GTA
All Springer/NP/PCP Air Gun Discussion General => "Bob and Lloyds Workshop" => Topic started by: rsterne on November 15, 2015, 03:17:15 PM
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Over the years, I have been trying to come up with a system for deciding how heavy a hammer has to be, and what the required stroke is, when building a PCP from scratch, or modifying one way beyond its original design FPE.... The variables are many, air pressure, valve throat area, barrel length and bullet weight, caliber, and so on.... However, I have always had it in the back of my mind that the overall governing factor was the FPE output of the gun.... Basically, it takes a certain amount of hammer energy and momentum to open the valve long enough to allow enough air to escape to produce the FPE you want.... If the pressure is higher, you need less dwell, but more of the hammer energy is used in cracking the valve off the seat.... If the bullet is heavier, you need more dwell, but the FPE goes up if you can deliver that.... I know it seems too simple to be true, but when I look at the combinations of hammer weight and travel I have used over the years, which will allow a gun to deliver the maximum power it is capable of, there seems to be a definite relationship there.... It just became a matter of quantifying that.... It turns out to be much simpler than I could have ever imagined.... Here is the formula I have come up with.... This is empirically derived (ie through experience), not calculated, hence the mixed units....
Hammer Weight (grams) = Energy Required (FPE) / Hammer Travel (inches)
If you have more travel available, you can use a lighter hammer.... If your travel is limited, you will need a heavier one.... While it is true that you can trade hammer travel for spring force, because hammer energy (ft.lbs) is force (lbs.) times distance (ft.).... there is a limit to how strong a spring you can use.... I have found that to be about 20 lbs. when cocked.... Any more than that, and it is simply too difficult to cock the gun.... Yes, you can use a side lever to increase the force available (and hence shorten the stroke), but isn't it nicer to use that to reduce the effort?.... So, if you make the force a constant, the energy becomes a factor of the distance that force acts through, ie the hammer travel.... Relating the hammer energy and momentum to the amount of work is takes to open the valve for a given time to create a certain level of FPE in the bullet just makes sense.... Here are a couple of graphs that show you what happens to the hammer weight, for three vastly different amounts of hammer travel....
(http://i378.photobucket.com/albums/oo221/rsterne/PCP%20Internal%20Ballistics/Hammer%20Weight%20vs%20Travel_zpsd78ut1ib.jpg) (http://s378.photobucket.com/user/rsterne/media/PCP%20Internal%20Ballistics/Hammer%20Weight%20vs%20Travel_zpsd78ut1ib.jpg.html)
The "Long Travel" option is (square root of the FPE) / 5.... The "Average Travel" is (square root of the FPE) / 10.... and the "Short Travel" is (square root of the FPE) / 15.... Other values for the travel can be used, but I think that covers pretty much the entire range of hammer travel and weight combinations that is workable.... One thing to note here, is that this relationship only applies to conventional "knock open" valves.... which, let's face it, rely on brute force to open them.... If you have a balanced valve, like the Cothran "force reduction" system, the balanced valve that Lloyd Sikes developed, or the one used by (QV)Tom Costin in the Slayer, then this formula doesn't apply.... because the forces required to crack the valve are MUCH less....
Here are some examples, using the "Average Travel"....
50 FPE.... Travel = 0.71".... Weight = 71 g. (works great in highly modded Discos at that power level)
100 FPE.... Travel = 1.00".... Weight = 100 g. (I have used this combination, or close to it, successfully in several PCPs)
200 FPE.... Travel = 1.41".... Weight = 141 g. (almost exactly the combination used in my .284 and .308 Hayabusas)
400 FPE.... Travel = 2.00".... Weight = 200 g.
Using the formula to calculate hammer weight....
150 FPE.... Travel = 1.20".... 125 g. (My .257 Haybusa uses 1.2" of travel, and a hammer that weighs 118 g. to achieve 160 FPE)....
500 FPE.... Travel = 2.00".... 250 g. (My .457 Hayabusa uses 2.0" of travel, and a hammer that weighs 237 g. to achieve 550 FPE)....
As you can see, the FPE I have achieved is slightly more than predicted.... but on the other hand, the gun is very hard to cock when tuned that way.... I'm not claiming that the formula is 100% accurate, and it should not be used to predict the FPE of the gun, that isn't the way it works.... It is intended to give you a combination of hammer weight and stroke that should work in a given FPE gun without requiring a ridiculous hammer spring that makes cocking next to impossible.... balanced valves excepted.... It should allow you to select a combination of hammer travel and weight capable of maxing your gun out at the pressure you are running.... driving the velocity all the way to the plateau, providing you have the correct hammer spring for your bullet weight in that gun....
Bob
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Hey Bob,
I think I recall a thread were you did something similar with regard to the projectile weight. If you happen to recall that, it would be interesting to see how that pans out. Perhaps averaging the two would result in a required stroke and weight with regard to desired FPE. Idea is to eliminate the all the trial and error or find a better starting place.
Bill G
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Way off the scale for my world within PCP's !!
Would you expand scaling on @ 12 ft fbs to 50 or so ?
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Wow, awesome. Thanks for posting that. Been thinking about lightening the hammer on my Noblesse. According to this formula it is almost 2X heavier than necessary.
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I would have thought you would be able to read the top graph would be OK.... I get about 17 g. for the long travel, 34 g. for medium travel, and 52 g. for short travel.... The corresponding travel distances, calculated from (sqr.rt FPE) which is 3.46 would be 3.46 /5 = 0.69", /10 = 0.35", and /15 = 0.23".... Any combination where hammer weight (grams) x hammer travel (inches) = 12 (FPE) should be about right....
Bob
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I would have thought you would be able to read the top graph would be OK.... I get about 17 g. for the long travel, 34 g. for medium travel, and 52 g. for short travel.... The corresponding travel distances, calculated from (sqr.rt FPE) which is 3.46 would be 3.46 /5 = 0.69", /10 = 0.35", and /15 = 0.23".... Any combination where hammer weight (grams) x hammer travel (inches) = 12 (FPE) should be about right....
Bob
Thank you Bob ... I'm a tad slow today :P
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Hey Bob,
More amazing stuff from the Professor 8) (aka desk clerk).
What does your wife think of this stuff?
Kirk
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She thinks I'm an addict.... what else is new?.... *LOL*....
Bob
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;D Bob , I have come to the conclusion that you do this research because the snow is way to deep for you to do much else. On the other hand maybe you do it so I have something to scratch my head over for a couple of days that it takes to sink through my thick skull and reach the grey matter underneath 8) 8)
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NAW,a half inch of snow on the ground doesn't slow me down.... but the minus 15*C this morning kept me inside.... I think by tomorrow morning it will be -18*C (0*F) here.... winter has definitely arrived....
Bob
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;D ;D Like I said way to much time to think and not enough trigger time ;)
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Never enough trigger time.... >:(
bob
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Hi Bob,
Thanks for the data and plots. I would like to find some way to better calculate hammer weight, spring K, and spring travel for a given valve.
I have some questions regarding your data and conclusions:
You gave some empirical examples of travel and hammer weight but I had a few questions for each of your setups:
1) at what K spring?
2) what preload (if any) do they have?
3) What are the valve seat diameters for these examples?
4) What is the seat thru hole diameter and stem diameters of the valves?
The energy from a spring is calculated by 0.5*k*x^2 so if you keep the end force constant (your personal given 20lbs max cocking effort) then using a lower k spring and a longer stroke you get more energy because of the square on "x".
For 20lb/in spring (k) and a 1" stroke
PE = 0.5*20*1^2 = 10 (in-lbs energy)
For 10lb/in spring (k) and a 2" stroke
PE = 0.5*10*2^2 = 20 (in-lbs energy)
For some reason my brain is in a knot about how preload effects this. One knows intuitively that for a given travel additional preload adds energy but I don't think its the same calculation as above. I'm sure there is calculus in there somewhere..... :o
Its so hard to figure out mathematically what really happens at the moment of impact. The compression of the material and bending modes in the valve stem all come into play just before the instant that the poppet releases from its seat. I've been using momentum calculations to try and see what weight hammer and what spring rate/travel I need but I've had to plug in "guesses" of what the time is between the hammer hitting and when the valve cracks. I've chosen 2ms but this is just pulled out of thin air.
Anyhow, its late so please read any formulas with a disclaimer :) Thanks for all of your help, Bob, your projects have inspired many of us! I'm sure as a collective we will crack this theoretical nut someday!
Ken
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Your formula is correct for the Kinetic energy released by the spring if at zero preload.... The 0.5 factor is what gives you that.... The Potential energy cocked is k*x^2, the Potential energy uncocked (with zero preload) is zero.... The average, therefore, is 0.5(k*x^2)....
I just look at the force.... The cocked force is k*x, and the uncocked force, for zero preload is k*0, so the AVERAGE force is half the cocked force = k*x/2, and the energy released is the Average force times the stroke distance x, which = k*x*x/2.... If the preload (uncocked force) is 50% of the final (cocked) force, then the average force is (0.5+1)/2 = 0.75 of the cocked force.... An example might help.... Let's use your 10 lb/in. spring with 2" of travel, but with 10 lbs. of preload....
Uncocked force = 10 lbs.... Cocked force = 30 lbs.... Average force = 20 lbs..... It is that force that is accelerating the hammer over the 2" stroke.... 40 in.lb. energy....
With no preload.... Uncocked force = 0 lbs.... Cocked force = 20 lbs.... Average force = 10 lbs.... It is that force that is accelerating the hammer over the 2" stroke.... 20 in.lb. energy....
My graphs are based mostly on empirical results (ie what works).... You are correct that the valve seat area and pressure are important (for opening), also the stem area and pressure (for closing), and the valve spring, and the drag on the head of the poppet.... However, all those things also produce FPE, so ultimately the FPE the gun produces governs how much hammer strike you need.... It may be over-simplified, but it works.... and that's why it's a "rough rule of thumb".... It should give you the ability to get a hammer to work without having more than about 20 lbs. of cocking force.... HTHs....
If you really want to calculate the energy required to lift the poppet from the seat (crack it), I'll give you a hint.... The force F (pressure times area = F in lbs.) holding the valve closed compresses the seat material a few thou, distance D (in ft.).... The energy it takes to decompress it is 1/2(FD) in ft.lb.... That energy is removed from the hammer before the valve starts to lift, decreasing the velocity of the hammer.... It is the RESIDUAL energy and momentum of the hammer (plus poppet), acted on by the closing forces, that determine the lift (a function of energy) and dwell (a function of momentum)....
Bob
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If the Valve only moves a few thou the energy to crack the valve would be tiny in comparison to what typical springs provide. does this mean that the extra energy is in fully opening and using momentum to resist closing? Interesting.
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Yes, by far the majority of the energy goes into creating lift, while the momentum creates the dwell.... Here is a thread you may find helpful....
http://www.gatewaytoairguns.org/GTA/index.php?topic=48892.0 (http://www.gatewaytoairguns.org/GTA/index.php?topic=48892.0)
Bob