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

$\displaystyle y = \frac{230}{tan\alpha}$

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- Aug 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 $\displaystyle y$ as the maximal viewing angle $\displaystyle \alpha$ changes.

$\displaystyle y = \frac{230}{tan\alpha}$ - Aug 23rd 2008, 08:44 PMscuzi
alpha equalling x?

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

- Aug 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?

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

So if $\displaystyle \alpha = \frac{\pi}{4}$, then $\displaystyle y = \frac{230}{tan\frac{\pi}{4}} = 230$ cm - Aug 23rd 2008, 09:14 PMscuzi
okay thank you, and just double checking, but tan can only be used on right angle triangles...

- Aug 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])

- Aug 24th 2008, 05:22 AMskeeter
let $\displaystyle \alpha$ be the angle of elevation from eye level to the bottom of the painting.

let $\displaystyle \theta$ 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 ...

$\displaystyle \alpha = \arctan\left(\frac{0.3}{x}\right)$

$\displaystyle \theta + \alpha = \arctan\left(\frac{2.3}{x}\right)$

so ...

$\displaystyle \theta = \arctan\left(\frac{2.3}{x}\right) - \arctan\left(\frac{0.3}{x}\right)

$

$\displaystyle \frac{d\theta}{dx} = \frac{\frac{-2.3}{x^2}}{1 + \frac{5.29}{x^2}} - \frac{\frac{-0.3}{x^2}}{1 + \frac{0.09}{x^2}} = \frac{-2.3}{x^2+5.29} + \frac{0.3}{x^2 + 0.09}$

setting $\displaystyle \frac{d\theta}{dx}$ = 0 yields the equation ...

$\displaystyle \frac{2.3}{x^2+5.29} = \frac{0.3}{x^2 + 0.09}$

solve for x ...

$\displaystyle x = 0.83$ meters - Aug 24th 2008, 05:58 AMscuzi
oh my gosh i love you so so much!!!(Rofl)

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