Ahh, mechanical advantage of levers...

Ahh, mechanical advantage of levers...

Joined: April 1st, 2009, 3:18 am

February 19th, 2011, 6:41 pm #1

...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."

Last edited by Duane30 on February 19th, 2011, 7:05 pm, edited 1 time in total.
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Joined: November 28th, 2002, 6:26 pm

February 19th, 2011, 8:06 pm #2

...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
Last edited by pneuguy on February 19th, 2011, 8:07 pm, edited 1 time in total.
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Joined: April 1st, 2009, 3:18 am

February 19th, 2011, 8:17 pm #3

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."

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Joined: April 1st, 2009, 3:18 am

February 19th, 2011, 9:45 pm #4

...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."

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Joined: November 28th, 2002, 6:26 pm

February 19th, 2011, 11:27 pm #5

...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
Last edited by pneuguy on February 19th, 2011, 11:32 pm, edited 1 time in total.
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Joined: June 5th, 2006, 12:49 am

February 20th, 2011, 12:44 am #6

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."

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.
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Joined: September 25th, 2006, 2:19 pm

February 20th, 2011, 12:46 am #7

...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."

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.
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Joined: April 1st, 2009, 3:18 am

February 20th, 2011, 12:50 am #8

...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."

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Joined: April 1st, 2009, 3:18 am

February 20th, 2011, 12:59 am #9

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."

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Joined: September 25th, 2006, 2:19 pm

February 20th, 2011, 1:22 am #10

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.
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.
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