# Why don't we use quater-circular dees instead of semi-circular dees in a Cyclotron

This is the setup, I have in my mind:

O1, O2, O3 and O4 are 4 oscillators.

The arrows in between the Dees represent the alternating EMF the Oscillators will generate.

I think we can easily adjust the frequency of the alternating EMF (T/4) in each oscillator and it seems to work. Any reason it wouldn't? and Why we only use semi-circular Dees?

Why 4 Dees?

Because it would increase the acceleration of the charged particle (not the final velocity); hence, it would take a lot less time to accelerate a particle to a great velocity as compared to that of a cyclotron with 2 Dees. It doesn't seems like a necessary requirement but my main question was is there a reason to use D shaped electrodes?

• No idea, but I guess it's cheaper that way (2 oscilators, instead of 4), and you can still get high speeds. Commented Jan 28, 2013 at 10:47
• If they are quarter circular, they'd be called $\Delta$s and not $D$s... :-) All joking aside, what is the problem you are trying to solve with the four quarter circular pieces instead of the two semicircular ones? Commented Jan 28, 2013 at 12:15
• Also, you realize that your O1 and O2 are hooked up to the same piece of metal, right? So at least as you drew it you will need to have the four oscillators timed relative to each other. Even if you remove two of the oscillators, the remaining two still need to be timed relative to each other. Unless you see somehow to get a huge benefit from this design, I guess the answer would be that the additional complication in the design is just not worth the effort. Commented Jan 28, 2013 at 12:21
• One can of course, separate the bending in a general particle accelerator into many different section as is commonly done in modern accelerators. But the cyclotron dates from early on (in fact it is the earliest recycling accelerator) and it's simple design is a consequence of being only as complex as needed. For out needs beyond what a cyclotron can easily provide we use other designs. Commented Jan 28, 2013 at 15:49
• @WillieWong I just added a possible advantage of my setup. Check the updated question. Commented Jan 31, 2013 at 7:52

$$r = \frac{p}{|q|B}.$$