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Eeach bunch travels at almost the speed of light (v = 0,999999991·c ) : v = 0,999999991 x 299792458   ⇒  v = 299792455 m/s So each one will go around 299792455 / 26659 = 11245 laps per second During the 20 minutes needed to reach the final energy, the protons cover a distance further than from Earth to the Sun and back. 11245 x 20 x 60 x 27 = 3.64·108 km The Frequency of the circular movement already calculated is: f = 11,245 kHz And the Period is: T = 1/f ≈ 9·10-5 s Protons will be  accelerated to v~c.  While travelling they will permanently be under a centripetal acceleration produced by Lorentz Force. Let´s calculate this centripetal acceleration: a ≈ c2/r   ⇒  ac ≈ 2.3·1013 m/s2 Therefore 2·1012 times the acceleration of gravity. It's been said before that the distance between the two bunches will be : d = 7, 48 m So the “time between bunches” -bunch spacing-  will be t = distance / velocity t = 7,48 / 3·108 t = 24,95·10-9 s  ⇒   t = 24,95 ns which is a very important parameter. Another interesting exercise is to calculate how many laps the protons travel before falling due to gravity. h = ½g·t2   ⇒   t =  (2h/g)1/2 where h  is the pipe radius (~ 28 mm) so: t = (2·0,028/9,81)1/2   t ≈ 76 ms Divided by the period: nº of laps = 76·10-3/8,9·10-5 n = 850 laps Therefore, we must not only correct the electromagnetic alterations but we must also correct those due to gravity. The magnetic multipoles will correct those alterations.