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We are doing a project where we hope to make people think about the fact that the earth is round.

We want to imagine if we in London could see others places in the world and how the earths curvature would distort them.

We want to make chairs.

How would we make this exact ? We are trying to find the degree of curvature of certain places in relation to london. We hope to distort a model of a normal chair in software like CAD by inserting a 'degree of curvature' .

Can you help with this ???

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    $\begingroup$ Can you clarify? Can you make a drawing of what you'd want it to look like? $\endgroup$ – Kvothe Feb 19 '14 at 13:26
  • $\begingroup$ The earth's curvature doesn't distort anything, at least not at human scale. At most these "chairs" will be at different tilts relative to you, assuming they are level at the other places you are trying to model. This question makes little sense. $\endgroup$ – Olin Lathrop Feb 19 '14 at 13:41
  • $\begingroup$ What's wrong with the seafaring observation? If you watch ships leave harbor, the top of the mast is the last thing seen. $\endgroup$ – Carl Witthoft Feb 19 '14 at 15:27
  • $\begingroup$ Or how about putting up poles at various latitudes and observing the shadow lengths for a sun at the equator? (sundial physics) $\endgroup$ – Carl Witthoft Feb 19 '14 at 15:28
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The curvature of the Earth is not enough to really 'curve' any object that's city-sized or smaller. The reason for this is that the radius of the Earth is really big.[citation needed]

To make this more precise, say you make your chair one metre wide, and you have the edges pointing towards the centre of the Earth. You can then form an isosceles triangle between the chair's top and the Earth's centre, whose sides are $w=1\,\text m$ and $R_\oplus=6,370\,\text{km}$, so that the angular deviation from parallel edges is, to very good accuracy, $$ \delta\theta=2\arctan\left(\frac{w}{2R_\oplus}\right)\approx\frac{w}{R_\oplus}\approx2.7\times10^{-9}{}^\circ. $$

If you want to get the curvature of the Earth to be noticeable, you need objects and distances that are comparable with $R_\oplus$. As you note, inter-city distances are often on this scale.

Thus, one thing that I would suggest is having a 'roadsign' that points to where the other cities actually are. For example, the distance from London to Sydney is, as the crow flies, $d=3,465\:\mathrm{km}$, and this means that they subtend an angle of $$ \theta=\frac d{R_\oplus}\approx 0.54\:\mathrm{rad}=31^\circ. $$ This means that if you're in London and want to point straight to New York, you need to point west-by-northwest but also about thirty degrees below the horizon.

enter image description here

That is, I think, your best shot.

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    $\begingroup$ I'm also tickled by the fact that I'm in London and my screen is facing South, so that the sign on my screen is, in fact, mostly pointing in the right direction. $\endgroup$ – Emilio Pisanty Feb 19 '14 at 14:05
  • $\begingroup$ thank you! maybe we could imagine a transparent globe, and if we look from London to Sydney , the perspective would distort what we see, then we could make literally a chair of what we see? $\endgroup$ – user40924 Feb 19 '14 at 14:36
  • $\begingroup$ I'm not sure that chairs are a fruitful avenue, and there will be no distortion of human-scale objects. You will, though, see the chairs in other cities tilted with respect to your vertical. $\endgroup$ – Emilio Pisanty Feb 19 '14 at 14:59
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KISS. Have a tall ship cruise toward shore from 50 or more miles out. The crow's nest appears then down the masts to the ship's deck. Calibrate mast length vs. distance and you have an empirical measure of curvature at sea level. The math, entry below, is explicit.

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  • $\begingroup$ Like I commented :-0 $\endgroup$ – Carl Witthoft Feb 19 '14 at 16:57

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