Paul,

Yes I have Ansdale book. and Kenichi Yamamoto, but non give the Rotor
POCKET
volume.


Thanks Rolf

But my question is not the Chamber swept volume, for this you do not
require
the Rotor POCKET volume, I wanted the pocket volume, of the Rotor, and
as I
say we used to use plasticine to fill it and smooth around the Rotor
Flank
with a steel rule, then take the mould out of the pocket and then measure
the fluid displacement to arrive at the CC of the pocket by computation..

I enclose a screen shot, whilst the FIM accepted the 1.5 rule in 1998 I
accept that this may no longer be true.

The wording was from a Guide for the ACU officials, for post race strip
downs and the like..

Bob



Ok Bob, here it is again from 30/12/2009
Rolf

 Hello Bob,

 Regarding the calculations for the CR, here is an example of how to go
 about it.

Definition of the CR = (displacement + volume at TDC) / volume at TDC,

Mostly known is the displacement calculation = 270.5 R e b

For the small volume at TDC we first calculate a rotor with flat flanks
Straight lines between centre point of apex seal contact.
Here is the calc for the small volume at TDC with straight flanks:

 Small volume = [Pi ((e/R)2 + 1/3) - 31/2 /4 (1 + 6 (e/R))] * R2 * b
     = [Pi * ((e/R)2 + 1/3) - 0.433 * (1 + 6  (e/R))] * R2 * b

Example:  RX8, e = 1.5 cm, R = 10.5 cm, b = 8 cm,
 e/R = 0.14286

 Displacement = sqrt (27) * 10.5 * 1.5 * 8 = 654.7 cm3

 Small volume = [ 3.1416 *(.14286)2 + 1/3) - .433 * (1 + 6 * .14286)] *
 10.52 * 8 cm  = [ 1.1113 - .8042 ] * 110.25 * 8 = 270.9 cm3
 (flat flanks)

 CR = (654.7 + 270.9) / 270.9 = 3.42 having flat flanks,

 Reducing the volume at TDC increases the CR.
 E.G. 270.9 - 180 = 90.9 thus: CR = (654.7 + 90.9) / 90.9 = 8.2:1,
(270.9 - 190 = 80.9 (654.7 + 80.9 ) / 80.9 =  9.1:1)

 We now add a radius to the flank to obtain the additional volume needed
 for the selected CR.

 Distance d from short axis point to flat rotor flank:
 d = (R-e) - (e+ R/2) = R/2 - 2e
 d = R/2 - 2e = 105/2 - 2*15 = 22.5

 Select clearance to = .5 mm, h = d - clr, h = 22.5 - .5 = 22 mm
 Length of ½ flank s = cos 30 * R = .866 * 105 = 90.93 mm

 Note : if the clearance is selected differently one must recalculate the
 radius and the volume of the circle segment.
 Also note that the radius will not have a constant distance to the apex
 of the small axis as the rotor turns. In other words, the flank may be
 constructed as a curve instead of a radius to minimize the clearance
 volume. However, the difference is small.

 Radius of arc R1 = (s2 + h2 )/ 2h = (90.932 + 222 ) / (2*22) =  198.92
mm, = 19.89 cm
 ½ angle = Asin (90.93/198.92) = 27.2 degrees,

 Circle sector = R12 * pi * 27.2° / 180° = 187.85 cm2
 Area triangle = (R12 - s2 )0.5 * s = 160.85 cm2
 Circle segment area = 27.0 cm2
 Volume circle segment = area * b = 27 * 8 = 216 cm3

 CR without a pocket and with .5 mm clearance radius:
 Small volume at TDC = (270.9 - 216) = 54.9 cm3
 CR = (654.7 + 54.9) / 54.9 = 12.93 :1

 For CR = 9.1 : 1 we need 190 cm3 , where we require a pocket of 216 -
190 = 26 cm3
 If the pocket is 0.8 cm deep and 4.5 cm wide on average, it will be
7.2  cm long,
or use any other combination to make up the 26 cm3 .

 For a CR = 8.2:1 we need a pocket of 216 - 180 = 36 cm3 using above
 radius and 0.5 mm clearance.

Select CR = 10:1. To find the clearance volume : Displacement / (CR - 1),
 Volume = 654.7 / (10-1) = 72.75 cm3, [(654.7 + 72.75 ) / 72.75 = 10 : 1]

 Pocket size CR 10:1 = 270.9 - 72.75 = 198.15, and 216 - 198.15 =
17.85 cm3
 Or: 216 - 270.9 + 72.75 = 17.85 cm3
 Or: 72.75 - 54.9 = 17.85 cm3

 Regards
 Rolf Pfeiffer