Even though the Earth is round, a ten foot 2x4 can lay on the ground flat because of the massive size difference between the two (20,925,656.17ft to 10.00ft by my calculations).

My question is: How long does a 2x4 have to be (given that there is no deformation of the 2x4 due to gravity or termites and a perfectly spherical Earth) before the Earth curves away at its ends? I am interested in the math to figure this out as well.

If the earth is assumed to be perfectly spherical and the 2x4 perfectly straight, any length will not lie flat. You need to specify the minimum deviation you can detect.

Well, assuming the earth is a perfect sphere (ha!) and that the board doesn’t bend under its own weight (double ha!), it’s easy to work out the formula – The required length of a beam (L)so that there is a clearance of Delta at each end when placed on a sphere of radius R is about two times the square root of (2 times R times Delta). For an Earth with R = 6,371.315 km and Deklta = 6", that works out to 2.8 km, or about 1 and three-quarters miles. That’s surprisingly long. There’s an old riddle about how much you’d have to add to the circumference of a steel belt hugging the earth in order to make it stand out 6" from the earth – it’s a LOT smaller than 1.75 miles (Figure it out).

Of course, if you tried to get a 2 X 4 to lay flat on the earth it wouldn’t – it would conform pretty well to the lie of the land. With willing suspension of disbelief you can get wildly different results – look up the case of Cyrus Teed/Koresh and his group, who “proved” that the surfacde of the earth is the concave INSIDE of a hollow ball by projecting a straight line along a beach until it plunged into the earth, insead of diverging from it. (You can find the story in Martin Gardner’s “Fads and Fallacies in the Name of Science.”

I really screwed up on that one but doing it again by trig I get a much smaller number this time, 1.18 miles from center to end. Not sure how many decimal places are used in the windows calc but rounding errors will creep in when comparing a 6" difference to the radius of the earth.

Cal could you explain how you arrived at this formula? I was trying to figure this on my own and compare it to what was posted.
[my thought process]
What we have is a right triangle.

The middle point of the board to the center of the Earth is one leg, R.

A line from end of the board to the center of the Earth would be 15cm (I know I said 6" orignally but I’m tired of converting back and forth) plus R which is the hypoteneus.

Second leg which is half the length of the board or 1/2X.

To solve for X we do 2*sqrt((R+15cm)[sup]2[/sup] + R[sup]2[/sup])
Using R=6371.315
I get something like 18000km
What am I doing wrong?

#$%^&* I get a different frigging answer every time. I took CalMeacham’s figure for radius and converted to inches. A right trangle with a hypoteneus of that and an adjacent side of that plus six inches give an angle of 0.0125318610465763513734081760373 degrees. The sine of that multiplied by the radius gives me, 0.8659… miles.

God, I love complicated, pointless algebra. It’s a good thing my job gives me a chance to use it all the time.

Try drawing out a picture. It’s what I always told my students to do, back when I was teaching. I like to solve everything algebraically, ending up with a single equation. Then convert everything to the same units, plug in, and crank.

My expression above neglects a term in Delta Squared, but, believe me, you’ll never miss it.

And that actually illustrates something about problems like this one. You will wind up applying a trig function to a value very near 1, or taking the square root of the difference between two nearly identical large numbers. Most calculators you might pick up will have sufficient precision to get your answer, but suppose we were trying for a 6 inch difference using the earth’s orbit about the sun (yes, I KNOW it’s actually elliptical. Assume a 93 million mile perfect circle for this example). The interesting problem is actually numeric, not analytical.

Problems of this type often have surprising answers. Here’s one to try:

Assume you have a mile of straight railroad track tied down at each end. Somebody welds an extra foot of track into it, causing it to bow up in the middle. How far out does it bow? You won’t believe it.

Not quite. That’s TOO high. A first order approximation may be obtained by calculating a leg of a right triangle like for the other problem (equivalent to having the track deform into a “tent” shape rather than a curve):

You want the other leg of a right triangle with a 1/2 mile + 6 inch hypotenuse and a 1/2 mile side.

Are you asking what happens to the height the center of the track should rise if we increase its length by less than 1/5000th of its original value? My guess is that the spring constant for the steel is sufficient to allow for a one foot compression over the course of one mile, and thus the damned thing wouldn’t move much. Assuming that the steel doesn’t compress, however, and we have a math problem.

Aproximate the bent track as an isoceles triangle with sides length 5280 feet, 2640.5 feet and 2640.5 feet.

The altitude of the triangle should be the maxium height the center of the track should raise. This would be the short leg of a right triangle whose hypotenuse was 2640.5 feet and whose long leg was 2640 feet.

Pythagorean theorem: 2640^2 +x^2 = 2640.5^2

in this case, x=51.38 feet (about). That’s pretty impressive, assuming a perfectly rigid piece of steel, which I’m not sure we can…