Interesting Problem, Please Help

**spandan** · August 2nd 2010, 10:20 PM

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.

Thanks in advance.

Spandan

**Vlasev** · August 3rd 2010, 02:14 AM

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!

**chiph588@** · August 3rd 2010, 05:53 PM

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.

Thanks in advance.

Spandan

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

**Vlasev** · August 3rd 2010, 06:01 PM

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.

**chiph588@** · August 3rd 2010, 06:34 PM

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

**Vlasev** · August 3rd 2010, 11:50 PM

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.

**chiph588@** · August 4th 2010, 11:32 AM

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.

**Vlasev** · August 4th 2010, 12:09 PM

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!

**chiph588@** · August 4th 2010, 01:56 PM

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.

**spandan** · August 4th 2010, 02:21 PM

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.

**Vlasev** · August 4th 2010, 05:12 PM

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?

**chiph588@** · August 4th 2010, 05:36 PM

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.

**spandan** · August 4th 2010, 11:19 PM

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.