1 / 11

Optimal Viewing Distance for the Statue of Liberty

Explore two proposed solutions to determine the best distance from which to view the Statue of Liberty: from the top of the Empire State Building or right underneath it. The problem is then approached mathematically, maximizing the viewing angle and finding the optimal distance through calculus. However, there is also a simpler formula provided for those who prefer an elegant solution.

trigg
Download Presentation

Optimal Viewing Distance for the Statue of Liberty

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Real Life ProblemsThe Statue of Liberty What is the best distance from which to view it?Two Proposed Solutions

  2. Liberty from the top of the Empire State Building …too far…

  3. From right underneath it - too close… heavily foreshortened

  4. The problem from my calculus book… What is the best distance to view the Statute of Liberty? Work it out…

  5. Problem devolves to maximising angle  – what distance? Too far T 46m  B 46m T G Obs 46m TooClose T About right… B 46m  B  46m 46m G Obs G X ? Obs G Can get the solution artlessly by simple substitution. But note; answer will not be analytic, and besides is rather inelegant…

  6. The Analytic Solution involves the differentiation of inverse trigonometric functions… so set up the problem T 1. Maximise  as x moves from 0 to 2. TG/GO = tan(+)3. (+) = arctan(TG/x), and4.  = arctan(TG/x) – arctan(BG/x) 46m B 46m   O x G  0

  7. The differentiation… d/dx = d/dx[(arctan(TG/x) – arctan(BG/x)] but d/dx(arctan(x)) = (1/(1+x2)) and d/dx(1/x) = -1/x2)) (Must use chain rule as function is in 1/x form) so that the differentiation devolves to 1 -TG 1 -BG d/dx = ----------------- * ------- - ------------------ * ------ 1 + (TG/x)2 x2 1 + (BG/x)2 x2

  8. The Calculation… Setting d/dx = 0 and substituting statue values we get0 = [1/(1+(92/x)2) * -92/x2] – [1/(1+(46/x)2) * -46/x2] 0 = -92/(x2 + 8464) + 46/(x2 + 2166)92x2 + 92*2166 = 46x2 + 46*8464collecting terms46x2 = 194672 ; x = sqrt(4232) => x  65.05m (at which  is 19.47o)

  9. Realisation:Need a zoom lens from the ferry to see it front-on!But if you can afford a zoom lens, you don’t have to do the maths… In fact, you don’t have to do anything if you have the money Above all – strive for elegance!

  10. A simpler formula.. For optimal viewing, the formula for the distance D to stand from a statue of height S and a plinth of height P isD = √(S x P) + (P2))Hence D = √(46*46) + (462))≈ 65.05mFor Nelson’s statue on Trafalgar square, S = 5, P = 49; D ≈ 51Source: “Why do buses come in threes?” Eastway and Wyndham, ISBN 1-86105-862-4Eastman and Wyndham give no derivation of the formula, probably because it’s a popular book and not meant to burden the reader with mathematical abstractions… I suppose their formula is derived from planar geometry rather than calculus.

  11. Thank you

More Related