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I'm trying to find some proof or any intuition behind the property explained in the following document. enter image description here

enter image description here

I find (c) hard to accept, what's the relationship between the amount of surface perceived at a certain angle and the component of the area as a vector in that direction? I understand it "feels" right but I can't quite find an explanation of why this is the case.

Original document here

Edit: I noticed that wikipedia also mentions this, but no explaination or proof is offered enter image description here

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I assume that you are OK with parts a and b. Now let's develop an intuition about part c. In the first instance, let's simplify the problem even more. Assume that the tilt is around the horizontal axis. So the normal to the plane makes an angle $\theta$ with the horizontal. The bottom part of the paper (say length $l$) will not change its apparent size when the rectangular piece of paper is tilted. The vertical side of the rectangle, $L$, when tilted an angle $\theta$ will change its apparent size to $L\cos\theta$. The apparent area of the rectangle is now $$dS'=lL\cos\theta=dS\cos\theta$$ Now let's get back to the original statement, by rotating the tilted piece of paper around the axis towards the eye. Notice that the normal to the paper is still at angle $\theta$ with respect to that axis (it moves over the surface of the cone with half angle $\theta$). Also, due to symmetry, the apparent area $dS'$ does not change with this additional rotation. Therefore part c is proved.

Edit:

enter image description here

In the figure above, initially the piece of paper is vertical (as seen from the side). The length $L$ is along the $OA$ axis, $l'$ is perpendicular to the figure (out of the image). The original normal to the paper points along $B$ (towards the eye). In the first case I've rotated the paper along the axis through $O$, perpendicular to the figure, by an angle $\theta$. That moved $B$ to $B'$ (direction of the normal changed by $\theta$) and $A$ to $A'$ (also by angle $\theta$). To prove that both angles $\angle BOB'$ and $\angle AOA'$ are the same, just use the fact that adding $\angle B'OA$ to both of them yields $90^\circ$. That means that the apparent size of $A'O$ is the vertical projection $A'O\cos\theta$.

For the last part, I took the above figure, and rotated around the $OB$ axis, in such a way that the $OB'$ describes the surface of the cone with vertex at $O$. No matter where on this cone you put $B'$, the angle between $OB'$ and $OB$ is unchanged.

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  • $\begingroup$ why is $\theta$ on the second term equal to the $\theta$ on the third term? unless im not visualizing correctly the first $\theta$ is between the plane of the projection and the plane observed, while the second $\theta$ is between the axis of rotation and the normal (thats always 90 degrees, right?) imgur.com/VQtKq8E $\endgroup$ – Joaquin Brandan Feb 13 at 4:33
  • $\begingroup$ Other than that the explaination makes perfect sense (if both angles are equal, that part I cant quite visualize) $\endgroup$ – Joaquin Brandan Feb 13 at 4:44
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    $\begingroup$ If you tilt the plane an angle $\theta$ away from the vertical direction, it's normal is also moving by angle $\theta$ (away from the horizontal). I will attach a figure to my answer in a few minutes $\endgroup$ – Andrei Feb 13 at 4:48
  • $\begingroup$ thank you so much. I will wait for the figure. $\endgroup$ – Joaquin Brandan Feb 13 at 4:49

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