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Question about the "Valve Ineffectiveness Chart"....

July 5 2012 at 11:18 AM
  (Login rsterne)
Crosman Forum Member
from IP address 66.244.241.34

Quite a while back I found Steve in NC's threads on the "Sonic Horizon Theory" and his "Valve Ineffectiveness Chart"....

[linked image]

It occured to me while pondering this chart that it should be possible to have the left axis of the chart read the amount of "usuable HPA" which (I think) should be double the percent of barrel length used (as a percent of bore volume).... As an example, at the 850 fps shown, the pellet is 33% of the way down the bore when the valve becomes ineffective (ie no more useful air is released by the valve).... At that point, assuming the pressure in the bore behind the pellet and the valve is equal (which would, of course, require no restrictions and 100% efficiency), the volume of air that can do useful work (either accelerating the pellet, or pushing the air that CAN accelerate the pellet out of the valve) would be twice the volume of the bore behind the pellet, ie 66% of the bore volume.... If the valve was a closed system (ie no other air could enter it during the shot) then we should be able to calculate the MINIMUM volume of the valve required, in combination with the pressure of the air inside it.... ie the MINIMUM bar-cc of air required....

In the simplest example, if the valve was 33% of the bore volume, and started at 3000 psi, then that air would have expanded to 66% bore volume (half in the barrel, half still in the valve) at 1500 psi at the moment of valve ineffectiveness.... Other combinations are possible, however, the valve could be equal to the bore volume and start at 2250 psi and still produce 66% bore volume at 1500 psi at valve ineffectiveness, as an example.... In the limiting case, the valve/reservoir volume would be infinite, and the pressure would not drop during the time the valve was "effective" (eg. 1500 psi in the example given).... In this case, the bar-cc would be 1500/14.5 x 66% of the bore volume.... By using this limiting case, the volume would be twice the value on the left axis of the chart.... while the MINIMUM pressure could be calculated with just a few variables....

The major variables are:

1. The barrel length and caliber (and hence the bore volume)
2. The muzzle velocity (and hence the position of valve ineffectiveness)
3. The pellet weight (which determines the force, ie pressure x area, required to reach the MV in the length of the barrel)

There are some minor variables which could affect the results:

A. The transfer port volume (ie wasted volume between the valve seat and pelelt base)
B. The drag of the pellet in the bore (friction losses)
C. Restrictions to airflow (and overall efficiency)
D. The mass of the air in the bore (at the point of valve ineffectiveness?)

Item "A." is usually pretty insignificant, and Item "B." is relatively small compared to the force available (it might use up ~50 psi).... Item "C." can be ignored if we just want to calculate the MINIMUM air required (more air will be needed to overcome it), but Item "D." will become quite significant as the pressure and volume increase....

I throw this idea out there just for discussion.... is there something I'm missing?....

Bob



    
This message has been edited by rsterne from IP address 66.244.241.34 on Jul 5, 2012 11:30 AM


 
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