This is exactly the problem I'm wondering and I can't find any solutions online. If you want to reflect anything across a plane, you need to be orthogonal to that plane (at 90 degrees), otherwise it's an inaccurate reflection.

@fuahuahuatime5196

27 күн бұрын

I found a method that requires a little bit of math, but you can compute them easily for common vanishing points (20&70, 30&60, 45&45) 1. Assume the height from SP to CVP is 1 (units don't matter. you can apply them afterward) 2. Be aware that anything above 0.63 (little less than 2/3's of total height) will put a line outside of the 60 degree COV 3. When you draw a line from a VP to the SP, the angle to that VP never changes. Therefore, when calculating the angle, the distance from the CVP to the VP is always fixed. This means that you must be aware (from 2) where within the 60 degree COV you want to place this line (height less than 0.63 means it's within 60 degree COV) Then from here, you just find the inverse tangent of the length from CVP to VP divided by the height ratio you decided (keeping 2 in mind) tan^(-1) (L/H) = angle These are approximate Length's I calculated: L @ 20 deg = 0.36 L @ 30 deg = 0.58 L @ 45 deg = 1 L @ 60 deg = 1.73 L @ 70 deg = 2.75 So, for example, if you want to find the angle to a 70 degree VP, where the height would be about 0.20 (within 60 degree COV), then the angle will be tan^(-1) (2.75 / 0.20) = 85.84 degrees 90 - 85.84 = 4.16 degrees On the paper, measure 4.16 degrees, and that's your line from within the 60 degree COV to the 70VP. Repeat for other lines. This really bugged me, so I had to try to work it out. I would appreciate if someone developed an easier method.

So theres no way to find the station point with the brewer grid method? Let alone the left and right measuring points?

@trustyourperspective

6 ай бұрын

I don't think so. Not that I know of.

@Lolwtfbbq12

6 ай бұрын

@@trustyourperspective don't think im using it then lol