Crude geometric estimation or am I missing something?

In summary, the conversation discusses a physics lab experiment in which the radius of the Earth is calculated by measuring the time it takes for the sunset to occur from different locations. The attached diagram shows the initial setup and suggests using the Pythagorean theorem to find the value of L. However, the person is unsure if multiplying the change in time by the circumference of the Earth divided by 24 hours will give the distance traveled along the circumference of the Earth or the value of L. They ask for clarification on this and for any geometric rules they may be missing. The equations used in the solution are also mentioned, along with the approximation of sin(theta) as theta. The final solution is given as R = (12 x 3600)/pi
  • #1
neolayman
8
0

Homework Statement



I'm doing a report for a physics lab experiment where we are calculating the radius of the Earth by measuring the time it takes to see the sunset from the base of a cliff looking out into the pacific ocean till when it sets in relation to an observer at the top of the cliff. The picture I attached shows the diagram and the initial setup on how to find the radius of the Earth with the pythagorean theorem, but in yellow where it suggests how to find the value L, I'm thinking that multiplying the change in time by the circumference of the Earth divided by 24 hours would really give me the distance traveled along the circumference of the earth, not L. Please let me know if I'm right about that and if so can you tell me if and how it might still be used as a crude estimate for L by my professor? If I'm wrong please let me know what geometric rule I'm missing to help me out.



Homework Equations



See attached

The Attempt at a Solution



Part 1 here covers that.
 
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  • #2
Sorry, I guess the attachment function doesn't work on this site. Here's the image:
zzz.jpg
 
  • #3
Without the approximations, if the time difference between sunsets is t, then the angle subtended by the sun is [tex]\theta=\omega t[/tex].

From the figure you get,

[tex](R+h)sin\theta=R[/tex]

[tex]R=\frac{hsin\theta}{1-sin\theta}[/tex]

You also know,
[tex]\omega=\frac{2\pi}{24\times 60\times 60}[/tex]

and that [tex]\theta=\omega t[/tex]

as [tex]\theta[/tex] is very small, [tex]sin\theta=\theta[/tex] (approximately).

This gives you (after substituting and rearranging)

[tex]R=\frac{12\times 3600}{\pi}-h[/tex] (approximately)
 
  • #4
The angle theta that I've taken is 90-theta in the figure.
 

1. What is crude geometric estimation?

Crude geometric estimation is a method used to approximate the value of a quantity or measure by using simple geometric shapes, such as rectangles or triangles, to represent the area or volume of the object. It is a quick and easy way to get a rough estimate of a measurement without needing precise calculations.

2. How accurate is crude geometric estimation?

The accuracy of crude geometric estimation depends on the complexity and irregularity of the object being measured. It is not as precise as using more advanced mathematical methods, but it can give a fairly accurate estimate when used correctly.

3. When is crude geometric estimation useful?

Crude geometric estimation is useful in situations where a quick estimate is needed, or when precise measurements are not necessary. It can also be helpful when dealing with irregularly shaped objects or when only basic measurements are available.

4. What are the limitations of crude geometric estimation?

Crude geometric estimation can only provide a rough estimate and is not suitable for precise measurements. It also assumes that the object being measured can be represented by simple geometric shapes, which may not always be the case.

5. How can I improve the accuracy of crude geometric estimation?

To improve the accuracy of crude geometric estimation, you can use more shapes to approximate the object, or use more precise measurements for the dimensions of the shapes. It is also important to carefully choose which shapes to use and to make sure they accurately represent the object being measured.

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