Help calculating the uncertainty in the Sun's rotational speed

[koots]

Homework Statement

Help calculating uncertainty when the equation includes sin functions

Relevant Equations

omega = A + Bsin^2(phi) + Csin^4(phi)

Hi everyone,

The equation is one we have been given to calculate the rotational speed of the sun for different latitudes. phi = average latitude. This shouldn't be a problem for me, but for some reason I just can't trust my error calcs.

We are given :
A = 14.713 ± 0.0491◦/d B = −2.396 ± 0.188◦/d C = −1.787 ± 0.253◦/d
and the latitudes I'm using have been taken from sunspot photos with a stonyhurst grid overlaid. They are:
31, 15, 5.5, 1.5, all with an uncertainty +/- 2.

I've so far used trigonometric identites, calculus, even calculating min and max values and halving the difference etc. My problem is that I end up with errors larger, and for the lower latitudes far larger than the result from the equation. Every method gives a slightly different result and I just can't carry on comfortably.

Could anyone suggest which method they would use for the above equation?

Cheers

 

[haruspex]

errors larger, and for the lower latitudes far larger than the result from the equation

I do not understand what you are saying there. Your wording implies the equation is for calculating an error, but you are calculating what should be the same error value by some other means and getting a much larger number.

It might help if you were to post details of an attempt (as forum rules require anyway) and show exactly what discrepancy you are seeing.

Remember, very few reading your post will have experience in this exact topic but many may be well able to assist if you explain clearly.

 

[koots]

Thanks haruspex.

The equation is for calculating the rotational speed of the sun. We are not given a method of calculating the error in the result. I did write up an example on my lunchbreak at work to post here when I got home after realising there were rules, but I can't for the life of me find it now.

I've since found a silly mistake I made using the calculus method earlier of taking the partial derivatives multiplied by the change in the variable, squaring, adding, and taking the square root and now it works out much nicer but I'm still not entirely sold on it. A quick run through the method is attached.

The reason I'm not sold on it is because as my latitude decreases, the uncertainty in that measurement increases as all measurements were +/- 2 degrees, however in the final answers the uncertainty becomes less at lower latitude values.

Any thoughts?

Cheers

 

[haruspex]

Looks good to me.
The reason the errors in latitude matter less at low latitudes is that $\sin(\phi)$ becomes very small.

 

[koots]

Thanks mate. Much appreciated. Makes perfect sense, I just had my guts telling me it was my fault..

Cheers