Quote of the Day

Many people want to leave a better world for their children. I'm trying to leave better children for my world.

— Carlos Slim

## Introduction

I have received a number of questions recently on how the curvature of the Earth affects building construction. In general, the effects of the Earth's curvature are ignorable because most man-made construction is on too small of a scale to notice the effects of the Earth's curvature. One well documented exception is the Verrazano-Narrows bridge, whose design took into account that the bridge towers are 1 5/8 inch farther apart at the top than at the bottom. In this post, I will show how to compute this value.

The calculations here are straightforward. My intent is to show that there are some structures that must take the Earth's curvature into account. There are two other examples that I know of: Stanford Linear Accelerator (Source), and Fermilab's neutrino communication experiments (Source). For more examples, see this comment.

## Analysis

Figure 2 shows how to compute the 1 5/8 inch of additional separation based on the drawing in Figure 1.

## Conclusion

The Earth's curvature will only have significant effects on massive structures that are sensitive to small errors. With the Verrazano-Narrows bridge, we are talking about a structure with a size on the order of a 1000 feet and the effect of the Earth's curvature is ~1 inch.

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Why must this adjustment be adhered to. What effect would it have leaving out. I would imagine that it has to do with the "pull" of the suspension cables.

I am not a bridge designer, but I have read from many sources (example) that the VN bridge design had to compensate for the curvature of the Earth. If I had to speculate, I would guess that the compensation has to do with tolerancing (e.g. making metal members with holes line up for fastening). For example, I personally have seen bridge decks that could not be fastened together until nighttime when temperatures cooled and the deck members shrunk enough for holes to line up. That being the case, both temperature and curvature must be taken into account.

One of the most interesting tales of compensating for the Earth's curvature involves an early H-bomb test. A 9000' long Krause-Ogle box (link) was built that extended from the bomb site to some test instruments. The box was filled with bags of hydrogen gas. The idea was to provide a portal for measuring the first radiation associated with the explosion before the instruments were destroyed by the blast. Early tests showed that the box did not provide proper alignment between the bomb and the instruments. It was quickly discovered that the box design had not corrected for the Earth's curvature. Once the design was corrected for the Earth's curvature, the box and instruments worked properly.

Temperature is a more common problem than the Earth's curvature. I often have to deal with temperature expansion and tolerancing. I used to work in the automotive metrology field (e.g. measuring transmission parts). All measurements were made to 0.0001" and temperature really matters (see this web page for an example). For an interesting historical situation where temperature compensation was important, see this example from the Trinity atomic bomb test. The plutonium core did not fit until all the temperatures were equalized.

mark

You know when you are installing fence posts and how you use that leveler tool and line the bubble up in the middle to set the post in exactly vertical? So when you do this on a suspension bridge of great length, even all your pillars are set vertical, they will not be exactly parallel. So if you don't account for this when calculating the dimensions of the suspension cables, I suppose you could end up with some problems. Your suspension cables might end up being slightly too short.

You got it. Plumb and level exist for a point on the surface. When you have more than one fence post, technically, each is pointing at a slightly different angle relative to the center of the Earth. Of course, the difference is insignificant for relatively small objects. However, there are large objects where it must be taken into account — think suspension bridges, atom smashers, tunnels.

mark

can you explain the H-bomb with plain words please

Bomb drop. Bomb go boom.

Not accounting for curvature will introduce forces on the material. Most likely tensile stress. For small structures, this stress is negligible. If a structure has a small safety factor, it needs to be accounted for.

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Wonder how much earth curvature "correction" the two LIDOS sites had to contend/correct for their four respective lasers.

It is "LIGO", not "LIDO".

Oops, I stand corrected.

You were referring to the LIGO near Pisa but thinking also of The Lido, near Venice

The Danyang Kunshan Grand Bridge in China is over 102 miles long. The structure like the Great Wall can be seen at tremendous high altitude. Not only was it engineered with curvature in mind, but it can't be seen from any altitude, or the photographs of it. Ergo no curvature detected. Now before anyone says something like the Earth is really big or some nonsense like that. The calculation given to us by the establishment breaks down from the Earth's circumference being roughly 25,000 miles, which provides us a calculation of 8 inches per mile squared. In the case of a bridge stretching out over 100 miles breaks down to over 7000 feet in curvature drop. So where is it at?

The earth is really big.

The 8 inches per mile squred formula would be applicable if the bridge was build along an imaginary horizontal line from one side of the bridge. That's not the case. A bridge is build along the sight line tangent to the horizon. If the bridge is 100 miles long, and buillt parallell to the curvature of the earth, then the bulge would be 264 feet not 7000 feet. The angle from the edge of the bridge up to the top of the bulge is about 0.0573°. Good luck finding that.

Oh boy. Your guy below is correct. it's the bulge you are looking for and not the 7000' difference from a level line drawn straight out into space.

Using a figure of 4000 miles for the Earth's radius, one significant digit, reduces the final answer of your calculations to one significant digit accuracy, so rightfully, you should state your answer with one significant digit (2 inches). You could have, however used a figure for the Earth's radius with up to seven significant figures. Such information is available from NASA. This would have allowed you to give your answer with three significant figures, considering that the height of the bridge is given to that accuracy. Of course, the Earth's radius varies at different locations because it is not an exact sphere, with the distance around equator being greater than the distance around the poles, so the radius at the location of the bridge would also have to be taken int account.

You are correct. I will correct the significant figures in a few days.

mark

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