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Real Life Problems The Statue of Liberty What is the best distance from which to view it? Two Proposed Solutions

Real Life Problems The Statue of Liberty What is the best distance from which to view it? Two Proposed Solutions

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Page 1: Real Life Problems The Statue of Liberty What is the best distance from which to view it? Two Proposed Solutions

Real Life Problems

The Statue of Liberty

What is the best distance from which to view it?

Two Proposed Solutions

Page 2: Real Life Problems The Statue of Liberty What is the best distance from which to view it? Two Proposed Solutions

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

Page 3: Real Life Problems The Statue of Liberty What is the best distance from which to view it? Two Proposed Solutions

From right underneath it - too close… heavily foreshortened

Page 4: Real Life Problems The Statue of Liberty What is the best distance from which to view it? Two Proposed Solutions

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

Page 5: Real Life Problems The Statue of Liberty What is the best distance from which to view it? Two Proposed Solutions

Problem devolves to maximising angle – what distance?

G

T

B

Obs

GT

B

G Obs

TooClose

Too far

Obs

T

B

G

About right…

46m

46m

46m

46m

46m

46m

X ?

Can get the solution artlessly by simple substitution. But note; answer will not be analytic, and besides is rather inelegant…

Page 6: Real Life Problems The Statue of Liberty What is the best distance from which to view it? Two Proposed Solutions

G

O

46m

46m

The Analytic Solution involves the differentiation of inverse trigonometric functions… so set up the problem

0

1. Maximise as x moves from 0 to

2. TG/GO = tan(+)

3. (+) = arctan(TG/x), and 4. = arctan(TG/x) – arctan(BG/x)

x

T

B

Page 7: Real Life Problems The Statue of Liberty What is the best distance from which to view it? Two Proposed Solutions

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

The differentiation…

Page 8: Real Life Problems The Statue of Liberty What is the best distance from which to view it? Two Proposed Solutions

Setting d/dx = 0 and substituting statue values we get 0 = [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*8464

collecting terms

46x2 = 194672 ; x = sqrt(4232)

=> x 65.05m (at which is 19.47o)

The Calculation…

Page 9: Real Life Problems The Statue of Liberty What is the best distance from which to view it? Two Proposed Solutions

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!

Page 10: Real Life Problems The Statue of Liberty What is the best distance from which to view it? Two Proposed Solutions

For optimal viewing, the formula for the distance D to stand from a statue of height S and a plinth of height P is

D = √(S x P) + (P2))

Hence D = √(46*46) + (462)) ≈ 65.05m

For Nelson’s statue on Trafalgar square, S = 5, P = 49; D ≈ 51

Source: “Why do buses come in threes?” Eastway and Wyndham, ISBN 1-86105-862-4

Eastman 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.

A simpler formula..

Page 11: Real Life Problems The Statue of Liberty What is the best distance from which to view it? Two Proposed Solutions

Thank you