Carl Quick's LiteWieght 2240
So for about the last 24 months, I've had her on a serious diet. Now I am happy to
report that she lost nearly 13 ounces and is feeling so much better. And let me
say one thing, when I pick her up with one hand now she is light as a feather.
Well, almost a feather.
Things I have done so far:
OEM main tube was lightened by me.
Barrel is OEM and trimmed down by me.
Main tube plug is made of Delrin by me.
Gas cap is aluminum, comes from HPA and bored out by me.
Hammer is made from Delrin by me with a modified 1701P striker.
Valve is an aluminum Disco rear end with a brass Pro-Top by Anthony.
Trigger grip frame is aluminum by Bluefork Design and then hogged out by Gregg.
Front barrel band is aluminum, made by Anthony, drilled by Rich and milled by me.
Re: It does make sense. However, the surface area...May 23 2012 at 6:06 PM
|Curt (Login curt44319)|
Crosman Forum Member
from IP address 188.8.131.52
Response to It does make sense. However, the surface area...
[quote]The same end result if the design was just a large diameter piston the same diamter of the larger telescopic piston.[/quote]
Depends. ( but definitely no IF the details get worked out )
As I alluded earlier, I haven't got the mechanics quite resolved, but it goes like this...
If the squish area is also matching concentric rings....
In other words, there are matching progressively smaller cylinders into which the telescoping piston parts travel. Probably opposite of what you're thinking based on the diagram.
The largest diameter compresses to its "shelf," reducing its squish to zero. At that point, the next smaller diameter continues to travel to its "shelf," further on in the stroke. So on and so on, until all that is left is the smallest diameter, proceeding to the valve inlet.
This way, as each larger diameter bottoms out, it effectively becomes the cylinder wall for the next smaller diameter, until the next smaller passes the shelf for the larger, at which point the shelf becomes the cylinder wall for that size, and the larger irrelevant, as it's now completely passed by the next smaller.
Spring pressure for the larger diameters needs to be higher than the highest expected cylinder pressure at that level of compression to insure the larger diameter ring in fact, bottoms on its self before the next inner diameter begins to move independently of the outer.
Minimum pumping effort would be the sum of all springs divided by mechanical advantage of the linkage.
Maximum effort would be determined by the highest expected PSI developed by the smallest diameter.
The smallest having no spring loading, and so being a direct mechanical connection to the pump linkage.
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