Labels:text | font | screenshot | document OCR: Stupid MathCAD requires that we declare variables as constants. := 0 Zp := 0 2q := 0 C := 0 v =0 Xp := 0 Xq '=0 A := 0 z := 0 ₱ Yp Yq Ok theory time. Let (u,v) be the coordinates on the projection plane, (p,q) the coordinates on the original plane, (Xp, Yp,Zp) and (Xq, Yq,Zq) the two vectors defining the plane, (A, B,C) the origin of th plane and the projection x=zu, y=zv. Then, the intersection of the projection ray and the plane in th plane coordinate system (p,q) in function of the projection plane coordinate system (u,v) is: z p.Zptq.Zq+C zu-p.Xp+ q.Xq +A (p.Zp +qZq +C).u p.Xp + q.Xq +A Equation A zv p.Yp+ q.Yq + B (p.Zp+qZq+C).v.p.Yp+q.Yq + B Equation B Now, we will solve for p in equation A and substitute in equation B then solve for q in equation B an lastly simplify. Then, we will do the opposite to solve for p. The resulting equations should be symetrical. (p.Zp +qZq +C).u-p.Xp+q.Xq +A p= (uZp -Xp) (p.Zp+qZq+C).v-p.Yp+q.Yq + B (uqZquC+qXq+A). 2p+ g.Zg +c .v=( uq.2quCfq.XqTA). yo+ g. Yq+ B (u.Zp - Xp) (u.Zp - Xp) q= (v.Zp.A __ v.C.Xp + Yp.u.C_ Yp:A - B.u.Zp + B.Xp). ("v.Zp.Xq+v.Zq.Xp - Yp.u.Zq + Yp.Xq + Yq.u.Zp - Yq. Xp) (when simplified) And now we do the same for p. (p.Zp +q.Zq +C).u-p.Xp+ qXq +A q= "(up:ZpuC __ p_Xp-A). (uZq -Xq) p.Zp + Cup:2puCfp:XpA). 2g +C |v=p.Vp+ Cup:Zp_uCfpxpA). Vg+B (u.Zq-Xq) (u.Zq -Xq) p= (v.Zq.A_v.C.Xq+ Yqu.CYq A -B.u.Zq + B. Xq)_ (once simplified) ("v.Zp.Xq+ v.Zq.Xp - Yp.u.Zq+ Yp.Xq + Yq.u.Zp - Yq. Xp)
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