Hello.

I came up with an interesting problem that I am now trying to figure out.

Imagine a plane in space, and within that plane two parallel lines. From the perspective of any point outside of the plane (i.e. if your eye was there), the parallel lines would appear to converge at some point. Keeping it simple for now, how would one go about finding the angle that the parallel lines appear to converge at from an arbitrary point? For the purpose of this problem, let's say that the point is equidistant from the parallel lines. So, as a function of the distance from the point to the plane, how would the angle change?

I am trying to work it out by myself, and it seems I've tried almost everything. If anyone can easily solve it, would you please push me in the right direction as opposed to directly telling me the answer? I want to try and work it out. Also, I'm a student with only a very basic knowledge of calculus.

Spandan

2. This is indeed a hard problem because it is difficult to define exactly what you want. Once we can put some boundaries on what we want, I think it'll be relatively easy to get the answer.

So, here's a method how you can do this in the real world. Say you are standing in the middle of that plane with your head right between the parallel lines and somewhat above them. Put the glass in such a way that the line from your eye to the point in the horizon where the lines meet is perpendicular to the glass surface (we gotta make some assumptions, right?). Now, while you keep the glass stationary and your head stationary, use a marker and sketch along the lines of what you see onto the glass. Once you are done doing this drawing, you should see a triangle on the piece of glass. Then you can take your proctractor and ruler and get to work. Soo, this is how you measure the angle.

Now, to turn that into math, I think a good place to start would be to consider the eye being a point with coordinates (0,0,h) (h for height). Then the parallel lines are along the x-y plane beneath your feet, running parallel to the x-axis. You can define there separation to be some number, but in reality, all you need is that the separation s = 1, since you can scale the problem. If you are unsure about this, just assume that s is arbitrary.

Now, for the line from your eye to infinity. This is tricky, since if you look directly at infinity, you will be looking straight ahead, i.e. parallel to the ground. If you don't understand this part, I can elaborate on it. Now, put the "glass" plane perpendicular to that line of sight, so it'll actually be vertical and parallel to the y-z plane. Put it at a distance d from you, where d is an arbitrary number. Now to draw the lines on this plane, this is exactly like taking the projection of the lines onto the plane, through the eye. You can look at this for a similar idea. Stereographic projection - Wikipedia, the free encyclopedia

I could help you with the math, but since you want to work on it yourself, I won't spoil it for you.

If you solve this, you can try something else, which I think will be more accurate. Instead of a plane, use a spherical glass. Then you'd have to project the lines onto the glass, and measure the angle between them right where they meet. Since this is a curved, surface you may need some calculus to find those. Good luck!

3. Originally Posted by spandan
Hello.

I came up with an interesting problem that I am now trying to figure out.

Imagine a plane in space, and within that plane two parallel lines. From the perspective of any point outside of the plane (i.e. if your eye was there), the parallel lines would appear to converge at some point. Keeping it simple for now, how would one go about finding the angle that the parallel lines appear to converge at from an arbitrary point? For the purpose of this problem, let's say that the point is equidistant from the parallel lines. So, as a function of the distance from the point to the plane, how would the angle change?

I am trying to work it out by myself, and it seems I've tried almost everything. If anyone can easily solve it, would you please push me in the right direction as opposed to directly telling me the answer? I want to try and work it out. Also, I'm a student with only a very basic knowledge of calculus.

Spandan
Wouldn't it depend on how far you can see? i.e. where the horizon is?

4. chiph588@, for simplicity we should assume that you can see all the way to infinity. Since we are on a plane (and not the surface of the earth, for example), the horizon turns out to be straight ahead at eye level.

5. But we can't see to infinity. If we could wouldn't there be no vanishing point?

6. A vanishing point is where the parallel lines seems to merge. This happens at a great distance, so we might as well assume infinity. Another way to look at it is that if you are at point (0,y) and you are looking down to a point (x,0) on the x axis. The angle between this line of sight and the the y-axis is just [LaTeX ERROR: Convert failed] . If you let x get bigger and bigger, you will get x/y getting bigger and bigger. If you let x go to infinity you are looking at the limit of the arctan function as its argument reaches infinity. That limit is [LaTeX ERROR: Convert failed] , which means that if you follow the x-axis to "infiinity", you will be looking parallel to it.

7. I think the angle will be zero degrees and the two lines won't appear to be straight. Intuitively it makes sense to me, but I'm probably wrong.

8. It's not zero. If you project the image on a plane it's quite simple. If you project on a sphere it's a little more complicated. It's fun!

9. What if you were height $h$ above the plane, right in between the two lines. Look straight down and the two lines appear to be distance $d$ apart.

Now look down "the road" at a $45$ degree angle. The two lines will appear to be distance $\displaystyle \frac{d}{\sqrt2}$ apart. This is under the assumption that if you move something twice as close to you then it looks twice as big. Not sure if that's true actually.

With these two distances we can then find the angle.

10. First of all, thanks for your help.

yes vlasev, this is very fun! I thought of that "pane of glass" method when I was hitting my head against the problem, but I abandoned it because I couldn't figure out at which angle to place the glass at. I tried to take into account the way the eye sees (you know, length is judged by the spacing of two points in our retina) and couldn't decide. Perpendicular makes sense, but technically speaking, if h is greater than zero, the eye couldn't see the intersection if it was looking parallel to the plane.

Instead of working with projections (sorry, my understanding of projections is very shaky, especially when i'm projecting something infinitely long), couldn't we possibly fix a height and find the relative apparent locations of two points on one of the parallel lines? Because it's a line, we could then proceed to find the apparent slope of the line and from there find the angle.

Actually, I kind of imagine this problem and others like it as a different form of math, one in which objects have an absolute property such as length, but also a variable property such as percieved length. A pen and a skyscraper have obviously different absolute lengths, but hold the pen closer to your eye and they can have the same relative length. Does this already exist? If you've heard of something like this, would you send me a link about it?

chiph588@, the horizon has to be at infinity, although the point at which they appear to converge may not be. If you think about it from a height of zero (i.e. the eyeball is inside the plane), the angle would appear to be 180 degrees. As the eyeball rises, the angle would change.

11. If you are looking parallel to the plane, the horizon will be right in the line of sight. This is how infinity works. Here is a short "derivation of this fact".

Let you be at point $(0,h)$ in the $xy$ plane. Look at point $(x,0)$ on the $x$-axis. The angle between your line of sight and the $y$-axis is just $\arctan(x/h)$. Now, via limits, we get

$\displaystyle \lim_{x \to \infty} \arctan\left(\frac{x}{h}\right) = \pi/2$

Hence the angle between your line of sight and the y-axis is precisely 90 degrees and you are looking straight ahead. It may seem paradoxical, but think of it this way. The parallel lines seem to meet at the horizon. By the same logic, your parallel line of sight should meet those lines at infinity.

EDIT: For the angle at infinity, there is a simple derivation without much insight and there is a more complicated derivation which offers greater insight. Do you want me to post both of these later?

12. I understand you'll be looking at a 90 degree angle, but in my earlier post I was talking about the angle the two parallel lines appear to make from your perspective.

13. vlasev, your explanation makes sense... to an extent. The two parallel lines seem to meet at infinity from the perspective of the eye, but the line of sight of the eye doesn't meet the lines. I guess, from a different perspective it would make sense, but thinking about it in terms of limits, the parallel lines approach each other (in this perspective). Maybe I'm thinking about this wrong. Post those infinity derivations for me please.

chiph588, I had left my computer on the site for a while before i submitted the last post, and as a result i didn't get to read your post about looking down the road. I think this is most like the solution i thought of, but I couldn't work out the function the percieved distance would change according to. I'm not sure if it's as simple as your 1:1 idea, and that's where i got stuck.

14. Alright, I'll put it as a spoiler down here:

Spoiler:
Let the eye be at $(0,0,h)$, and the lines be on the $xy$ plane and parallel to the $x$ axis, each at distance s from the $x$-axis. Let the eye be facing towards the positive $x$-axis. Let the plane $D$ be at distance $d$ from the eye. Let each point on that plane have coordinates $(u,v)$ (they are analogous to $(y,z)$ actually).

Here is a more informal derivation. The lines intersect the plane $D$ at $(d,s,0)$ and $(d,-s,0)$. Since the vanishing point is directly straight ahead, the projections of these lines on the plane $D$ must intersect at the point $(d,0,h)$. Since the lines are on a plane and we are projecting them on another plane, their projections must be lines also. Thus we have that the projections of these lines on plane $D$ are the lines starting at $(s,0)$, $(-s,0)$ and intersecting at $(0,h)$ in $(u,v)$ coordinates. From these you can use simple trig to deduce that the angle between these two projections must be $2\cos^{-1}\left(\frac{s}{\sqrt{h^s+s^2}}\right)$ on the plane.

Surprisingly, this angle does not depend on the distance we have chosen for the plane. It seems counter-intuitive.

Here is the more complicated derivation that uses some vectors. The idea is to write down an equation of the line going from the eye at $(0,0,h)$ to a point on one of the lines $(p,s,0)$. Then we need to find the intersection of this line with the plane $D$. This will be at coordinates $(d,u,v)$.

To get the equation of a line, we need a position vector $\vec{R_o}$ and a direction vector $\vec{V}$, which we'll multiply by a parameter $t$ to get the vector equation for the line to be $\vec{R_o}+t\vec{V}$. So, choose the position vector $\vec{R_o} = $ and the direction vector be from that point to the eye. By the rules of vector addition, this is $\vec{V}=<0,0,h>-\vec{R_o} = <-p,-s,h>$. Thus the vector equation for the line is

$\displaystyle{\vec{R_o}+t\vec{V} = }$

Since we want the projection onto the plane $D$, we need to have the x-coordinates match, which means $p(1-t) = d$. From this we get that $(1-t) = d/p$ and $t = 1-d/p$. Subbing both of those in the equation of the line, we see that the intersection point is $(d,\frac{sd}{p},(1-\frac{d}{p})h)$

Now, if we set $p = d$, that is we are looking at the intersection between the line and the plane $D$, we will simply get $(d,s,0)$. However, the interesting part is when we let it go to infinity.

$\displaystyle{\lim_{p\to \infty}(d,\frac{sd}{p},(1-\frac{d}{p})h) = (d,0,h)}$

This is the same result as before! Only this time we have it more rigorously. Furthermore, since the coordinates of the projection of the line on the plane $D$ are $(d,\frac{sd}{p},(1-\frac{d}{p})h)$, then in $(u,v)$ coordinates, this is just $(\frac{sd}{p},(1-\frac{d}{p})h)$. These are just parametric equations of the line in parameter $p$. They form a system of parametric equations

$\displaystyle{u = \frac{sd}{p}}$
$\displaystyle{v = (1-\frac{d}{p})h) }$

From these we get that $d/p = u/s$ and $v = h - \frac{h}{s}u$, which is the equation of the projection of the line onto the plane $D$. You can see that it is a line! the other line has equation $v = h + \frac{h}{s}u$. With trig you can get that the angle between one line and the vertical axis is $\theta = \cos^{-1}\left(\frac{s}{\sqrt{h^s+s^2}}\right)$ and the angle between the lines is just $2\theta$

Here is the reason why I think this doesn't depend on the distance $d$ of the plane $D$. Say we are looking at two parts of the line. One at distance $p_1$ and the other at distance $p_2$. The projections on the plane $D$ will BOTH involve the parameter $d$, but since we are looking at the relation between those two distances $p_1$ and $p_2$, we will have a sort of cancellation of the parameter $d$. I think this is a neat result.