Without knowing the viewing capacity of the eye, the only answer you can give is a function of the optimal viewing distance as the maximal viewing angle changes.

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- August 23rd 2008, 08:38 PMSpec
Without knowing the viewing capacity of the eye, the only answer you can give is a function of the optimal viewing distance as the maximal viewing angle changes.

- August 23rd 2008, 08:44 PMscuzi
alpha equalling x?

- August 23rd 2008, 08:58 PMSpec
You can call it whatever you like as long as you know what it stands for. In this case, or is the maximal viewing angle.

- August 23rd 2008, 09:02 PMscuzi
okay, so how does knowing the viewing capacity affect the equation? i think the equation is ment for people with 20/20 perfect vision, as it hasnt been specified that the view has imperfections. does this help?

- August 23rd 2008, 09:08 PMSpec
Well if you know what the maximum viewing angle is, then you simply input it into the function and you'll get the optimal viewing distance .

So if , then cm - August 23rd 2008, 09:14 PMscuzi
okay thank you, and just double checking, but tan can only be used on right angle triangles...

- August 23rd 2008, 09:46 PMscuzi
just double checking, but the derivitive of http://www.mathhelpforum.com/math-he...51840086-1.gif is y=1/(cos^[2][230\x])

- August 24th 2008, 05:22 AMskeeter
let be the angle of elevation from eye level to the bottom of the painting.

let be the angle from the bottom of the painting to the top of the painting, i.e. the viewing angle.

let x be the horizontal distance to the wall in meters.

then ...

so ...

setting = 0 yields the equation ...

solve for x ...

meters - August 24th 2008, 05:58 AMscuzi
oh my gosh i love you so so much!!!(Rofl)

- August 24th 2008, 08:13 AMskeeter
remember ... it helps a great deal if one knows the original problem statement.