18" x 2 x pi / 3 = 37" = ~3', that's an average force of only 14.6lbs.

18" x 2 x pi / 3 = 37" = ~3', that's an average force of only 14.6lbs.

Joined: March 1st, 2002, 12:22 am

August 6th, 2012, 6:45 pm #1


(Trying not to hijack James post about his awesome gun.....)

Steve, now that yer back on earth, I ran that same equation thru my calcualtor and got 38.04. Which I supoose is close enough that it probably doesnt matter. But,  as I started breakng it down trying to figure it out, it occured to me maybe I didnt follow the "order of operations" when running thru the equation. But IIRC from my programming class days, Multipication and division carry the same "wieght" and in this case just work it left to right. The other thing I think my have skewed my answer is not using enough places to the right of the decimal point (I used 3.17.... cant mremember more than that...)

I did figure out a couple things on my own: the "/3" referes to the arm moving thru a 120 degree arc, and 120 is one third of 360 degrees of the circle. Still dont understand what the 2 between 18 and pi is, or how you got from 36 inches to 14.6 pounds of effort. the other thing I may not have accounted for is the inches aprt of that... does that have an influence on the arc of travel the pump arm moves thru to the energy needed?

(Edit: Wikipedia says pi is " 3.14159". Which brigns my 38.04 down to 37.69908, and thats much closer to your answer.)


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Joined: July 4th, 2012, 7:51 pm

August 6th, 2012, 7:25 pm #2


When Steve responded to my post he explained that the pumping effort wasn't uniform throughout the stroke (something I should have known but had a brain burp).

So, you start with the compression energy of 46ft*lb and set the pump effort the same.

We can calculate the distance travelled of the pump (using extreme end). It travels in an arc.

the formula for the distance = 2piR for a complete circle. Thats where the "2" comes from. You got the 120deg out of 360deg factor right.

Running the numbers we get about 3'. Then we take the 46 ft*lb vave energy and divide by 3 => 14.6 pounds.



WyoMan
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Joined: July 29th, 2012, 9:21 pm

August 6th, 2012, 7:28 pm #3

(Trying not to hijack James post about his awesome gun.....)

Steve, now that yer back on earth, I ran that same equation thru my calcualtor and got 38.04. Which I supoose is close enough that it probably doesnt matter. But,  as I started breakng it down trying to figure it out, it occured to me maybe I didnt follow the "order of operations" when running thru the equation. But IIRC from my programming class days, Multipication and division carry the same "wieght" and in this case just work it left to right. The other thing I think my have skewed my answer is not using enough places to the right of the decimal point (I used 3.17.... cant mremember more than that...)

I did figure out a couple things on my own: the "/3" referes to the arm moving thru a 120 degree arc, and 120 is one third of 360 degrees of the circle. Still dont understand what the 2 between 18 and pi is, or how you got from 36 inches to 14.6 pounds of effort. the other thing I may not have accounted for is the inches aprt of that... does that have an influence on the arc of travel the pump arm moves thru to the energy needed?

(Edit: Wikipedia says pi is " 3.14159". Which brigns my 38.04 down to 37.69908, and thats much closer to your answer.)


dr_subsonic's pneumatic research lab

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*Edit: Damn. Ninja'd.


The factor of "2" is there because he's calculating the circumference of the circle the lever would trace if it swung the full 360°.

Circumference is 2 x pi x radius, so his calculation is giving 1/3 of that circumference in order to get the distance your hand would move for a 120° lever stroke.

Energy equals force x distance, so he's getting his 14.6 AVERAGE force by dividing the 46 foot-lbs of energy needed for the compression example he gave by the 37.7 inches (3.14 feet) the end of the lever moves throughout the stroke.

(46 ft-lb)/(3.14 ft) = 14.6 pounds of force.
Last edited by Hudson12tum on August 6th, 2012, 7:31 pm, edited 1 time in total.
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Joined: March 1st, 2002, 12:22 am

August 6th, 2012, 8:11 pm #4

(Trying not to hijack James post about his awesome gun.....)

Steve, now that yer back on earth, I ran that same equation thru my calcualtor and got 38.04. Which I supoose is close enough that it probably doesnt matter. But,  as I started breakng it down trying to figure it out, it occured to me maybe I didnt follow the "order of operations" when running thru the equation. But IIRC from my programming class days, Multipication and division carry the same "wieght" and in this case just work it left to right. The other thing I think my have skewed my answer is not using enough places to the right of the decimal point (I used 3.17.... cant mremember more than that...)

I did figure out a couple things on my own: the "/3" referes to the arm moving thru a 120 degree arc, and 120 is one third of 360 degrees of the circle. Still dont understand what the 2 between 18 and pi is, or how you got from 36 inches to 14.6 pounds of effort. the other thing I may not have accounted for is the inches aprt of that... does that have an influence on the arc of travel the pump arm moves thru to the energy needed?

(Edit: Wikipedia says pi is " 3.14159". Which brigns my 38.04 down to 37.69908, and thats much closer to your answer.)


dr_subsonic's pneumatic research lab

the Lunatic Fringe of American Airgunning
Southwest Montana's headquarters for Airgunning Supremacy
Proud Sponsor of team_subsonic
I see where the 2 falls in now. Thanx Guys! I think its syntax....piRsqaured is 2*pi*R. Not used to seeing it described that way.

But where did you get the 46 lbs? Is that from James' direct measurement in/at the valve?


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Joined: March 1st, 2002, 12:22 am

August 6th, 2012, 8:26 pm #5

(Trying not to hijack James post about his awesome gun.....)

Steve, now that yer back on earth, I ran that same equation thru my calcualtor and got 38.04. Which I supoose is close enough that it probably doesnt matter. But,  as I started breakng it down trying to figure it out, it occured to me maybe I didnt follow the "order of operations" when running thru the equation. But IIRC from my programming class days, Multipication and division carry the same "wieght" and in this case just work it left to right. The other thing I think my have skewed my answer is not using enough places to the right of the decimal point (I used 3.17.... cant mremember more than that...)

I did figure out a couple things on my own: the "/3" referes to the arm moving thru a 120 degree arc, and 120 is one third of 360 degrees of the circle. Still dont understand what the 2 between 18 and pi is, or how you got from 36 inches to 14.6 pounds of effort. the other thing I may not have accounted for is the inches aprt of that... does that have an influence on the arc of travel the pump arm moves thru to the energy needed?

(Edit: Wikipedia says pi is " 3.14159". Which brigns my 38.04 down to 37.69908, and thats much closer to your answer.)


dr_subsonic's pneumatic research lab

the Lunatic Fringe of American Airgunning
Southwest Montana's headquarters for Airgunning Supremacy
Proud Sponsor of team_subsonic
nt

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Joined: July 4th, 2012, 7:51 pm

August 6th, 2012, 8:47 pm #6

I see where the 2 falls in now. Thanx Guys! I think its syntax....piRsqaured is 2*pi*R. Not used to seeing it described that way.

But where did you get the 46 lbs? Is that from James' direct measurement in/at the valve?


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WyoMan
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Joined: July 29th, 2012, 9:21 pm

August 6th, 2012, 8:47 pm #7

I see where the 2 falls in now. Thanx Guys! I think its syntax....piRsqaured is 2*pi*R. Not used to seeing it described that way.

But where did you get the 46 lbs? Is that from James' direct measurement in/at the valve?


dr_subsonic's pneumatic research lab

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not pounds of force. Steve gave 46 foot-pounds as an example of the energy needed for compressing 7.1 cubic inches of swept volume to 3000 psi.

The same amount of energy as raising a 1 pound weight 46 feet, a 46 pound weight 1 foot, or accelerating a 14.3 grain pellet to 1200 fps.
Last edited by Hudson12tum on August 6th, 2012, 8:50 pm, edited 1 time in total.
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Joined: July 29th, 2012, 9:21 pm

August 6th, 2012, 8:52 pm #8


WyoMan
Nah, just integral calculus. nt
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Joined: July 29th, 2012, 9:21 pm

August 6th, 2012, 8:54 pm #9


WyoMan
Whoops, duplicate post. nt.
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Joined: November 28th, 2002, 6:26 pm

August 6th, 2012, 10:39 pm #10

(Trying not to hijack James post about his awesome gun.....)

Steve, now that yer back on earth, I ran that same equation thru my calcualtor and got 38.04. Which I supoose is close enough that it probably doesnt matter. But,  as I started breakng it down trying to figure it out, it occured to me maybe I didnt follow the "order of operations" when running thru the equation. But IIRC from my programming class days, Multipication and division carry the same "wieght" and in this case just work it left to right. The other thing I think my have skewed my answer is not using enough places to the right of the decimal point (I used 3.17.... cant mremember more than that...)

I did figure out a couple things on my own: the "/3" referes to the arm moving thru a 120 degree arc, and 120 is one third of 360 degrees of the circle. Still dont understand what the 2 between 18 and pi is, or how you got from 36 inches to 14.6 pounds of effort. the other thing I may not have accounted for is the inches aprt of that... does that have an influence on the arc of travel the pump arm moves thru to the energy needed?

(Edit: Wikipedia says pi is " 3.14159". Which brigns my 38.04 down to 37.69908, and thats much closer to your answer.)


dr_subsonic's pneumatic research lab

the Lunatic Fringe of American Airgunning
Southwest Montana's headquarters for Airgunning Supremacy
Proud Sponsor of team_subsonic
...that's French for: Geeze Louise whatta' knit-picker.

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