Use a sunset photo to estimate the Earth's radius

Source: Internet
Author: User

Do you believe that using only one sunset photo, you can get the Earth's radius! In a recent paper, Robert vandrebei of Princeton University analyzed a sunset photo taken at Lake tranquility, which not only confirmed that the earth is round, the Earth radius is estimated based on the content in the photo. I will describe the general process of computing for everyone to worship.

The cause of the incident is the usual sunset photo above, and a problem that everyone is not very concerned about at ordinary times: the sun's exposure to the water should be a standard bow, but why do we see the sun in the same shape as rugby at sunrise and sunset? You may soon think that the lower half of the luminous body is actually caused by sunlight reflection on the water surface. Another problem arises: why is the lower half of a table smaller than the upper half?


This is because the answer to this question is not easy-the earth is round. It is a proportional exaggerated version of which people stand on the earth to watch the sunrise, where o is the center of the earth, a is the position of human eyes, AB is the video line, and B is the junction of the Water Day. Since the sun is far away from us, we regard the sun as an ideal parallel light. The angle between the sunlight directly injected into the human eye and AB is recorded as α, and the angle between the light and AB that passes through the C reflection on the water is recorded as beta. From the figure, we can see that the angle β is smaller than α, that is, the image of the sun on the water is smaller than itself.

 

What is Beta smaller than Alpha? Accurate Measurement of the photo shows that the diameter of the sun is equivalent to 317 pixels in the photo, while the height of the exposed part of the water is 69 pixels, and the reflection in the water is only 29 pixels. We all know that the sun's visual diameter (view the sun's angle of view) is 0.5 degrees, so we get α = 0.5*69/317 ≈ 0.1088 degrees, β = 0.5*29/317 ≈ 0.0457 degrees.
If we know that the vertical distance from the human eye (or camera) to the water surface is 2 meters, then we can estimate the Earth's radius based on the data. The AOB should be recorded as Phi, the AOC should be recorded as θ, the AB distance from the human eye to the water and sky should be recorded as D, and the AC distance from the human eye to the reflection point should be recorded as D, the incident angle and reflex angle are counted as gamma, and R is used to represent the Earth's radius. At this time, we have a total of six unknown numbers. To solve these six unknowns, we need to look for six different equations. The six equations can be obtained from the following six equal relations:

 
1. The inner angle of the Quadrilateral obac is 360 °, that is, (?-θ) + 90 ° + β + (180 °-Gamma + 90 °) = 360 °, simplifiedEquation (1) Phi + β = θ + γ

2. The interior angles of the two parallel lines are added to 180 °, that is, (α + β) + (180 °-2 gamma) = 180 °, that isEquation (2) α + β = 2γ

3. Since Ao = H + R and Ao = AD + DO = D · sin Phi + R · cos PhiEquation (3) H + R = D · sin Phi + R · cos Phi

4. BD can be equal to d · cos PHI and R · sin Phi.Equation (4) d · cos Phi = R · sin Phi

5. Since Ao = H + R and Ao = AE + eo = D · sin (gamma + θ) + R · cos θEquation (5) H + R = D · sin (gamma + θ) + R · cos θ

6. CE can be equal to d · cos (gamma + θ) and R · sin θ.Equation (6) D · cos (gamma + θ) = R · sin θ

 
A series of complex algebra operations (with hundreds of characters omitted) finally tell us:

R = H/(√ 1-2 · cos β · cos gamma + cos2 gamma/sin β-1)

Gamma = (α + β)/2. It can be obtained by substituting known α, β, and H. The earth's radius R is about 7.29312*106 meters, that is, 7293 km.

What is the error of this estimation? In fact, the Earth's radius is more than 6300 kilometers, and the visible error is not very large. However, considering that our estimation is only based on a picture, we can estimate the order of magnitude to be equivalent to B. In addition to measurement accuracy, there are many potential factors that may cause errors. At present, it seems that the main source of the error is not completely calm-a small wave will have a huge impact on the alpha and beta values.

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