A Burning Man Art installaton. Followed by a math problem. • videosift.com

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A Burning Man Art installaton. Followed by a math problem.

posted by eric3579

9 months 1 week ago • 586 views

Perspective - An art installation from Burning Man 2007. The letter "A" is six inches tall. The "E" is ten feet tall. This led me to a question, I was hoping someone could answer.

If "A" is six inches tall and 10 feet from the camera,and "E" is ten feet tall, how far away do you place"E" to make the letters appear to be the same size from the cameras perspective?

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Whoa... damn that's cool. No idea on the math problem though, not my bag... would be interested to find out what the answer is though.

written by raven

| 9 months 1 week ago | CH

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The "E" is too big. Whoever did the math, did it wrong.

written by K0MMIE

| 9 months 1 week ago | CH

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Its not unnecessarily the math thats wrong, but the position of the person taking the camera. you need to know more than the hight of the A and E, but also the observer point your looking from (where the eye is of the person looking). All thats assuming you want them not just further away, but lining up horizontally as well. most of the letters arnt trying to achieve that, which helps, as it makes the impression work despite the vertical hight of the observation point (how tall the person looking is). infact, most of the letters are atleast a little off, from where the photo was taken, but its because the E looks so close the A at tis base, that the error for it is more noticeable. where it vertically miss aligned more, it would be harder to tell. but also look less cool i expect.

Anyway, if you knew the distance of the observer, and the height of the A that would let you work out all aspects of the triangle as you could assume it was a right angled one (works best if you place the eye line on the floor). then, knowing the hight of the E, and the angle your looking up at, you can calculate the distance from the observer to E, and so on. The height variance makes it different, but not impossible, you just need to also know the intended optimal hight of the observer, of the A and of all the other letters too.

If you want an example, in either case, poke me. Its all just geometry and triangles.

written by Thylan

| 9 months 1 week ago | CH

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Ah a challenge, suppose that the angle to the top of A and to the top of E need to be the same. The angle of triangle A is 120in long by 6in high, or arctan(6/120). Triangle E is 120in high and x long and has an angle (arctan(120/x)) equal to triangle A. Skipping all the trig and using proportions means that 6/120 = 120/x, x= 2400in. So E is 200 feet away.

written by thesnipe

| 9 months 1 week ago | CH

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^ yeah, that works too. not sure why inverting the A triangle to aim at the E didnt occur to me. haven't used trig math in many years.

Gets more complicated though if you dont assume same base heights, but only in as much as the difference in base heights needs to be removed form E's hight first, so this method is still quicker.

written by Thylan

| 9 months 1 week ago | CH

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