Dot Physics

Some time ago, I wrote about the awesome things the Greeks did in astronomy. Basically they calculated the size of the Earth, distance and size of the moon and distance and size of the sun. The value obtained for the distance to the sun was a bit off, but still a bang up job if you ask me. (where bang-up is meant as a good thing) If the greeks were in my introductory physics lab, they would need to include uncertainties with their measurements. What would the uncertainty in the final value look like?

In my introductory physics lab course, I have students measure things and estimate the uncertainty in these measurements. I also have them calculate stuff with these measured quantities and estimate the uncertainty in that. It appears that I have failed to previously post about measurements and uncertainty, so let me give a VERY brief example. Suppose I want to determine the surface area of a rectangular table. To do this, I measure the length and the width. Pretend I get the following values:

i-81d162aead96d5fe4a046bf3e4741628-2009-12-29_calculating_the_distance_to_the_sun_with_uncertainty_dot_physics_1.jpg

If that looks weird, let me tell you what it means. If I try to measure the length of the desk, there are two problems. First, how would you define the actual length of the desk? It is surely not a perfect desk such that the length at different points is different. Also, the edge may be rounded and not well defined. Finally, the instrument that I use to measure the desk has limitations. All this combined gives me what is called the uncertainty in the length. It is typically designated with a +/- following the best estimate of the value. This gives a range in which the actual value resides. For the length above, this means that the length is almost certainly between 133.0 cm and 133.4 cm. The uncertainty in L is typically denoted as delta L. How do you get the uncertainty? For now, just assume it is an estimate.

Ok, now how about the surface area? To calculate the surface area of the table, you would simply multiply length times width, right? Yes, but what about the uncertainty in the area? If you are not sure about the length and not sure about the width, the area is not certain either. Here is a diagram that shows the uncertainties for the area:

i-3f0717fb9bb69f493b2c3eeb4bef255f-2009-12-29_area_1.jpg

Great, but how do you calculate the uncertainty in the area? The answer depends on how formal you want to do it. The easiest method calculates Amin = LminWmin and Amax = LmaxWmax. Do not think that Amax is the same distance above A as Amin is below (but it could be). For this method, I could find the uncertainty as:

i-d5cba33dc5c7f1cd12cc45b4abb81a34-2009-12-29_delta_a.jpg

If you are going to use this method, be careful. For some calculations, to find the minimum value you may need to put in the maximum for a variable. For instance, suppose you are calculating the density from measurements of the mass and the volume. To calculate the min density, you would do the following:

i-128b3e5077e434058a21d60755dc5b91-2009-12-29_density.jpg

Since the mass is divided by the volume, a larger volume will make a smaller density. Ok, moving on. Let me just write down a more sophisticated way of finding the uncertainty of a calculated quantity (often called propagation of error). Suppose I want to calculate something, say f. Where f is a function of the measured values x and y. If I know the relationship between f and x and y, and I know the uncertainties in x and y, then the uncertainty in f would be:

i-575296f7f11c79262204f56b10e86b2f-2009-12-29_deltaf.jpg

If that looks complicated, no big deal – it is essentially the same idea as the area example. If you don’t know what a partial derivative is, again no big deal. It is essentially saying “how does f change with x?” Ok, I think that is enough about uncertainty to do some good. Back to the Greeks and astronomy.

Measuring the size of the Earth.

The story says that Eratosthenes used the angle difference between two shadows a given distance apart. Here is a diagram:

i-7b14d0bd8b345fe185aeee58c9a32d2f-2009-12-29_measure_earth.jpg

I will assume that the sun was directly overhead in Syene (so no measurement) and he just needed to measure the angle at Alexandria and the distance between these two. I am not going to work with numbers right now, but the following would be the radius of the Earth:

i-6aa3d512ba0e61b1c966360fe3906025-2009-12-29_r_2.jpg

Where this angle is measured in radians. I guess the Greeks might have measured angles in degrees, so that would make it:

i-5e8bc3d90f883e8fe801cc837e5ffab2-2009-12-29_r_3.jpg

I am not really sure how the Greeks measured angles (or distances between cities) but I will proceed anyway.

Distance (and size) of the moon

As I posted previously, I am not exactly sure this is how the Greeks found the distance to the moon, but this should work. Since the moon rotates around the center of the Earth and not a point on the surface, you should see it in a slightly different location. (of course the moon’s orbit is not completely circular – but as long as you can say where it “should” be and where it is that is fine)

i-df91c83e578e36ede89850eaa44497d0-2009-12-29_distancemoon.jpg

From this diagram, if I know the radius of the Earth and the angle between where the moon should be and where it is (I will call this angle alpha) then the distance to the moon (from the center of the Earth) would be:

i-a740b2a1c1b60b74422ca36cdd1caa2e-2009-12-29_distance_34.jpg

You can see that the distance to the moon depends on the angle measurement AND the radius of the Earth. Combining these two formulas:

i-f2af8a5f4a86560c60b883d21f606464-2009-12-29_dmoon_2.jpg

Distance to the Sun

For this calculation, the Greeks used the distance to the moon and the angle between the Sun and Moon during a quarter phase moon. Here is a diagram:

i-62d66bead89735c5da45b55db4b1e5ec-2009-12-29_sunmoonearth.jpg

From this right triangle, I can calculate the distance to the Sun. I will denote the angle between the Sun and moon as beta. This will give:

i-990ed0d45b0835b609be5d3d33c37062-2009-12-29_calc_distance_sun.jpg

And, again putting in an expression for the distance to the moon:

i-cb9cbfc15afbfe3919d64ee5096b569d-2009-12-29_dsun_3.jpg

So to calculate the distance to the sun, I would measure:

  • The distance between two cities (s) in whatever distance units you like. The units for this will be the same units as the distance to the sun.
  • The angle between the two shadows at the two cities at the same time (theta) measured in degrees.
  • The angle between the predicted location of the moon (assuming you are at the center of the Earth) and the actual location of the moon (alpha). Technically, you could use any units here, but it turns out to be simpler if I use radians because of the trig function.
  • The angle between a quarter moon and the sun (never look at the sun. Although Bad Astronomy says you won’t go blind, still don’t do it just to be safe and so you won’t sue me for saying you can.) This angle will be beta, again measured in radians.

Ok, now what about the uncertainty?

Of course you notice that I have not given any values for anything yet. Well, I will. But first, let me find the uncertainty in the distance to the sun.

i-17b0eeb8472c90c9658fab17e0fb9c65-2009-12-29_deltadsun_1.jpg

So, all I need to do is to calculate the partial derivatives and estimate the values and their uncertainties. If you do not like calculus, avert your eyes (even though I am not going to show you how I did it).

i-5c7a5f920a492a2c86c87163ccb3b785-2009-12-29_delta_1.jpg
i-b07333ca8612cdb5adc9cf4db054f99c-2009-12-29_delta_2.jpg
i-8c08732e36339d5c4d34d61344df70b4-2009-12-29_delta_3.jpg
i-5dd2a5d00ec43b62fa953213cbf66fee-2009-12-29_delta_4.jpg

If I made an error, I am sure someone will point it out. Now, before I put this all together, let me guess at some values with uncertainties.

  • s = 800,000 +/- 5,000 m
  • theta = 7.5 +/- 0.2 degrees
  • alpha = 0.02 +/- 0.005 radians (completely guessing on this one – I will fix it later)
  • beta = 1.57 +/- 0.005 radians (close to being perpendicular)

Now, what to do? I am going to do all my calculations in a spreadsheet so that you can change the values if you want. Remember the point is not to get the correct value of the distance to the sun, but rather to see how the error in the measurements impacts the value.

Here you can change all the values you want and it will give you the calculated values with uncertainty. Because I wanted to give both the Radius of the Earth with the distance to the moon, I calculated their uncertainties as well. When I calculated the uncertainty for the distance to the sun, I used the uncertainty of the angle measurement and the uncertainty in the distance to the moon.

I cheated. I knew the accepted values of the distances, so I adjusted my angles to give me approximately that value. Also, I completely guessed at the uncertainties. With these values, it still shows my point. Look at the distance to the sun:

i-ea576faf6ac8a0c2b910f210eb4ef964-2009-12-29_dsun_34.jpg

Yes. I know I am breaking my own rules here. The rule is that there really should only be one significant figure in the uncertainty. How could you say the time was 5.1234 seconds +/- 0.2324 seconds? If you know the uncertainty to that many significant figures, wouldn’t the uncertainty be smaller? Also, the decimal place of the value should match that of the uncertainty. It wouldn’t make since to say “I will meet you in 30 seconds +/- 0.000001 seconds”. So, this is how I should have written it:

i-0513ac4531a0bc40bca3f77cc6de504c-2009-12-29_dsun.jpg

That looks bad, doesn’t it. It basically says the distance to the sun is …..something? Why is the error in the distance to the sun so large? It has to do with the formula with is inversely proportional to the cosine of the angle. Here is a plot of 1/cos(beta) for angles close to pi/2:

i-e0785cb428336ec4b7b3c9c83be13a91-2009-12-29_pi_plot.jpg

Forgive me for using Excel (it makes very ugly graphs), but it was open at the time. Here you can see that when the angle gets near pi/2, the function explodes. With such a steep slope, a small change in angle makes a huge difference. That is why this is a difficult measurement and why the uncertainty is so large.

Comments

  1. #1 Kevin Reid
    December 30, 2009

    You’ve got some “?”s where it looks like various Greek symbols should be.

    Also, in the length uncertainty diagram, shouldn’t the arrows ∆W and ∆L be half the length they are?

  2. #2 Rhett Allain
    December 30, 2009

    @Kevin,

    You are correct. When I copied and pasted from my old site, some things tried to copy as text. I think I temporarily fixed most of them. Thanks for pointing this out.

  3. #3 Allison
    December 30, 2009

    “I am not really sure how the Greeks measured angles (or distances between cities) but I will proceed anyway.”

    According to Holton & Brush…

    The angle measurements were done as percentages of whole circles, according to Eratosthenes, the angle measurement was exactly one fiftieth of a whole circle. The distance between Alexandria and Syene was measured by timing a well-trained group of soldiers whose rate of marching was known and fairly constant to be 5000 Stadia. One stadia ~ .10 mile.

    Now I wonder what the random uncertainty associated with timing soldiers is…

  4. #4 Anonymous Coward
    December 30, 2009

    Same comment as the last time you posted this:

    Your error analysis for the distance to the sun is wrong. Orders of magnitude wrong.

    The formula that follows “OK, now what about the uncertainty” is valid most of the time, but you should expect it to fail if any of the partial derivatives are infinite or discontinuous over the range of the uncertainty of the measured quantities.

    The measured quantities you list put you exactly in that limit.

    (If you’re concerned about accurate error propagation there are actually more restrictions than the one I mention above, but I’ll leave it at that for now.)

    Going a little further along these lines, one should expect that – for angles near pi/2 – starting with symmetric error bars on your measured quantities will result in highly asymmetric error bars on d_sun. This – and why your error analysis formula breaks down – should be apparent from examining the final graph of the post.

  5. #5 Robert
    January 1, 2010

    According to Carl Sagan in “Cosmos” the distance between Alexandria and Syene was approximated as 800 km because Eratosthenes hired a man to pace it out. How accurate would pacing be? Sticks and feet were the tools used.

  6. #6 H Bos
    January 1, 2010

    We are more developed but less in touch than the Greek were.
    many things web do become aware of by study were common concepts in the past. The sun and moon were universal time-pieces.
    The curvature of the Earth will become very apparent to road and canal builders!
    Measuring the waterlevels in a visually straight canal will show the curvature of the Earth and so its diametre.
    Builders did use rope levels and used triangulation, thus builders could have calculated the size of the Earth as well.
    However people rarely extend their knowlegde beyond practical use let alone record it.

  7. #7 Rhett Allain
    January 3, 2010

    @Robert,

    I put the uncertainty at 5000 meters (complete guess). However, I put that value in the embeded spreadsheet so that you could change it and see what happens.

  8. #8 Rhett Allain
    March 16, 2010

    @Anon,

    I think you have a valid point about the propagation of error in this case. Sorry it took so long to address this, but I will (at some point).

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