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Sorry for the long paragraph, I'm not too good at describing things: Firstly, to explain the situation, I need to know how the area of a circle can change as you view it from different angles. For example, if you were viewing a circle at a 0 degree angle, it would look like a circle and it's area would be $\pi r^2$. However, as the angle of observation increases or decreases, one side of the circle will "collapse" inwards, and become smaller. It would create an ellipse, where the variable 'a' (longest distance from centre to side) stays the same, and where variable 'b' (the smallest distance from centre to side) decreases. I want to know the best way to find the area, based on the angle theta. For clarification, theta refers to the angle from a line perpendicular to the circle. The best way to approach the issue seemed to be the diagram in the link.

Diagram

In the diagram, the 2 cm long line, is the circle. Therefore, the radius is 1 cm. The best approach I could find, was by using a cos ratio in the triangle highlighted in red as whenever I try any trig rules, it always gets too complicated with multiple variables. The angle closest to the centre of the circle would have to be 90-theta and, if the triangle is right angle (which it was made to be this way), the opposite angle must be theta. Hence by using $\cos \theta = a/h$, where h is the 1cm line, 'a' (which refers to adjacent not the other longer 'a') must be cos theta. This only works if the entire line is 2a. But if it's not and either side is different, then I have a problem. So, I wanted to know, am I right with the above working, or am I wrong. And if I'm wrong, how should I approach this problem differently. Thanks.

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  • $\begingroup$ It sounds like a good approach, what formulas have you come up with for the area of the ellipse according to the angle of observation? $\endgroup$ – abiessu Aug 7 '17 at 2:49
  • $\begingroup$ Your doing. Well. You have an elipse. The full width of the elipse is 2r. The full height of the elipse is 2r cos theta (you were right about that.) Area of ellipse is .... googles... $\pi*r^2\cos \theta $. Seems fine to me. $\endgroup$ – fleablood Aug 7 '17 at 3:03
  • $\begingroup$ The apparent width of the ellipse also depends on the aperture angle, which varies with the distance of the viewpoint to the center of the circle. $\endgroup$ – amd Aug 7 '17 at 19:41
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As you’ve observed, the projected line segment is not symmetric about the view line to the circle’s center—the center shifts. Another reason that you might be having some trouble working this out is that the apparent width of the ellipse also depends on the visual angle subtended by the circle, which varies with both the distance of the viewpoint from the center of the circle and the circle’s radius.

enter image description here

(Diagram provided by the OP.)

This problem can be reduced to finding the intersection of pairs of lines, which is easily done using homogeneous coordinates. We place the circle on the $x$-$y$ plane, centered at the origin, and Let the viewpoint be $V=(0,d\cos\theta,d\sin\theta)$, ($d>0$). We’ll first look at what’s going on in the $y$-$z$ plane. In the above diagram, the horizontal direction represents $z$ and the vertical $y$. To reduce clutter, the $x$-coordinate will be suppressed and, at the risk of introducing some confusion, the others will be designated $x'$ and $y'$, respectively.

The line perpendicular to $\overline{OV}$ that passes through the upper point $A(0,r)$ is $y'=r-x'\cot\theta$, which we can rewrite in normal form as $x'\cos\theta+y'\sin\theta-r\sin\theta=0$. The homogeneous vector that represents this line consists of the coefficients of the latter equation: $$\mathbf l=(\cos\theta,\sin\theta,-r\sin\theta).$$ The line through $V$ and the lower point $B(0,-r)$ is represented by the cross product of the homogeneous coordinates of these points: $$\mathbf m=(d\cos\theta,d\sin\theta,1)\times(0,-r,1)=(r+d\sin\theta,-d\cos\theta,-dr\cos\theta).$$ Their intersection is $$\mathbf l\times\mathbf m=(-dr\sin2\theta,dr\cos2\theta-r^2\sin\theta,-(d+r\sin\theta))$$ which in Cartesian coordinates is $$C=\left({dr\sin2\theta\over d+r\sin\theta},{r^2\sin\theta-dr\cos2\theta\over d+r\sin\theta}\right).$$ The minor axis length of the circle’s image is $AC$, which you can compute using the standard formula for the distance between two points as $${2dr\cos\theta\over d+r\sin\theta}.$$ Compared to your initial guess, there’s an extra factor of ${d\over d+r\sin\theta}$ that accounts for the asymmetry of the view and the “angular size correction” factor. As $d\to\infty$, so that the projection becomes closer and closer to parallel, this correction factor approaches unity.


This isn’t the whole story, though. Rays from the viewpoint through points on the circle converge as they get closer to the viewpoint, so the other semi-axis of the circle’s projection isn’t going to be equal to $r$, either. To work out what this is, first project the center of the ellipse back to the $x$-$y$ plane. This center is the midpoint of $A$ and $C$ and its pre-image can be found by intersecting lines again: $$\begin{align}D &= \frac12\left((0,r)+\left({dr\sin2\theta\over d+r\sin\theta},{r^2\sin\theta-dr\cos2\theta\over d+r\sin\theta}\right)\right) \\ &=\left(\frac{d r \sin (\theta ) \cos (\theta )}{d+r \sin (\theta )},\frac{r \sin (\theta ) (d \sin (\theta )+r)}{d+r \sin (\theta )}\right)\end{align}$$ and its back-projection is $$\overline{VD}\times\overline{OA}=(V\times D)\times(-1,0,0)=\left(0,\frac{d r^2 \sin (\theta ) \cos (\theta )}{d+r \sin (\theta )},\frac{d^2 \cos (\theta )}{d+r \sin (\theta )}\right)$$ which becomes $\left(0,\frac{r^2}d\sin\theta\right)$ in Cartesian coordinates.

Going back to 3-D coordinates, this is the point $\left(0,\frac{r^2}d\sin\theta,0\right)$. The $x$-coordinates of the points on the circle with this $y$-coordinate are $\pm\sqrt{r^2-y^2}=\pm r\sqrt{1-\left(\frac rd\sin\theta\right)^2}$. I’ll leave finding the projections of these points and the resulting semi-axis length to you (hint: you can use similar triangles).

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  • $\begingroup$ Your point on parallel projection is well taken. I've deleted my answer. Thank you. $\endgroup$ – Cye Waldman Aug 8 '17 at 17:50
  • $\begingroup$ Sorry, I hadn't thought about the fact that you wouldn't see my response if I didn't comment. As for the reason I didn't comment, it isn't because I can't but rather because I can't add an image to a comment (or at least don't know how if it's possible). $\endgroup$ – J. Doe Aug 10 '17 at 12:19
  • $\begingroup$ @J.Doe BTW, if you want to see a similar problem worked entirely via projective geometry, check out math.stackexchange.com/a/2392962/265466. $\endgroup$ – amd Aug 15 '17 at 18:00
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An elipse has a horizontal radius $R $ and a vertical radius $r$.

The area of the elipse is $\pi*R*r$. For a circle $r=R $ and thus we get our familiar circle area formula.

You correctly figured out that if the tilting angle is $\theta$ then the vertical radius is $r*\cos \theta $. (Just look at the image... it's clear that the vertices radius is the side opposite the angle $90 - \theta $ on a right triangle with hypotenuse of $r $. I.e. $r*\sin (90 -\theta)=r*\cos (\theta) $)

So the area of the elipse is $\pi *r^2*\cos \theta $.

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I'm currently responding to fleablood's answer, however I couldn't seem to attach a picture when only commenting, so I'm responding to my own question. What you have said is exactly how I initially believed when I viewed the problem; however, after observing a change in angle using a graphing program, I noticed that the found value for R may not be right. This is because the total length of the line doesn't seem to necessarily be 2R. This can be seen visually by the link. It can be seen that the red line, which is the R, as found above, is much longer than the other side of the line. The above equation works but only when assuming the red line is equal to the blue line. Obviously I have no mathematical proof of this claim, but from moving around the view point using GeoGebra, a graphing software, the lines seem dramatically different at big angles. Any idea on how to approach this given the new diagram. Thanks.

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    $\begingroup$ You can add text (and pictures) to the end of your question by editing it. Then you can comment under an answer, "I considered this, but ..." and direct them to the edits in your question. Much neater (and more effective) than posting non-answers as answers. $\endgroup$ – David K Aug 14 '17 at 21:36
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I'm responding to amd's response without commenting for the same reason as with fleablood, I'm going to attach a photo: Here

Thanks for the answer! In the diagram, I've labelled a bunch of points for easier questioning. I believe I have assigned all the variables discussed to the image so if any are in the wrong place please tell me. For the line which you state is perpendicular to OV, and is written as "y=r−xcot⁡θ", are you referring to AC or the line from A to the 'd' line? Not sure if my uncertainty about that is the reason, but how did you obtain the equation "y=r−xcotθ"? Why is there a z direction when the diagram is only dealing with changes on a 2d plane? Also, on a diagram what does this 'l' value represent?

Also, I hadn't learnt what homogeneous meant so I did some research and found this website: http://www.tomdalling.com/blog/modern-opengl/explaining-homogenous-coordinates-and-projective-geometry/

It talks about letting w=1 which is where I'm assuming you got the z coordinate for the next part of your answer. However I have a similar problem with 'm' as I did with 'l' where I don't understand how/why you chose the value. If we're talking vectors, m would give a vector coming out/into the screen at a slight angle (to account for the +1 in the z) perpendicular to the vector OV and OB.

For the next part of your explanation I understand how you reduced the z direction back down to 1, then turned it into a 2d vector/point. However, I haven't had time to go through the "algebraic manipulation", which I'm assuming at the moment is the difference between this vector, and viewpoint V vector. Finally, for the equation d/d+rsinθ I can't get this far as I haven't understood l and m.

Sorry for so many questions, I'm in year 12, Math C and B so my knowledge only goes so far. If you're wondering what I mean by C and B, go to the following link which briefly showcases what kind of information is learnt in each class: https://en.wikipedia.org/wiki/Mathematics_education_in_Australia#Mathematics_C

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  • $\begingroup$ You should do something to get more reputation so that you can comment on answers. If you don’t post your responses to answers as comments, people are unlikely to see and respond to them. $\endgroup$ – amd Aug 9 '17 at 9:17
  • $\begingroup$ Projective geometry is a deep and beautiful subject, but for this problem, all you really need to know about homogeneous coordinates is in the wikipedia article, in particular the intro and the section on its use in computer graphics. Using cross products to compute line intersections (in the plane) is Cramer’s rule in disguise. $\endgroup$ – amd Aug 14 '17 at 19:30

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