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Post by Admin on Nov 20, 2016 7:11:59 GMT
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Nov 21, 2016 22:40:57 GMT
Vasco,
I have gone over your solution to the exercise on p. 318. This exercise puzzled me because [36d] looked wrong with the label $2\rho$ being drawn as an apparent chord and not as a diameter. I like your derivation and it all looks correct to me, but shouldn't Needham have included some qualification, such as "$2\rho ds$ is a good approximation of $1-|z|^2$ for larger values of $2\rho$"? I have a similar reservation regarding the equality of $ds$ and $d\widehat{s}$ in [36b].
My copy shows no "E". Perhaps it was cut off by the plot range. I take it that the E is the point intersecting the circle after extending AB.
Gary
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Post by Admin on Nov 22, 2016 6:46:27 GMT
Vasco, I have gone over your solution to the exercise on p. 318. This exercise puzzled me because [36d] looked wrong with the label $2\rho$ being drawn as an apparent chord and not as a diameter. I like your derivation and it all looks correct to me, but shouldn't Needham have included some qualification, such as "$2\rho ds$ is a good approximation of $1-|z|^2$ for larger values of $2\rho$"? Gary I agree. I would just say, as I do in my document, "As $\rho\rightarrow\infty$, $ds\rightarrow 0$ and $2\rho ds\rightarrow 1-|z|^2$". I'm not quite sure what you mean here. I don't think Needham is saying that $ds$ and $d\widehat{s}$ are equal. I have corrected this. Thanks Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 22, 2016 15:39:34 GMT
I'm not quite sure what you mean here. I don't think Needham is saying that $ds$ and $d\widehat{s}$ are equal. Vasco, Sorry, I meant to write "regarding $d\widehat{s} = \frac{2}{1-|z|^2}ds$" as applied to [36b]. I think this figure needs a little more explanation. Gary |
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