torsten at inetw.net
Wed Feb 12 14:44:10 PST 2003
On Wed the 12 Feb 2003 at 15 hours EST
>Alas! torsten spake thus:
>> >Suffice it to say, when you multiply by a constant, you also multiply
>> >the error by that constant... and k just happens to be 8.99 * 10^9.
>> What's your error on k? Since you are calculating error on F, and
>> if k is assumed to be a constant, the the sum of errors won't include
>> the error on k.
>> You'll then find out, your balls are where you want them to be.
>No, that's not how it works.
>In the equation 'z=kx', where x varies, k is constant, and you want to
>know the error in z, the equation is so:
Somehow I'm thinking I used to use
>(where 'dz' is the error in z, and 'dx' is the error in x). x itself
>just happens to be 'q1q2/r^2'.
>The error that we calculated for 'x' turned out to be fairly reasonable
>(on the order of millimeters), but we had to factor k in, as it's part
>of the equation. When you multiply millimeters by 8.99 * 10^9, you get
>hundreds of thousands of kilometers ;)
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