...what joy. First, someone correct me on this if I am wrong. The mechanical advantage of the Benji pump lever using the futherst point rearward on the lever alone with no wood (effort arm) should be 2:1, right? And the ratio should get better as the stroke progesses inward, right?

It is 8" from rearmost point on the arm to the link/rivet pivot (fulcrum) and 4" from there to the lever pivot in the front, center-to-center on holes. I used the traditional first class lever math...

Something is missing here I feel, as air rifle levers are toggle types. Are they considered second class levers?

Hit me up!

"Well, I thought it was a rabbit but it turned out to be Bear Grylls in a rabbit hide."

...on the full distance between the points of load/effort and the fulcrum.

But that's really beside the point when the subject is a simple-in-appearance but complex-in-behavior machine like the Benji pump linkage, because the mechanical advantage changes throughout the operating cycle as the various angles between arm and lever and lever and piston change.

For example, as the pump arm and link become parallel to each other near the end of the stroke, the (theoretical) advantage goes to infinity.

Steve

I put 2, didn't I? My fault. I meant 3. So for air rifle levers we use the total overall lever distance divided by where the link arm pivot pint is at, correct? Hmmm, on to something here, bare with me....

So, if my lever is 23.5" overall (rearmost to front pivot hole) and the link hole is 4.25 back from the front hole, my MA is 5.52:1. Credit, as you stated, the MA is infinate towards the end of the stroke.

"Well, I thought it was a rabbit but it turned out to be Bear Grylls in a rabbit hide."

...have you figured already at some point what the actual effort/resistance is when a Benji piston is compressed through the pump cycle, i.e. 1st pump, 2nd, 3rd, so on? Not the user effort/resistance, but the, I suppose, weight required to compress the piston itself.

I know you have. What is the formula to determine actual effort (again, minus lever) to compress a piston of X length, of X tube diamter, and of X volume size?

Would I use the formula of compession ratio spred over the piston surface area squared each pump stroke? For example: stroke is 15 cu-in, valve volume is .5 cu-in, tube diameter is 1.5". So, the ratio is 30:1, which is 441psi (this is one stroke). Now take 441psi and divide in to the surface diameter of the tube diameter (which is 1.76 square inches), so 441/1.76 = 250.56 psi for stroke one????

"Well, I thought it was a rabbit but it turned out to be Bear Grylls in a rabbit hide."

...hypothetical piston would be 441psi x 1.77si = 779lbs for stroke #1. 1559lbs for #2, etc.

The actual numbers for a Benji pump are about 3ci for the pump, and 0.3ci for the valve, for a 10:1 ratio, with a piston area of about 0.47si. Figure 68lbs of piston force for stroke #1, 136lbs for #2, and so forth.

An interesting sidelight is that the checkvalve in a stock Benji has a cracking pressure of about 200psi, roughly doubling the pumping effort of the first couple of strokes simply because Crosman is too cheap to use separate springs to close the check and exhaust valves.

Steve

you should consider the valve volume plus tube volume(sweep) divided by valve volume to determine compression ratio. NO? Then your ratio would be more at 31:1.

...getting at is MA of lever systems. The most effective/efficient system without going with a compund link system. If I have .5 cu-in of valve volume and 15 cu-in of sweep, ratio is 30:1, or 441psi with one sweep.

"Well, I thought it was a rabbit but it turned out to be Bear Grylls in a rabbit hide."

...that initial 1atm off the top. So the valve pressure after one stroke 31 atm absolute, but only 30atm gauge = 441psig.

Steve

The mechanical advantage of a the pump arm itself can be changed by changing where you
apply the force. A 1:1 ratio would be if you applied pumping force right at the 2nd pivot.
Simply a longer wooden handle can improve your MA.

But the mechanical advantage of the remaining lever system is fixed. This is where the real
MA comes in - as Steve says, approaching infinity at the end of the stroke (as piston
progression approaches zero).

It's quite easy to work out the mechanical advantage at any stage of piston progression.
The pivot points form a triangle with the base along the piston's axis. The force acting
back on the piston acts along this axis. With some basic trigonometry you can make up
a spreadsheet of MA versus piston progression (or pump arm angle). You'll see how the really high MA
only kicks in right near the end - and if you plot compression ratio with piston progression
you get a similar looking curve at the end, so these nicely coincide to give what feels
like a fairly constant force required.

...stating a fixed point for reference more than anything else. You're right, MA can depend on where a person places the hand.

Here is my lever that is getting a waterjet cut job. It is 3/8" thick and 24 inches long, 4130 chrome-moly:



"Well, I thought it was a rabbit but it turned out to be Bear Grylls in a rabbit hide."

will be plenty strong for sure. What about pivot bearings? I'm thinking of trying some small
roller bearings that I have for hammer pivots.

...but they were only rated for 270KPG load. I suspect much more than that will be on the front during peak compression stroke. Dunno, thought just of using A2 steel steel press fit bushing as the slight benefit of bearing addition propbably would not be be noticed.

"Well, I thought it was a rabbit but it turned out to be Bear Grylls in a rabbit hide."

nt

Often times the high MA at the very end of the stroke is more than we need. So will utilizing a long link between the pump handle and the piston move the high MA more into the stroke arc for ease of pumping. At the cost of increasing the pump tube length, of course.

But then the upshot to increasing the tube length also allows for a longer handle to fit the scheme.

you'll see there's another wasted bit of MA, the part at the beginning where it drops down
from infinity (providing the pump arm opens 180 degrees). Some kind of shifting pivot linkage
could use this initially high MA to great advantage to compress a lot of easily compressible
air in the beginning, then shift to a smaller pivot spacing at the end.

I know what you're saying about the wasted MA at the end too. Since the MA is a function of
the pump handle's rotational position, to bring that MA on earlier you'd need a longer stroke
set-up, using a smaller pump arm angle, to get the same swept volume. But the catch is,
you lose MA by spreading the pivots further and having a longer 2nd lever.

The pump arm is a 2nd class lever - the reaction force from the piston/linkage acts between
the fulcrum (tube pivot) and the applied force (pump handle).

Class 1: The fulcrum is located between the applied force and the load, for example, a crowbar or a pair of scissors or a seesaw.

Class 2: The load is situated between the fulcrum and the force, for example, a wheelbarrow or a nutcracker.

Class 3: The force is applied between the fulcrum and the load, for example, a pair of tweezers or the human mandible.

Steve

...will be okay with 1/8" thick steel pump connetor on the lever? The flat piece of steel that travels from lever to piston. What I'm doing is slotting the lever like what the Crosman guns has at 1/8" slot so the steel can fit between and fuction (the Mac1 billet lever for example). This leaves me 1/8" on either side of steel to fasten/rivot/Chicago-screw (like the FX Indy) to link to. Will that be enough meat?

Also, what should the piece of flat steel be made out of from lever to piston? Tim stated he uses only mild steel for his levers. During our conversation I spaced asking him the type of metal that piece is. It certainly seems much, much harder and stronger than mild steel, as all levers inspected do.

Slotting the lever is the slim-line way I'd like it to be done, but could go double by placing the steel strips on the outside of lever, but this makes for a wider pump slot, hich is what I'm trying to prevent. Want strong for sure! This is a side stroker gun.



"Well, I thought it was a rabbit but it turned out to be Bear Grylls in a rabbit hide."

...is pure compression - no bending, no shearing.

But it's pretty meaningless to discuss materials and dimensions until you have at least a rough idea of the magnitude of the forces (e.g., psi x piston_area) that will be involved.

Steve

Okay, say the piston is 1.5" diameter = 1.76" piston surface area. Say the valve will be at 1,500 psi on three strokes. Now, the formula?, 1.76" piston area (1500)= 2,640 Pounds(?).

You're a great help man! I have a new name for ya': "Steve_in_NC the Physics Emcee" Got the idea from "Bill Nye the Science Guy"

Figure 1/8" thick hardened 4130 should do the trick.

"Well, I thought it was a rabbit but it turned out to be Bear Grylls in a rabbit hide."

http://www.matweb.com/search/DataSheet.aspx?MatGUID=284aa856470341acbf3c16bc90849ea1&ckck=1

Therefore for a 1/8" thick strip to support 2640lbs with a 3:1 safety factor, the pivot pin hole will need to be at least 2640 x 3 x 8 / 161000 = 0.39" in diameter. The total width of the strip should probably be at least 3/4".

Steve

The tube sent to me is 1.5" I.D. and the wall is .125, overall diameter 1.75". You see, I had ordered 1.495" I.D. with a wall of .065. I was sent the wrong tube! Go figure. Even tripple checked my order slip, it states what was ordered...

Anyway, do you think .065 wall will be alright if re-order it? All else, sure. My concern is the 4130 tube with a wall of .065 strong enough to retain the valve and front plug under compression stroke loads?

The valve and the front plug will be secured with screws radially, of course. .065 on Crosman tubes is fine, but not sure about an air rifle with 1.5" pump bore.

"Well, I thought it was a rabbit but it turned out to be Bear Grylls in a rabbit hide."

...e.g., no fewer than three #10 high-strength pan socket head screws with the heads - not threads - in shear, then a 0.065" 4130 wall thickness should be adequate for 1500psi.

http://www.network54.com/Forum/275684/message/1245904688/A+little+follow-up+to+Fireball%27s+understatement-+%26quot%3BWalter+builds+them+a+little+stronger-%26quot%3B

Steve

- the primary mode of failure I'd suspect. I've had a Crosman stamped lever fail because
it began bending sideways and unevenly distributing the load. In a perfect world there'd
be no lateral force but the human element introduces it.

[IMG][/IMG]

...of a 6 x .75 x 1/8" steel strip would be about 4000lbs = only about half its tensile strength assuming 0.39" (i.e., 10mm) pivot pins.

So designing the lever to provide lateral support to the link near the end of the stroke would be a very good idea.

Steve