involved in the intersection and their visibility is shown accordingly.
To see the step-by-step solution, click the forward arrow below.................
The cutting plane method is now used on line AC.
Projection lines for the cutting plane are drawn from view H into view F.
The intersection of line AC and plane QRS is found at point J in view F, which is then projected into view H.
The visibility is determined and the construction lines have been removed for clarity.
Now we will find a virtual intersection point.
We will consider plane QRS to be the unbounded plane. Draw projection lines from line BC in view H down to view F. Point K is the virtual intersection point.
The virtual intersection point K is projected into view H.
The line between points J and K is what the intersection line would look like if plane QRS was unbounded.
Notice that this virtual intersection line crosses the actual boundary of plane QRS.
The intersection between line JK and QR finds point I, which is the actual intersection point of line QR with the plane ABC.
Visibility is determined for line QR.
Line IJ is the actual intersection between the two planes ABC and QRS.
The solution is now complete. To see this solution again, click the forward arrow below.
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Intersection of two planes (vir
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Intersection of Two Planes (Virtual Intersection) Quiz - 1/1Q
Quiz 9.23
A virtual intersection is located on a bounded plane.s
false, or "Inconclusive" (if applicable) if insufficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.23. A virtual intersection is located on a bounded plane..........
Button
answer
False. A virtual intersection is located off a bounded plane. Part of the definition of a virtual intersection is that the bounded plane is extended to infinity........
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Intersection of two planes (virt
Distance (LinMthd) - 1
Development (Cylin) - 2
Distance between lines (plane m
Intersection of a Plane and a So
Animation
example
Descriptive Geometry Problems -- Intersections
Intersection of a Plane and a Solid
The basic approach to finding the intersection of a plane and a solid is similar to that
of two planes. However, to simplify the problem, you must use some principles of virtual
intersection, spatial reasoning abilities, and common sense.
The procedure is as follows:
1. Identify all lines. It helps to write down a list so that you don't forget to consider any
of the lines.
2. Identify the intersection area.
3. Find the intersections, or virtual intersections, of the solid edges on the plane. You
will treat the solid surface as a group of planes.
4. Establish intersection lines, bounded by the physical limits of the objects.
5. Establish visibility.
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Click on the background to return to Real Intersections.
Intersection of a Plane and a So
Development of Cones and Cylinde
Intersection of two planes (vir
10 Basic Geometric-5
Fp>j>j>
Int (Pln & Sld) - 1
Descriptive Geometry Problems -- Intersections
Intersection of a Plane and a Solid
Example 1. Find the intersection between the rectangular plane
and the triangular prism. Do not show lines within the prism...............................
Dashed construction lines are drawn in both views to complete plane ABCD.
These are not the final lines, but they are needed to find the intersections between the prism and the plane in the following steps.
backward
forward
To see the step-by-step solution, click the forward arrow below.ard arrow below.
First, lines 1 and 2 of the prism are dashed in black. Their presence will be needed to show the visibility of lines AB and CD.
Note that AB and CD are not involved in the intersection, as both are shown in their entirety in view H..
The visibility of lines AB and CD is determined using the visibility test and shown in view F.
Notice that point A is behind the prism in view F, while edge CD is in front of the prism.
The cutting plane method is used with line 1, beginning in view H.
Projection lines are drawn into view F, stopping at the appropriate plane edges.a
Point X is found from the intersection of the cutting plane and line 1 in view F. Then it is projected into view H.
X is the intersection point between line 1 and the plane ABCD.
The visibility of line 1 is determined on both sides of the intersection point.
The construction lines have been removed for clarity.
Now the cutting plane method is begun with line 2.
However, line 2 does not intersect plane ABCD in view F, so we must find the virtual intersection point.
Point Y is the point where line 2 would have intersected plane ABCD if the plane was unbounded.
This virtual intersection point was first found in view F, and then projected into view H.
Since line 2 does not intersect plane ABCD, visibility is trivial to determine. In view H, line 2 is behind the prism. In view F, it is only hidden where it is covered by the edge of plane ABCD.
Point Y is marked for later use in finding the intersection lines.
Now the cutting plane method is used with line 3.
Projection lines are drawn from view H into view F, stopping at the edges of plane ABCD.
The real intersection point Z is found in view F, and then projected into view H.+
The visibility around point Z is determined, and the construction lines are removed.
We now have three points of intersection, X, Y, and Z. (Don't forget that Y is a virtual intersection point.)
The intersection of the bounded plane ABCD with the prism is found, and the visibilities can be found intuitively or using the visibility test.
To see how this intersection was found, you may want to go back and look at the previous step again. The solution is now complete.
A triangular plane is formed by points X, Y, and Z. This triangular plane would be the intersection of the prism with plane ABCD if the plane was unbounded.
However, since the plane is bounded, one more step is needed to show the final intersection and correct visibility.
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Int (Pln & Sld) - 2
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Descriptive Geometry Problems -- Intersections
Intersection of a Plane and a Solid
Example 2. Find the intersection of the
plane ABC and the tetrahedron
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To see the step-by-step solution, click the forward arrow below.ard arrow below.
There are 9 different lines to consider in this problem.
Two of the edges of plane ABC, AB and BC, are not involved in the intersection. Their visibility is found and shown.
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Three of the lines on the tetrahedron, RS, ST, and RT can also be seen in their entirety outside the intersection area, so they do not intersect the plane.
Their visibility is found by a visibility test on the tetrahedron.
Temporary construction lines are drawn to complete plane ABC.
These lines will be needed when we use the cutting plane method to determine the intersections of the lines of the tetrahedron with the plane.
The cutting plane method is first done with line QS. The cutting plane is created in view F, and projection lines are draw into view H.
Line QS in view H is drawn temporarily as a dashed black line.
The intersection of the cutting plane and line QS in view H finds the intersection point, 1, of QS with plane ABC.
Point 1 is then projected back into view F.
&&>%#&
The visibility of line 1 is found in both views and the construction lines are removed for clarity.
The position of point 1 remains marked so that it can be used later in finding the complete intersection..
Now the cutting plane method is begun with line QT.
The cutting plane is created in view F and projection lines are drawn into view H.
The cutting plane is projected into view H and intersection point 2 is found.
The visibility of line QT is found in both views.
The cutting plane method is begun for line QR.
However, line QR does not intersect the plane, so a virtual intersection point must be found.
Point 3 is found in view H and projected back into view F.
As a virtual intersection point, point 3 is the point where line QR would have intersected the plane ABC if the plane was unbounded.
Line QR is completely visible in both views, but point 3 remains marked for use in finding the complete intersection.
The intersection would appear as a triangular plane between points 1, 2, and 3 if plane ABC was unbounded.
In order to get the final solution, this triangle is bounded by plane ABC.
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This is how the final solution appears. All the lines within the intersection are removed, and the correct visibility of all the lines is shown.
Try to picture this intersection in three dimensions. Click the forward arrow to see the tutorial again.
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Intersections of Planes and Sol
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Intersection of a Plane and a Solid Quiz - 1/2C
Quiz 9.24
The intersection region of a plane and a tetrahedron can be a triangle................." (if applicable) if insufficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.24. The intersection region of a plane and a tetrahedron
can be a triangle..........
Button
answer
True. The previous example problem showed that this is possible. If the plane ABC had been larger, the intersection would have been a triangle....
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Intersection of a Plane and a Solid Quiz - 2/2C
Quiz 9.25
The intersection region of a plane and a tetrahedron can be a quadrilateral....................plicable) if insufficient information is provided to determine if the
statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.25. The intersection region of a plane and a tetrahedron
can be a quadrilateral.eral.
Button
answer
True. At certain angles, a plane can intersect with all four faces of a tetrahedron.on.{
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Intersection of Two Solids
Descriptive Geometry Problems -- Intersections
Intersection of Two Solids
This type of problem is usually very complicated. It involves multiple solutions
of the plane-solid intersection problem.
Tactics:
1. Get the easy ones first.
2. Use methods of real and virtual intersections
3. Work with one plane at a time.
4. It may be useful to turn extruded prisms on edge view, to use cutting planes.
Animation
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Click on the background to return to Real Intersections.
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Distance between lines (plane m
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Distance Between Lines (Plane Method) Quiz - 1/4e
Quiz 9.15
The horizontal connector appears horizontal in the front view...................clusive" (if applicable) if insufficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.15. The horizontal connector appears horizontal in the
front view..........
Button
answer
True or Inconclusive are acceptable answers. The connector can be horizontal, or in exceptional cases, a point. It would only be a point if the lines were intersecting............
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Distance Between Lines (Plane Method) Quiz - 2/4E
Quiz 9.16
The location of the shortest connector between two skew lines is found in a view where
both lines are in true length......................... determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.16. The location of the shortest connector between two
skew lines is found in a view where both lines are
in true length..........
Button
answer
True. This was shown in examples 1 and 2 of this section....
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Distance Between Lines (Plane Method) Quiz - 3/4S
Quiz 9.17
The true length of the shortest connector between skew lines is found when one of the
skew lines is in true length.....................
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.17. The true length of the shortest connector between skew
lines is found when one of the skew lines is in true
length..
Button
answer
False. The true length of the shortest connector can only be found if one of the lines is in point view or if both skew lines appear to be parallel...
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Distance Between Lines (Plane Method) Quiz - 4/4E
Quiz 9.18
The location of the shortest horizontal connector is found in a view 90 degrees from the
horizontal view.....information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.18. The location of the shortest horizontal connector is
found in a view 90 degrees from the horizontal view...........
Button
answer
The location of the shortest horizontal connector is found in a view that is orthogonal to both the horizontal view and a vertical view where both skew lines appear to be parallel................
This is the simplest kind of intersection. We will also be looking at intersections involving
plane with plane, plane with solid, and solid with solid. All these more complicated
intersections are based on the line-plane intersection, so you must be absolutely clear
on this type of intersection before you continue.
Procedure:
1) First you must determine if that line and plane actually intersect. By using the visbility
test at the apparent interesections between the line and the plane's edges, you can
determine this. If the line is visible or invisible at both edges, it did not intersect the
plane.
2) Find the a view with the plane in true shape, then create an adjacent view parallel to
line which intersects the plane. This will give you the true length of the line and the
edge view of the plane. Not only will this allow you to find the intersection point, but
also the angle between the line and the plane.
3) Note that, using Descriptive Geometry, this is a long process involving the creation of
at least three new views. Later you will learn a "shortcut method" called the cutting
plane method that is the only practical way to solve more complicated intersections.
Show example
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step1
1) First you must determine if the line and plane actually intersect. By using the visibility
test at the apparent intersections between the line and the plane's edges, you can
determine this. If the line is visible or invisible at both edges, it does not intersect the
plane...............................
step2
2) Find a view with the plane in true shape, then create an adjacent view parallel to the
line which intersects the plane. This will give you the true length of the line and the
edge view of the plane. Not only will this allow you to find the intersection point, but
also the angle between the line and the plane...............................
step3
3) Note that, using Descriptive Geometry, this is a long process involving the creation of
at least three new views. Later you will learn a "shortcut method" called the cutting
plane method that is the only practical way to solve more complicated intersections.
Click on the button "Show Steps" to see the steps of the solution for these problems.
Show Steps
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Descriptive Geometry Problems -- Intersections
Intersection of a Line and a Plane
Example 1. Determine if the bounded plane ABC and the line DE intersect?
To see the solution, click the forward arrow below..............ow.
backward
forward
Create projection lines 1 and 2 from the apparent intersections in view H. Using visibility, we find that the plane edge AB is on top of DE, but the plane edge BC is below DE.
This means that line DE must pass through the bounded plane. Therefore, they must intersect.
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Int (Ln & Pln) - 2
' (4)2*
Descriptive Geometry Problems -- Intersections
Intersection of a Line and a Plane
Example 2. A line DE intersects plane ABC. Determine the point of intersection and
the true angle between the line and the plane................................................................................................................................................................................
To see the step-by-step solution, click the forward arrow below.ard arrow below.
backward
forward
First create a line on plane ABC in view H parallel to the view line.
Draw in the projection lines...
Draw the new line in view F. Note that it is now in true length. The only reason we have created this line is to allow us to get the edge view of the plane quickly.
Create view 1 perpendicular to the line in true length in view F.[%
Draw in the projection lines...
Construct the edge view of plane ABC and line DE in view 1.
Point P is the point of intersection between the line and the plane.
~'2'{'
Create view 2 parallel to the edge view of plane ABC in view 1.
Draw in the projection lines...
Construct the true shape of plane ABC and line DE in view 2. Plane ABC is now in true shape because view 2 is parallel to view 1, where the plane is in edge view.1)
Create view 3 parallel to line DE in view 2. Remember that any view adjacent to a plane in true shape will give the edge view of the plane.
Draw in the projection lines...
The final construction in view 3 gives the edge view of plane ABC and the true length of line DE. The point of intersection is labelled P, and the true angle can be measured in this view.
The solution is now complete.
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The Cutting Plane Method
Descriptive Geometry Problems -- Intersections1
The Cutting Plane Method
This is a shortcut method for finding the intersection point between a line and a plane
without creating any additional views as done in the previous example.
Technique:
Given a plane and a line intersecting it, consider a "cutting plane" perpendicular to one of
the original views (so it appears in edge view) and also containing the intersecting line.
That is, the cutting plane is drawn in edge view directly over the line in one of the original
views.
The intersection point between the original line and plane is also contained in the
intersection between the original plane and the cutting plane. Project the intersection of
the cutting plane and the original plane (called the "cut line") into the other view. The
solution is the intersection of the cut line and the original line, since both contain the
intersection between the original line and plane. This is point P.
Project P into the adjacent view and establish visibility.
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The Cutting Plane Method
Point View Ex.
Cutting Pln Mthd - 1
Descriptive Geometry Problems -- Intersections
The Cutting Plane Method
Example 1. Find the intersection point P between line m and plane ABC................................
Default &Color
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To see the step-by-step solution, click the forward arrow below.ow.
backward
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Draw the cutting plane (1,2) in edge view in view F.
Draw its projection lines into view H, intersecting with the correct edges of plane ABC.
Draw the cutting plane in view H. Point P is the intersection of the plane ABC and line m.c
Project point P into view F......
Show the visibility in view H. The red dashed line indicates where line m is under the plane ABC.
This is the invisible portion of the line becuase a projection line from the apparent intersection of line m and the edge AC would hit AC first in view F.
Show the visibility in view F. The red dashed line indicates where line m is under the plane ABC.
The solution is now complete.
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Cutting Pln Mthd - 2
L!*'O
Descriptive Geometry Problems -- Intersections
The Cutting Plane Method
Example 2. Find the intersection point P between line m and plane ABC. This is
identical to the previous problem, but shows an alternate solution method........................
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To see the step-by-step solution, click the forward arrow below.................
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First, create the cutting plane. In this example, we will make it in view H.
Draw in the projection lines for the cutting plane, terminating them at the correct edges of plane ABC.
Construct the cutting plane in view F. The intersection of the cutting plane, line m, and plane ABC is at point P.
Project point P into view H.
Determine the visibility of view H.
A projection line drawn from the apparent intersection of edge AC and line m in view H will hit AC first in view F, thus the part of line m to the left of P in view H is invisible.
Determine the visibility of view F.
A projection line drawn from the apparent intersection of edge AB and line m in view F will hit line m first in view F, thus the part of line m to the right of P in view F is invisible.
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Cutting Plane Method Quiz - 1/1
Quiz 9.19
Consider the case of a line intersecting a plane. In one view, the edge view of a cutting
plane is coincident with the line. In an adjacent view, the intersection of the two planes
forms a line that contains the intersection of the original line and plane...utting plane intersects another plane with a line in one view and contains it in an adjacent view.........n an adjacent view..........
Button
answer
True. This is the definition of a cutting plane....
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Cutting PLane Method (1)
Cutting PLane Method (1)
Intersection of Two Planes -- Re
Descriptive Geometry Problems -- Intersections1
Intersection of Two Planes -- Real Intersection
There are two methods for solving problems involving the intersection of two planes. This,
the first, solves the problem by real intersection. Virtual intersection is covered in the next
section.
In these problems, you are usually given the outlines of two planes. You are asked to find
the shape of the intersection (which should be a line), and to establish the visibility of all
the lines.
Procedure:
1. Identify all the lines (edges of the planes).
2. Identify the "intersection" area.
3. Determine if the lines intersect the other plane (i.e. if on the same side of the plane in
the intersection area).
a. If it doesn't intersect - establish its visibility
b. If it does intersect - find the intersection point
- establish visibility of the lines
- establish the intersection line on the plane
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Int (Pln & Pln) Real - 1
Descriptive Geometry Problems -- Intersections
Intersection of Two Planes -- Real Intersection
Example 1. Determine the intersection of planes ABC and XYZ.
Show correct visibility.
To see the step-by-step solution, click the forward arrow below.................
backward
forward
There are six lines to consider in this problem: AB, BC, AC, XY, YZ, and XZ.
From view H, it is obvious that AB and XZ are not involved in the intersection. From view F, one can see that AC, XZ, and YZ are also not involved.
The visibilities of these four lines are determined.
Two lines remain: XY and BC. For now, these lines are dashed in gray.
The following steps will determine the points of intersection of these lines, and their visibility will be shown..A
The cutting plane method is begun on line XY.
Projection lines are drawn from the apparent intersection of XY with ABC in view H.9
The intersection of line XY and plane ABC has been found. The visibility was determined, and line XY was dashed at the invisible portions..
The cutting plane method is also used with line BC.
The point of intersection of line BC with plane XYZ is shown and visibility is determined.
The last step is to draw in the intersection line. Simply draw a line between the intersection points we have just found.
This intersection is shown in red, and the unnecessary construction lines have been erased. The solution is now complete.
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Edge View and True Shape of a Pl
Descriptive Geometry Problems -- Basic Examples
The Edge View and True Shape of a Plane
Finding the edge view and true shape of a plane is very simple when you have mastered
the principles behind finding the true length and point view of a line.
While solving these types of problems, keep in mind:
1) The point view of a line is also the edge view of any plane parallel to that line.
2) Note that the edge view is not unique. There are an infinite number of edge views for a
given plane.
3) Thus, to get the edge view of a plane, find the point view of any edge of the plane. To
do this, create a view showing the edge in true length, then create the point view.
Creating a plane parallel to the edge view will give the true shape.
There is a shortcut, which is shown on a few of the example problems. Note that the
method above requires the construction of two additional views to get the edge view. There is a way to get the edge view with the construction of only one additonal viewing plane. In one of the two views of the original plane, construct any line that is parallel to the other viewing plane on the original plane. Project this line to the other view, where it is now in true length. Now create the point view of this line, which also gives the edge view of the original plane... plane.
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EV & TS of Plane - 1
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Descriptive Geometry Problems -- Basic Examples
The Edge View and True Shape of a Plane
Example 1. Find the edge view and true shape of plane ABC.
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Draw in the projection lines from view F and view H.
Construct view 1 parallel to line AC of the plane in view H.
Draw the projection lines...
Draw the projection lines in view 1. The lengths of these projection lines are the same as those in view F.
From the projection lines just drawn, create plane ABC in view 1. Note that the edge AC of the plane is now in true length.
Draw view 2 perpendicular to line AC. Since the view is perpendicular to a line on plane ABC, it is also perpendicular to the plane itself.
Draw in the projection lines...
Draw the projection lines in view 2. Note that points A and C appear on top of each other.
Construct the edge view of the plane in view 2.i$
Make view 3 parallel to the edge view of plane ABC.
Draw in the projection lines...
Draw the projection lines in view 3. Note that points A and C lie along the same line, but at different distances from the view line.
Connect the ends of the projection lines to show plane ABC in True Shape (TS) in view 3.
The solution is now complete. Click the forward arrow below to see the solution again.
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EV & TS of Plane - 2
Descriptive Geometry Problems -- Basic Examples
The Edge View and True Shape of a Plane
Example 2. Find the edge view and true shape of plane XYZ.
backward
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This solution uses a simple trick to find the solution without creating as many views as in the previous problem.
To see the step-by-step solution, click the forward arrow below.
below.
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Construct a line WZ in the plane parallel to the view line in view F.y
Draw in the projection lines.
Line WZ is now in true length because it was parallel to the view line in the previous view.
View 1 is drawn perpendicular to WZ in view H.
Draw in the projection lines...
The projection lines are drawn in view 1. (The lengths were obtained from the projection lines in view F.)
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The edge view is constructed in view 1. The plane is in edge view because line WZ, which lies on plane XYZ and was in true length in the previous view, is now in point view.
View 2 is constructed parallel to the edge view of plane XYZ.
The projection lines are drawn in...
The projection lines are drawn in view 2. These lengths are obtained from view H.
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Finally, the true shape of plane XYZ is constructed. It is not necessary to show line WZ, since it was only created to assist in the solution.
lick the forward arrow to see the solution again.
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EV & TS of Plane - 3
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Descriptive Geometry Problems -- Basic Examples
The Edge View and True Shape of a Plane
Example 3. Find the perpendicular connector between point P and plane ABC. Show
the connector in all views.r in all views...............................
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To see the step-by-step solution, click the forward arrow below.ard arrow below.
First, construct line AD in view H parallel to the view line.i
Draw in the projection lines...
Draw line AD in view F. Notice that it is now in true length.
Draw view 1 perpendicular to line AD in view F.C
Draw in the projection lines...
Draw the projection lines in view 1.e
Construct the edge view of plane ABC and point P in view 1.
Construct the perpendicular connector in view 1. Its end points are P and Q and it is in true length in this view..
Find line PQ in view F. It is parallel to the view line between F and 1 because it was in true length in view 1.
Finally, find line PQ in view H. The solution is now complete.
.Click the forward arrow below to repeat the solution.
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The Edge View and True Shape of a Plane Quiz - 1/4E
Quiz 9.10
A plane is in true shape when two lines in that plane are in true length...............h.(if applicable) if insufficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.10. A plane is in true shape when two lines in that
plane are in true length..........
Button
answer
Inconclusive. Two parallel lines may be in true length, but the plane containing them may not be in true shape. (Think 3-D!)
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The Edge View and True Shape of a Plane Quiz - 2/4E
Quiz 9.11
Any view orthogonal to a viewing plane showing a plane in edge view will show that plane in true shape.
edge view of a plane will show that plane in true shape..t is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.11. Any view orthogonal to the edge view of a plane will
show that plane in true shape..........
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answer
Inconclusive. The view orthogonal to the edge view of the plane must be parallel to the edge view in order to show the true shape.
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The Edge View and True Shape of a Plane Quiz - 3/4E
Quiz 9.12
There is only one true shape view of a plane.tement is
false, or "Inconclusive" (if applicable) if insufficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.12. There is only one true shape view of a planeeeeeeeeee
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The Edge View and True Shape of a Plane Quiz - 4/4E
Quiz 9.13
There are three edge views for a triangular bounded plane. false, or "Inconclusive" (if applicable) if insufficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.13. There are three edge views for a triangular bounded plane..........
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Dihedral Angle
Descriptive Geometry Problems -- Basic Examples
Dihedral Angle
Finding the Dihedral angle of two intersecting planes is an extension of the edge view
procedure. In this kind of problem, you need to find the edge view of both the intersecting
planes.
Procedure:
1) Find the line which is the intersection of the two planes, that is, the line which belongs
to both planes.
2) Create a view which gives the true length of this line.
3) Create a view perpendicular to the true length line, which will give the edge views of
both planes.
4) Measure the dihedral angle between the two edge views.
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Click on the background to return to Real Intersections.
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Dihedral Ang - 1
Descriptive Geometry Problems -- Basic Examples
Dihedral Angle
Example. Find the angle between planes ABC and ABD (the dihedral angle).............................
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To see the step-by-step solution, click the forward arrow below.ard arrow below.
First, draw in the projection lines.
Create view 1 parallel to line AB in view H. We choose line AB because it is the line of intersection of the two planes ABC and ABD.
Draw in the projection lines...7
Draw the projection lines in view 1. (The lengths are obtained from view F.)
Construct the planes in view 1. Notice that line AB is now in true length.
Create view 2 perpendicular to line AB.C
Draw in the projection lines...
Draw the projection lines in view 2.
Construct the planes in view 2. Now line AB is in point view, and you see the two planes ABC and ABD in edge view.I
The dihedral angle is measured between the two planes in edge view.
The solution is now complete. To see the solution again, click the forward arrow below..^!
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Dihedral Angle Quiz - 1/1
Quiz 9.14
The dihedral angle is shown when the intersecting line between two planes is in true
length...h.insufficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.14. The dihedral angle is shown when the intersecting
line between two planes is in true length..........
Button
answer
False. The intersecting line must be in point view, which puts both planes in edge view...
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Distance Between Lines -- Plane
Descriptive Geometry Problems -- Basic Examples
Distance Between Two Lines -- Plane Method
We have already covered finding the distance between two lines by the line method. The
plane method allows you solve the problem without creating as many views. It will help
you to find the shortest connector faster than the previous method allows.
This method involves finding a view where both lines appear parallel, then measuring the
perpendicular distance. This works because if the connector is in true length, and both
lines are perpendicular to it, then those two lines should appear perpendicular to the
connector. This means that the two original skew lines must appear parallel in this view.
(Note that this view is not unique.)
This is called the plane method because you must create a plane from one of the lines
that is parallel to the other line. Finding the edge view of that plane will give you a view
where both lines appear parallel. Pay careful attention to the example problems!
Problems involving finding the distance between two lines may be more specific than
simply asking for the shortest connector. They may ask for something more specific
than the shortest horizontal connector, or the shortest connector 40 degrees from the
horizontal.
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Distance (PlnMthd) - 1
Descriptive Geometry Problems -- Basic Examples
Distance Between Two Lines -- Plane Method
Example 1. Find the minimum distance between the skew lines AB and CD.................................
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backward
forward
To see the step-by-step solution, click the forward arrow below.............
Create a line (in green) in view F parallel to the view line.
Make another line in view F parallel to CD. The intersection is point E. We have just created a plane ABE parallel to line CD.
Line EA (in green) in view H must also be parallel to line CD in view H. We do not know its correct length yet.
Draw in the projection line from point E to determine the length of EA in view H.
From point E in view H, we can now draw in line BE, which is in true length (since it was drawn parallel to the view line in view F.)
Draw in the projection lines...!
Draw in the projection lines...
View 1 is created perpendicular to line BE in view H. Since line BE lies on the plane ABE, view 1 is also perpendicular to that plane.
Construct the edge view of plane ABC and line CD in view 1. Notice that line AB and CD appear parallel.
Now the minimum distance can be found as the perpendicular connector in this view.
The solution is now complete. To repeat this solution, click the forawrd arrow.
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Distance (PlnMthd) - 2
Descriptive Geometry Problems -- Basic Examples
Distance Between Two Lines -- Plane Method
Example 2. Find the shortest connector between lines AB and CD. Show it in all views...............................
backward
forward
To see the step-by-step solution, click the forward arrow below.ard arrow below.
First, draw a line in view H parallel to the view line. We will be creating a plane from this line.
Draw a line in view H parallel to CD in view H. Where it intersects the first line is point E. We have just created plane ABE, parallel to line CD.
Next, draw the edge of the plane parallel to CD in view F.
To determine the length of this edge, we draw the projection line. Thus we have located point E in view F.i
Connect points E and B to form the line EB, which is now in true length. It is in true length because EB was drawn parallel to the view line in the previous view.{
Draw in the projection lines...
Create view 1 perpendicular to line EB in view F. We do this because we want to get the edge view of plane ABE in view 1.
Draw the projection lines into view 1...
Construct lines AB (edge view of plane ABE) and CD in view 1. Notice that the lines appear parallel in this view.
The answer to the problem, the shortest connector, is found in this view.)
View 2 is created parallel to line AB and CD in view 1. It is not necessary to create this view to solve the problem, but it is interesting to note that the shortest connector appears in point view in this view..
Now we create projection lines from view 1 back into view F in order to show the shortest connector in all the views, as the problem requests..u
The shortest connector is drawn in view F.
Projection lines for the shortest connector are drawn into view H.
Finally, the shortest connector is drawn in view H.
The solution is now complete. Click the forward arrow below to see it again.
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Distance (PlnMthd) - 3
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Descriptive Geometry Problems -- Basic Examples
Distance Between Two Lines -- Plane Method
Example 3. Find the shortest horizontal connector from AB to CD.
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To see the step-by-step solution, click the forward arrow below.ard arrow below.
First, draw a line in view F parallel to the view line. We will be creating a plane from this line.
Draw a line in view F parallel to CD in view F. Where it intersects the first line is point E. We have just created plane ABE, parallel to line CD.
Next, draw the edge of the plane parallel to CD in view H.
To determine the length of this edge, we draw the projection line. Thus we have located point E in view H.
Connect points E and B to form the line EB, which is now in true length. It is in true length because EB was drawn parallel to the view line in the previous view./&
Draw in the projection lines...
Create view 1 perpendicular to line EB in view H. We do this because we want to get the edge view of plane ABE in view 1.
Draw the projection lines into view 1...
Construct Lines AB and CD in view 1. We see the edge view of plane ABE and lines AB and CD appear parallel.
In order to find the horizontal connector, we create view 2 perpendicular to view H. (Thus the horizontal connector should appear in point view in view 2.)
Draw the projection lines...
Draw the projection lines into view 2.
Construct lines AB and CD in view 2. The horizontal connector is in point view at the apparent intersection of these lines.
Project the horizontal connector into view 1. Since it is in point view in view 2, the connector lies along the projection line in view 1.;-
Project the horizontal connector into view H. The line is created by the intersections between the projection lines and lines AB and CD in view H.
The same is done for view F. The horizontal connector should appear horizontal in this view.
The solution is now complete. Click the forward arrow to see the solution again.f/
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Distance (PlnMthd) - 4
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Descriptive Geometry Problems -- Basic Examples
Distance Between Two Lines -- Plane Method
Example 4. Find the shortest connector 20 degrees from the horizontal between AB and
CD, going downward from AB to CD................................
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To see the step-by-step solution, click the forward arrow below.ard arrow below.
First, draw a line in view F parallel to the view line. We will be creating a plane from this line.
Draw a line in view F parallel to CD in view F. Where it intersects the first line is point E. We have just created plane ABE, parallel to line CD.
Next, draw the edge of the plane parallel to CD in view H.
To determine the length of this edge, we draw the projection line. Thus we have located point E in view H.q&
Connect points E and B to form the line EB, which is now in true length. It is in true length because EB was drawn parallel to the view line in the previous view.
Draw in the projection lines...
Create view 1 perpendicular to line EB in view H. We do this because we want to get the edge view of plane ABE in view 1.
Draw the projection lines into view 1...
Construct Lines AB and CD in view 1. We see the edge view of plane ABE, and lines AB and CD appear parallel.
In order to find the connector 20 degrees from the horizontal, we create view 2 at 20 degrees from the perpendicular to view H. q+
Draw the projection lines...
Draw the projection lines into view 2.
Construct lines AB and CD in view 2.
Project the connector into view 1.
Project the connector into view H.
In view F, the connector appears 20 degrees from the horizontal. The solution is now complete.%/
Another Example: A tetrahedron ABCD. Show the visibility of the edges................................................
Draw in the edges of the tetrahedron.
Draw in a projection line from the apparent intersection in view H into view F.
Since the projection line crosses AC first in view F, line AC must be on top in view H.
Since AC is on top, line BD is dashed.
Next, a projection line is drawn from the apparent intersection in view F.
It hits line AD first in view H, so AD is on top in view F.
Line BC is dashed in view F.
The solution is now complete.
Click the button below to see the example again.
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(Click on the forward button to see the next step after each new construction and message.)
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Basic Problem Solving - 1
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Basic Problem Solving
Now some simple problems will be solved using the 2 basic principles and 10 basic
geometric relationships.
Example 1. Locating a point. Point A is given in views 1 and 3. Find it in view 2............................
First draw a projection line from the point in view 1 into view 2.
Remember that the projection line must be orthogonal to the view line between planes 1 and 2.
To see the step-by-step solution, click the forward arrow button below.ow button below.
Another orthogonal projection line is drawn from point A in view 3.
Point A in view 2 is located where the projection lines cross.
This must be true in order to satisfy the first principle of orthogonal projection.
The solution is now complete.
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Basic Problem Solving
Example 2. Plane ABC and line DE are parallel. Both objects are given in view F. Find D
in view H...................
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First, draw a line parallel to line DE anywhere on the plane ABC in view F.
Since the problem states that line DE and the plane are parallel, you know that the line you have just drawn is also parallel line DE.lane.u
Next, draw projection lines from the intersection of the new line and the edges of the plane. This will allow you to find the line in view H.
Draw in the parallel line on the the plane in view H where the projection lines intersect the edges of the plane.S
Draw line DE in view H from the point E, which was given.
Remember that it must be parallel to the line you just drew on plane ABC in view H.c
Draw a projection line from D in view F to locate the end of the line in view H.
Finally, trim the line DE to its proper length.
The solution is now complete.lick the forward arrow button below.
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Basic Problem Solving
Example 3. Given line PQ and point R, construct a line through R that is
perpendicular to PQ at an undetermined point S. Find the length of the
connector RS...
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To see the step-by-step solution, click the forward arrow button below.ow button below.......k
Draw view 1 parallel to PQ in view F. That way, you can get the true length of PQ in view 1.
You will need the true length of PQ in order to find the perpendicular line.
Draw the projection lines from view H into view F and up to the view line between F and 1.
Remember that projection lines are always perpendicular to view lines.
Construct the projection lines in view 1.
By the second principle of orthogonal projection, the projection lines in view 1 must have the same lengths as the projection lines in view H.
Draw in line PQ (which is now in true length) in view 1. Point R is also located.
Line SR, which has not yet been constructed, will appear perpendicular to PQ in this view.)#
The view line to view 2 is drawn perpendicular to line PQ in view 1 so as to find PQ in point view in view 2.
Draw projection lines up to the view line between views 1 and 2...........
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Line RS is drawn. Point S is on the line PQ.
RS must be in true length in this view because it is perpendicular to PQ, which is in point view. You can now measure the length of RS in view 2.
Using the lengths between the points in view F and the view line between F and 1, the lengths of the projection lines are found in view 2.
Point S is located at the intersection of the projection line and line PQ in view F. The connector RS is found in view F.
The point view of line PQ and the point R are located.
The next step is to draw in the connector between R and PQ.
You solved the problem in the last step, but you should show the line RS in all views.
Since RS is perpendicular to PQ (which is in true length in view 1), it will appear perpendicular in view 1. That is the only information you need to construct RS in view 1.
The projection line for point S is drawn from view 1 into view F.
The projection line from view F into view H is drawn.c,
Finally, RS is drawn in view H.
Notice that constructing line RS by "working backwards" was very easy. Since we knew that point S was on line PQ, it could simply be found at the intersections of its projection lines and PQ......
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DG Problems - 1
forward
es (Real
Int (Ln & Pln) - 2
Cutting Pln Mthd - 1
Development (Cone) - 1
Point View of a Line
Descriptive Geometry Problems -- Basic Examples
Point View of a Line
Finding the point view of a line is very important in solving many descriptive geometry
problems.
The point view of a line occurs when both endpoints of a line project to points on top of
each other on the orthographic plane. This problem's importance will become more clear
as we continue with more advanced problems.
Method:
1. First you must find the true length of the line. This is done by creating an orthographic
plane parallel to the line.
2. To get the point view, you must then create an orthographic plane perpendicular to the
true length of the line.
Remember: You must find the true length before you can find the point view!
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Point View Ex.
Descriptive Geometry Problems -- Basic Examples
Point View of a Line
Example. Find the point view of Line AB.......
Arial
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forward
To see the step-by-step solution, click the forward arrow button below.utton below.
First draw in the projection lines from view H to view F.
Draw the view line between views F and 1 parallel to line AB in view F.
Put in the projection lines...
Construct the projection lines in view 1. The projection lines have the same lengths as those in view H.
Draw line AB in view 1. Remember that it is now in true length.1
Draw view 2 perpendicular to AB in view 1.
Draw in the projection line...i
Find the length of the projection line into view 1.
Notice that the lengths from points A and B in view 1 to the view line between F and 1 are the same. (Which is why there only appears to be one projection line.)
Finally, draw in a point representing the point view of the line AB. The solution is now complete.
Click the forward arrow to continue.
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Point View of a Line (2)
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Point View of a Line Quiz - 1/2
Quiz 9.6
Any view orthogonal to the point view of a line will show the line in true length...............plicable) if insufficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the ques
Quiz 8.6. Any view orthogonal to the point view of a line will
show the line in true length.................
Button
answer
True. All views perpendicular to the view with the line in point view are also parallel to the line itself.self....
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Point View of a Line Quiz - 2/2
Quiz 9.7
Given the front and horizontal views of a line, two additional views are required to
establish the point view of the line................... determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.7. Given the front and horizontal views of a line, two
additional views are required to establish the point
view of the line..........
Button
answer
- If one view shows the line in true length, only one additional view is required.
- If one view shows the line in point view, no additional views are required................
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Distance Between Lines -- Line M
pause
repeat
Repeat
Descriptive Geometry Problems -- Basic Examples
Distance Between Lines -- Line Method
This method is the first way you will learn to find the distance between two lines.
The shortest connector is one that is perpendicular to both lines.
Finding the distance between two lines is an application of the point view of a line. This
method involves putting one of the original lines in point view, so the connector will appear
in true length. Since the connector is in true length, it will also appear perpendicular to
the second line.
Finding angles between lines by the line method:
To find the angle between two skew lines, put both lines in true length. Any view off the
point view of a line yields the true length of the line. So create a view from the point view
of the first line which also gives the true length of the second line. You will see more
clearly how to do this by looking at the following example problems........
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Distance (LinMthd) - 1
%L)p-
Descriptive Geometry Problems -- Basic Examples
Distance Between Lines -- Line Method
Example 1. Determine the minimum distance between skew lines AB and CD.. AB and CD.
nt to
Dot &
lines
Default
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Courier
To see the step-by-step solution, click the forward arrow below.ard arrow below.
First, draw in the projection lines in views H and F.
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Draw in the projection lines.W
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Draw in line CD and the point view of line AB.
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The solution is now complete.
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Distance (LinMthd) - 2
Descriptive Geometry Problems -- Basic Examples
Distance Between Lines -- Line Method
Example 2. Determine the angle between skew lines AB and CD.......
backward
forward
ic Geomet
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Are you s
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To see the step-by-step solution, click the forward arrow below..rd arrow below.
|#6#y#
First draw in the projection lines from view H to view F.
Construct view 1 parallel to line AB in view F.A$
Draw in the projection lines...
Construct the projection lines into view 1. The lengths of the these projection lines are the same as those in view H.
Draw in lines AB and CD in view 1. Note that AB is in true length because view 1 was constructed parallel to AB in view F.
Construct view 2 perpendicular to line AB in view 1.
Draw in the projection lines...
Draw the projection lines into view 2. They have the same lengths as the projection lines in view F.
Construct lines CD and AB. Notice that Line AB is in point view.W)
Make view 3 parallel to line CD in view 2.
r*F*o*
Draw in the projection lines...
Create the projection lines in view 3. Note that points A and B are along the same line.M+
Construct lines CD and AB in view 3. Now both lines are in true length.
The answer to the problem is the acute angle shown when both lines are in true length.
Click the forward arrow below to see the solution again.
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Distance Between Lines (line Me
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Distance Between Lines (Line Method) Quiz - 1/2
Quiz 9.8
The minimum distance between two skew lines can be determined from a point view of
one of the lines...ficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.8. The minimum distance between two skew lines can be
determined from a point view of one of the lines..........
Button
answer
True. When one of the lines is in point view, the shortest connecter is a line from the point to the other line. The connnector is also perpendicular to the latter line....
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reference
Distance Between Lines (line Met
Distance Between Lines -- Plane
LinVis - 4
Distance (PlnMthd) - 2
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Distance Between Lines (Line Method) Quiz - 2/2
Quiz 9.9
The true angle between two intersecting lines can be seen when one line is in true length.....................nsufficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.9. The true angle between two intersecting lines can
be seen when one line is in true length..........
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answer
Inconclusive. This is only true if the angle happens to be 90 degrees. Otherwise, both lines must be in true length.
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10 Basic Geometric
Int (Pln & Sld) - 2
10 Basic Geometric - 6
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10 Basic Geometric RelationshipsQ
6. Skew lines are lines that are neither parallel nor intersecting.
Clicking the button below rotates the skew lines around the y axis to show
that they do not actually intersect (it is obvious that they are not parallel).ot parallel.)
Skew lines may appear to intersect, but one is always in front of the other at the apparent
intersection point. Skew lines do not exist in two dimensions.......
Geometric6
Rotate Skew Lines
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Skew lines may appear to intersect, but one is always in front of the other at the apparent intersection point. Skew lines do not exist in two dimensions..............llel.)
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10 Basic Geometric - 7
10 Basic Geometric RelationshipsI
7. If two lines are perpendicular, they will appear to be perpendicular in any view
in which one or the other is in true length.....
Geometric7
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10 Basic Geometric - 8
10 Basic Geometric Relationships
8. A plane is defined by three points, two intersecting lines, 2 parallel lines, or a
point and a line (if the point does not lie on the line).
The four pictures below show how these objects can define a plane:::I
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10 Basic Geometric - 8
Distance (LinMthd) - 2
10 Basic Geometric - 9
10 Basic Geometric RelationshipsQ
9. A line is parallel to a plane if it is parallel to any line that lies on that plane.
To find a line called "m" parallel to the plane ABC and show it in views H and F, click the
First, create a line, called "a," parallel to line "m" on the plane ABC in view H. Then find line "a" in view F. Finally, draw line "m" in view F so that it begins at the given point and is parallel to line "a" in view F..... H. Then find line "a" in view F. Finally, draw line "m" in view F so that it begins at the given point and is parallel to line "a" in view F.e "a" in view F.
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10 Basic Geometric - 10
10 Basic Geometric Relationships
10. A line is perpendicular to a plane if it is perpendicular to two intersecting
lines on that plane.
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Orthogonal Projection and Geometric Relationships Quiz - 1/5C
Quiz 9.1
A plane is defined by two lines.
Quiz 8.1. A plane is defined by two lines.
ble) if insufficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.1. A plane is defined by two lines.
real exam, you would be required to write this down.)
Go to the next page after you have tried answering the question.
Quiz 8.1. A plane is defined by two lines.
1. A plane is defined by two lines.
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Orthogonal Projection and Geomet
LinVis - 2
Dihedral angle (1)
Dihedral angle (1)
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Orthogonal Projection and Geometric Relationships Quiz - 2/5G
Quiz 9.2
A line is perpendicular to a plane if it is only perpendicular to one line contained in
that plane.sufficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.2. A line is perpendicular to a plane if it is perpendicular to
one line contained in that plane..........
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Orthogonal Projection and Geometric Relationships Quiz - 3/5I
Quiz 9.3
Two parallel lines have no points in common..atement is
false, or "Inconclusive" (if applicable) if insufficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.3. Two Parallel lines have no points in common..........
Button
answer
True. Parallel lines can be viewed as two lines, two points, or one line, but they never have a point in common....u
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Orthogonal Projection and Geometric Relationships Quiz - 4/5[
Quiz 9.4
Two skew lines have one point in common... statement is
false, or "Inconclusive" (if applicable) if insufficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.4. Two skew lines have one point in common..........
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answer
False. Skew lines have no points in common.n.I
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Orthogonal Projection and Geometric Relationships Quiz - 5/5O
Quiz 9.5
A line is parallel to a plane if it is parallel to one line in that plane.............ve" (if applicable) if insufficient information is provided to
determine if the statement is true or false.
Go to the next page after you have tried answering the question.
Quiz 8.5. A line is parallel to a plane if it is parallel to one line in
that plane..........
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answer
True. This is Rule 9 of the 10 Basic Geometric Relationships....
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Line Visibility - 1
Line Visibility
Line Visibility can easily be determined using descriptive geometry. Using only graphical techniques one can determine which line segments in a drawing are visible and which are not.
Lines m and n are shown in two views below. Do they intersect????????????????????????????????????????????????????????
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The projection line starting at P in view H crosses line n before line m in view F. This means that n is above m in view H.s
Answer
Lines m and n do not intersect.
By taking the projection at the apparent intersection points, we can see that at P, n is above m. At Q, n is in front of m.
Answer
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