Recent content by 1Keenan

Hi, Well, I solved with a good quality PTFE, I could apply up to +/-50 on both electrodes. Then I had discharges but not through the insulator. The happened between the electrodes and a metal reinforcing pillar in the vacuum chamber. So I made a chassis with PTFE that also reduces the gap, and...

Yes, you did. I forgot to tell you that theta is not small to approximate sin(theta) = theta. I think I'm messing up with the constants. Or do you see any huge mistake in the derivative?

ok, I did calculation and the most reasonable result is attached here. There is a problem, it doesn't work at all... any help?

Tan or tg is the same... I'll redo the calculation tomorrow, I think I made a mess with all the other constant terms

no, I don't want you to do it for me, off course! I didn't posted them because they are wrong. My strategy is to derive it as a sqrt((a+f(x))^2) but I guess it doesn't work properly...

Hello, I would need some help in calculating the derivative of the function T_el in the attached image. I want to calculate d T_el /d yd, where yd is the variable and it appears in the term I called A_elSide. Its expression is again in the image. Numbers you see are not important.Just to...

sure, you are right. Diameter has to be smaller than b to have "reflection" And actaully you expressed the second transition in the best way. If I read it correctly you mean that the output direction is the same as the diagonal, is that correct? Anyway, here, if the detector is extremely long...

I put values and it works... but now I know why I thought it was wrong: some particle will be reflected back (if R<b) some on the detector placed on the side (if R>b and R<110.6mm) and some would go on the front detector. It is like in the pic I posted. So, now these expressions for y and...

I don't know, I'm not really sure. I have the feeling that I made some mistake. I did the math a few times and I get the same expression again and again... so maybe there is no mistake. Can you please doublecheck?

hmmm, I'm not really sure this is the correct way...

hmm... what if: y = R sin (theta) and b= y tg (theta/2) if this is the way I can say tg (theta/2) = b/y= b/R sin (theta) = b / R(2t/(1+t^2)) so now I have t=b / R(2t/(1+t^2)) and, if I am not making a mess with calculation, I can say that t= +/- sqrt(b/(2R-b)) does it work?

well, the triangle has 2 sides equal to R and the 3rd is the chord, so one angle is Theta, the other two are both equal to 90+theta/2. I need to figure out how to find theta/2... thinking

ok, my memories of geometry are quite bad... any hint?

Simplified field geometry works fine when you know effective length of the magnet. So maybe the easiest would be to estimate the effective length of the dipole along the path of the eletron. I mean, if input and output are opposite edges we usually consider effective lengthof the magnet. In...

Well, sounds a bit too far from a magnetic dipole actually... That direction could lead to a totally wrong way