Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 1, 2017 1:19:09 GMT
Vasco,
After several days of trying Ch 8, Ex 2 with no success, I took a look at your answer. It appears that the key to your solution was the decision to convert sine and cosine to their series. That allowed $\phi$ to be defined in terms of DP, with the given value $\frac{1}{2}\hspace{.3em}\Delta\hspace{.3em}\sec(\theta)$. I wonder how you knew that a series was needed. Were you aware that you needed an $A_C$ defined in terms of $\phi$ as in (2)? I found that I always came back to the necessity to use $\phi$, and I defined it in terms of $DP$, but I had nothing good in which to substitute it.
Gary
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Post by Admin on Sept 1, 2017 6:00:27 GMT
Vasco, After several days of trying Ch 8, Ex 2 with no success, I took a look at your answer. It appears that the key to your solution was the decision to convert sine and cosine to their series. That allowed $\phi$ to be defined in terms of DP, with the given value $\frac{1}{2}\hspace{.3em}\Delta\hspace{.3em}\sec(\theta)$. I wonder how you knew that a series was needed. Were you aware that you needed an $A_C$ defined in terms of $\phi$ as in (2)? I found that I always came back to the necessity to use $\phi$, and I defined it in terms of $DP$, but I had nothing good in which to substitute it. Gary Gary In equation (1) of my solution $A_C=R^2(\phi-\sin\phi\cos\phi)$, I decided I needed to approximate $A_C$ and I wanted to get an expression in terms of $\phi$ only, not in terms of $\phi$, $\sin\phi$ and $\cos\phi$. Usually the first order approximations to the trig functions are sufficient: $\sin\phi\approx\phi$ and $\cos\phi\approx 1$. But if we substitute these into the expression for $A_C$ we obtain a first order approximation to $A_C$ which vanishes, which is not much use for further calculation. So I decided that I needed a second order approximation for $\sin\phi\cos\phi$ to get a non-vanishing second order approximation to $A_C$. I think this is the key. Vasco
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