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Mathematics Challenge For You

Posted: Sun Feb 26, 2012 4:19 am
by bristolflyer
So I was watching an old episode of Top Gear this week where James May was driving the Bugatti Veyron at top speed on the (straight) VW test track. At one point he is standing at one end and said that it's not possible to see the other end due to the fact that the road follows the curvature of the earth. There's often sone dubious facts spouted on that show...I was wondering if anyone has the inclination to do some math to see if this is true.

I'll make a start but then I run out of talent. The 'highest' point will be the mid point of the road at 2.5 miles; assuming James May is around 6' tall, is there more than 3' of 'crown' at the mid point? If so, you wouldn't be able to see the ground at the other end.

I found this website that had a similar calculation but I'm still scratching my head.

http://www.wired.com/wiredscience/20...e-radius-of-the-earth-with-a-lake/

Anyone?

RE: Mathematics Challenge For You

Posted: Sun Feb 26, 2012 4:23 am
by Fly2HMO
Quoting bristolflyer (Thread starter):
At one point he is standing at one end and said that it's not possible to see the other end due to the fact that the road follows the curvature of the earth

It doesn't take that much distance to see the curvature. But in this specific case you would need to get the exact elevation figures along the track to get any sort of accuracy. That's my WAG anyways.

RE: Mathematics Challenge For You

Posted: Sun Feb 26, 2012 4:54 am
by QFA380
Its highly unlikely the road would be perfectly flat but its perfectly likely you can see the curvature. The horizon is the point at which you can no longer see things on the other side (unless they're elevated). According to wiki for an eye level of 1.7m the horizon is about 5km away, also according to wiki the straight is 9km long, so they're correct that you can't see the other end, due to the curvature of the earth.

I tried doing some simple trig but it stopped working.

Read the wiki articles on figure of the earth and the horizon if you want all the complicated maths.

RE: Mathematics Challenge For You

Posted: Sun Feb 26, 2012 4:56 am
by dc9northwest
A very rough first estimate gives me a "crown" of 1.25m at the center (bigger than the 0.9m required)

Seems like James May is right: you can't see the end of the runway.

values used: runway length=8km, radius of earth 6380km

Some more accurate values for this location would be: length 9km, radius 6365km... which gives me a value of 1.6m for the crown--you'd have to be 3.2m (or 10 ft) tall to see the end of the runway.

[Edited 2012-02-25 21:09:21]

RE: Mathematics Challenge For You

Posted: Sun Feb 26, 2012 7:43 am
by vikkyvik
Quoting bristolflyer (Thread starter):
The 'highest' point will be the mid point of the road at 2.5 miles; assuming James May is around 6' tall, is there more than 3' of 'crown' at the mid point? If so, you wouldn't be able to see the ground at the other end.

I did a quick Solidworks sketch to actually measure. Values:

Track length: 5mi
James May Height: 0.0011mi
Radius of Earth: 3980mi

I basically did it by making a curve of radius 3980, that starts out horizontally at James May's feet, and ends 5 miles away below horizontal. Then drew a straight line that starts at May's head (6 feet above his feet) and ends at the same point as the curve. Then drew tangent lines to both the line and the curve, starting the at the far ending point of the line and curve.

Based on those tangent lines, the curve has a larger angle from horizontal than the line does, meaning they intersect somewhere in between the start of the track and the end of the track.

However, the difference is only ~0.02 degrees. Variation in flatness of the track, heat haze, air quality, etc. likely have a larger effect than the curvature of the earth.

EDIT: here's a screenshot so you can see what I'm talking about:

Big version: Width: 1336 Height: 619 File size: 24kb


[Edited 2012-02-25 23:49:38]

RE: Mathematics Challenge For You

Posted: Sun Feb 26, 2012 8:00 am
by Flighty
At a height of 1 centimeter, how far can you see?   

RE: Mathematics Challenge For You

Posted: Sun Feb 26, 2012 9:25 am
by EY460
In marine navigation there is a formula which allows you to calculate the distance you can see an object (usually a lighthouse, an aerial or a mountain) from the ship. The formula is:

D = 2.04 * (SQRT e + SQRT h)

Where D is the distance in nautical miles, e the elevation of the eye in meters and h the height of the object, again in meters. For instance if my eye level is 36m I can see an object 500m high at about 57.8 nautical miles.

With this formula you can calculate the distance of the horizon when the elevation of the eye change. In this case h = 0 and the formula becomes:

D = 2.04 * SQRT e

If I am in a lifeboat and the height on my eye is 2m I can only see about 2.9 nautical miles. If the eye level is 1 cm you can see about 0.2 nautical miles.

1 nautical mile is 1852 meters.

Of course this is the theoretical formula and it may vary slightly depending on the weather conditions.

RE: Mathematics Challenge For You

Posted: Sun Feb 26, 2012 1:01 pm
by FingerLakerAv8r
When I joined Airliners.net I was told there would be no math. This is most disturbing.

RE: Mathematics Challenge For You

Posted: Sun Feb 26, 2012 1:27 pm
by bristolflyer
Great answers all, thanks a lot.

Quoting dc9northwest (Reply 3):
A very rough first estimate gives me a "crown" of 1.25m at the center (bigger than the 0.9m required)

So based on this, if May had a twin brother standing at the other end you could see his head but not much else (heat haze etc notwithstanding).

Quoting vikkyvik (Reply 4):
I did a quick Solidworks sketch

I was thinking that this would be easy to do on a CAD program but I don't have one installed. As much as I'd like AutoCAD it's pretty darn expensive.

RE: Mathematics Challenge For You

Posted: Sun Feb 26, 2012 3:45 pm
by photopilot
Here you go. A website where you can input the height and get the distance to the horizon. Also includes showing you all the math in the calculation.

http://www.ringbell.co.uk/info/hdist.htm

So at 6 feet height, the horizon would be 3 miles away.

RE: Mathematics Challenge For You

Posted: Sun Feb 26, 2012 4:12 pm
by dc9northwest
Quoting bristolflyer (Reply 8):
So based on this, if May had a twin brother standing at the other end you could see his head but not much else (heat haze etc notwithstanding).

Yes, I'd say that's about right, unless the approximations I used are incorrect (I think they work out alright though).

Of course, you could've believed it because May said it. Wouldn't trust Clarkson or Hammond though! 

RE: Mathematics Challenge For You

Posted: Sun Feb 26, 2012 11:18 pm
by bristolflyer
Quoting photopilot (Reply 9):
Here you go. A website where you can input the height and get the distance to the horizon. Also includes showing you all the math in the calculation.

Thanks!

Quoting Fly2HMO (Reply 1):
But in this specific case you would need to get the exact elevation figures along the track to get any sort of accuracy.

The altitude would be the same all the way along the track if the track followed the curvature of the earth.

RE: Mathematics Challenge For You

Posted: Mon Feb 27, 2012 2:19 am
by vikkyvik
Quoting bristolflyer (Reply 8):
I was thinking that this would be easy to do on a CAD program but I don't have one installed. As much as I'd like AutoCAD it's pretty darn expensive.

Yeah, I started to try and remember how to do the math, but then remembered that I have Solidworks. Makes quick calculations like this ridiculously easy.

Of course, I may not have paid for it....

RE: Mathematics Challenge For You

Posted: Mon Feb 27, 2012 3:56 am
by Fly2HMO
Quoting bristolflyer (Reply 11):
if the track followed the curvature of the earth.

Keyword: IF

I'm sure its far from perfect.