Please help with calculation involve solid angle.by yungman Tags: angle, calculation, involve, solid 

#1
Jan913, 07:55 PM

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This is only an example from Kraus Antenna 3rd edition page 404. The question is really a math problem involves calculation of ratio of solid angles. Just ignore the antenna part. this is directly from the book:
Example 121.1 Mars temperature The incremental antenna temperature for the planet Mars measured with the U.S. Naval Research Lab 15m radio telescope antenna at 31.5mm wavelength was found to be 0.24K(Mayer1). Mars subtended an angle of 0.005 deg at the time of the measurement. The antenna HPBW=0.116 deg. Find the average temperature of the Mars at 31.5mm wavelength. The answer from the book: [tex]T_s=\frac {\Omega_A}{\Omega_S}\Delta{T_A}\;\;\approx\;\frac {0.116^2}{\frac{\pi}{4}0.005^2}(0.24)=164K[/tex] I don't understand where the [itex]\frac {\pi}{4}[/itex] in the denominator comes from. This is just a simple problem where the ratio of the area of two disk one with Θ=0.116/2 and the other with Θ=0.005/2. I try to use the following to calculate: [tex]\Omega=\int_0^{2\pi} d\phi\;\int_0^{\theta/2}\sin {\theta} d \theta[/tex] So [tex]\frac{\Omega_A}{\Omega_S}\;=\;\frac{\int_0^{2\pi} d\phi\;\int_0^{0.116/2}\sin {\theta} d \theta}{\int_0^{2\pi} d\phi\;\int_0^{0.005/2}\sin {\theta} d \theta}\;=\;\frac{2\pi\int_0^{0.116/2}\sin {\theta} d \theta}{2\pi\int_0^{0.005/2}\sin {\theta} d \theta}[/tex] [tex] \frac{\Omega_A}{\Omega_S}\;=\;\frac{(1 0.999999487)}{(10.999999999)}=538.2[/tex] But using the answer from the book: [tex]\frac{0.116^2}{\frac{\pi}{4}0.005^2}\;=\;685.3[/tex] I just don't get the answer from the book. Can it be the limitation of my calculator to get cos 0.0025 deg? Can anyone use their calculator to verify the number? please help. Thanks Alan 



#2
Jan913, 08:30 PM

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I'll take a guess at this. It seems like the same idea as "geometric form factors" in radiative heat transfer.
The telescope is "seeing" a piece of space that you can take as a flat disk (diameter 0.116 degrees) But the surface of Mars that is radiatiing is not a flat disk, it is a hemisphere. The "brightness" as seen by the telescope will be maximum in the center and less at the edges. That will give you another factor of ##\sin\theta## or ##\cos\theta## in the integral for Mars. Probably the "correction factor" of ##\pi/4## is a well known result, since most astronomical objects are approximately spherical. 



#3
Jan913, 10:44 PM

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Thanks for you time. I don't mean to say you are not right as it might well can be. On the first pass, I don't think that's what the book meant, the reason is, this example is right at the beginning of the chapter and the book did not mention anything about the concaveness of the star. Also the answer of the book:
[tex]T_s=\frac {\Omega_A}{\Omega_S}\Delta{T_A}[/tex] Which does not contain any compensation for the concaveness of the planet. So far, this book has been pretty straight forward and concise. I type out the exact words from the book in my post and nothing about this. I scan through the later part of the chapter and there's nothing about this either. Basically it only talked about the ratio of the solid angle for object smaller than the half power angle. Again thank you for your time. I manage to go on line and use a cosine calculator that don't flag error of 0.0025 deg and gave me 2 more digits in the decimal place than my calculator, the answer is even more off!! So I don't think the small angle is the cause of the problem. 



#4
Jan1013, 12:18 AM

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Please help with calculation involve solid angle.yungman, is it possible that HPBW is to be interpreted as a square, not a circle? 



#5
Jan1013, 12:35 AM

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That I am pretty sure, the main lope a very directional antenna is like a long balloon or sausage!!! But still don't explain the pi/4 even if you look at it as square. 



#6
Jan1013, 01:52 AM

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#7
Jan1013, 10:30 AM

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The physics question here is whether each point on the surface of Mars radiates energy in all directions like a "point source", or whether some effect (e.g. electrical conductivity of the Martian material) means sonething different happens. It's not selfevident to me that EM radiation with wavelength 31.5mm would behave the same way as visible light (the wavelengths are different by a factor of 10^6) but I don't know whether it does or it doesn't. 



#8
Jan1013, 02:45 PM

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The question is whether I am doing it right with my approach assuming the Mars has uniform temperature on the surface. I really believe the book is wrong to put the ∏/4 in. The square of the degree will be good already as the ratio is the same using Sr or degree. I just try taking my result (538) divide by ∏/4, the answer is 685 which is the same as the book!!! 



#9
Jan1013, 04:01 PM

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#10
Jan1113, 07:37 AM

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Yes, is the only way to for it to make sense. I just don't know of any antenna that have a square cross section beam. Mostly has round or at least roundish cross section. I guess we never know what the author think. Anyway, thanks for the help.




#11
Jan1113, 02:41 PM

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#12
Jan1113, 02:48 PM

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Thanks for your help.
Alan 


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