Recent content by Charles Link

I thought @hutchphd 's of post 49, (see also post 52) was simpler. I think this latest (post 73) I would call "a" most efficient solution, but not "the" most efficient. :) Edit: Notice it does indeed give the general result that ## x^2=(a^2+ab+b^2)/3 ##.

He was one of the smarter ones on here. We will miss him.

@DaveC426913 I am still waiting to see most of the geese fly south. I don't think most of them have gone south yet. If they did, I missed them.

The above is a photo of a 4'6" snowman that I made today, Monday 11-10-25, in Chicago. It is the first one for this season. We had about 3" of snow overnight, but at first the snow was too cold to pack. I waited a couple hours until things warmed up above freezing around noon, and it packed...

In that case, one would need additional spectral detail, such as measuring with two or more bands.

I was able to use it to determine the temperature of a calibration blackbody source with known aperture size at close range, ( distance of one foot). In general, bandpass measurements with the Planck blackbody function are necessary.

I think it is probably impractical and perhaps almost impossible to use Stefan's law, (which is the power radiated over the complete spectrum), except over very short distances because of atmospheric absorption bands, but using the Planck blackbody function integrated over a wide bandpass...

Perhaps we should include a little more detail. Stefan's law is ## M=\sigma T^4 ## where ## M ## is ## \pi ## times the integral of the Planck function over the entire spectrum: ## M= \pi \int\limits_0^{\infty} L(\lambda,T) \, d \lambda ##. I think generally they don't use Stefan's law like...

I don't know that I can answer what you are asking, but for a laboratory type blackbody at approximately 1000 degrees Centigrade, one of the standard methods to measure its temperature about 30-40 years ago was to use an optical pyrometer where the color/brightness was matched between an...

@neilparker62 My solution with your diagram is simple. Let RL=red line. Then ## (RL)^2=2^2+4x^2-2(2)(2x) \cos{\theta} ## and ## (RL)^2=11^2+x^2-2(11) x \cos(180-\theta)=11^2+x^2+2(11)x \cos{\theta} ## (both are law of cosines) with ## \cos{\theta}=2/(2x)=1/x ## We set the two ## (RL)^2 ##...

@neilparker62 Yes, the law of cosines used twice with your diagram solves it. Very good. :)

@Gavran Still I am a little disappointed that the PF members who can respond to these posts don't seem to have looked carefully over post 41 by @daverusin to get the same kind of adventure from it that I got. I tried to summarize what he did once I figured out his method. He didn't present a...

a copy and paste of mine from the thread "a trigonometry problem of interest": I'm finding it somewhat amazing how the ai overview is able to summarize for me the steps that are needed for the method of finding rational roots of a conic section with linear parametrization using the rational...

I'm finding it somewhat amazing how the ai overview is able to summarize for me the steps that are needed for the method of finding rational roots of a conic section with linear parametrization using the rational root theorem. I would like to copy and paste their summary which I think even did...