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WTT3 bow photos and FD curves (images on page 3)

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Re: WTT3 bow photos and FD curves (images on page 3 added May 9)

Postby badger5149 » Thu May 14, 2009 12:16 pm

Pretty sure your formulas are correct. Simply covert inch pounds to ft pounds, then compare that to final draw weigth for a percentage. Formula for kinetic energy is vXvX m/450240=KE. 98% just seems to high as the string stays with the bow and uses some energy on it's own, also if the limbs have any movement whatsoever after the arrow is released they are using some energy. If the energy in the limbs is acting more like an energy wave than flopping it could indeed be very efficient, just 98.8 sounds unreasonably high. Steve
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Re: WTT3 bow photos and FD curves (images on page 3 added May 9)

Postby jwillis » Thu May 14, 2009 1:24 pm

I'm going to play with my charts as soon as I can get back to them. Last night and tonight I'm creating the t-shirt summary to post. The spreadsheet I used is supposedly Norb Mulaney's spreadsheet, but I modified it some, adding the values on the lower left corner for ARROW KE and DYNAMIC EFFICIENCY. In fact, in the "about this file" information, I saw where it said the spread sheet was created by Norbert Mulaney. I also created the embedded graph of the FD curve and styled it to be bolder and more attractive. The formulas for SE and SE/PDF were already written into the program.

PDF=MAX(B20:B40)
The MAX command finds the largest value in the range. So in this formula it finds the the maximum value in the FORCE POUNDS column. For traditional bows this is usually the same as the last draw weight measured.

SE=(MAX(F17:F40))/12
Finds the maximum value of the SUM AREA INCH LB column and divides it by 12 to convert it to ft-lbs.

SE/PDF=S.E./MAX(B20:B40)
Divides the stored energy value calculated in the above equation in cell D43 by the maximum value in the FORCE POUNDS column.

ARROW KE=(H6*H7*H7)/450240
Multiplies the arrow speed value in cell H6 times the velocity in cell H7 times the velocity in cell H7 divided by 450240

DYNAMIC EFFICIENCY=(D45/S.E.)*100
Divides the value calculated for ARROW KE in cell D45 by the value calculated for SE in cell D43 then multiplies by 100. This represents the kinetic energy of the arrow divided by the stored energy in the bow as a percentage.
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Re: WTT3 bow photos and FD curves (images on page 3 added May 9)

Postby wantoknow » Thu May 14, 2009 1:56 pm

Jim I saw that too.
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Re: WTT3 bow photos and FD curves (images on page 3 added May 9)

Postby Kirkll » Fri May 15, 2009 11:19 pm

hey guys .... i'm no engineer, but i've always been told velosity squared times mass divided by 450240 = KE this is saying its speed times velosity squared here. something doesn't sound right with that.... what's up with that?
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Re: WTT3 bow photos and FD curves (images on page 3 added May 9)

Postby jwillis » Sat May 16, 2009 9:12 am

Kirkll wrote:hey guys .... i'm no engineer, but i've always been told velosity squared times mass divided by 450240 = KE this is saying its speed times velosity squared here. something doesn't sound right with that.... what's up with that?

Kirk, this is a very good question! I don't completely understand it. I did a little online research and came up with the following explanation that I copied from (get this) a physics forum. If you go to this forum and read it, you'll quickly see just how good the question is if even these mathematicians have a hard time explaining it. I like the explanation below because it is not so mathematical.

http://www.physicsforums.com/showthread ... 484&page=2

Scroll down to Matthew Keating's post...

"Wk= Work done in accelerating object

Work done = Force x distance

If body a is travelling v m/s after 1 second, it is intuitive that it would be travelling with an average speed of v/2 m/s, so in one second it will travel a distance of v/2 m.

Now if body B was accelerated under the same force, it should also be intuitive that it will be travelling at twice the velocity after 2 seconds (velocity 2v). Over the 2 seconds it will have an average velocity of v m/s, and it follows that it will travel a distance of 2v m in that time. So body B has travelled 4 times the distance of body A in order to achieve twice the velocity, with the same applied force, thus the work done on body B is 4x the work done on body A. Now the work done on each body equates to the kinetic energy gained by each. Assuming no other external forces.

So.... doubleing the velocity causes the Ke to increase by a factor of four. You could show the same sort of arguement for a body travelling at 3v, 4v etc and you will arrive at the result that KE is proprtional to the square of the velocity.

Sorry if this is too non-mathematical"
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Re: WTT3 bow photos and FD curves (images on page 3 added May 9)

Postby wantoknow » Sat May 16, 2009 12:04 pm

(H6*H7*h7)/450240 Looking at the chart H6 is arrow weight in grains and H7 is the arrow speed. Looks OK.
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Re: WTT3 bow photos and FD curves (images on page 3 added May 9)

Postby cupcake » Mon May 18, 2009 2:59 pm

When I first got the Excel spreadsheet from Jim I thought that the equations to sum up the stored energy were incorrect. I was wrong, they are correct. I spent some time sketching out what the S.E. summation was doing and I think it is correct. I am an engineer so I am particular about knowing what I am doing.

The arrow kinetic energy calculation might need a bit of improvement and I will share the origin of the 450240 divisor.

We have to convert the arrow weight, in grains, to pounds mass, which are called slugs in physics. There are 7000 grains in one pound, so arrow weight (gns)/7000 = arrow weight in pounds. To get the mass (in slugs) of the arrow in pounds the weight in slugs has to be divided by the acceleration of gravity (32.174fps^2). The weight of the arrow is mass times gravity acceleration so we have to get rid of the gravity part.

Taken together the arrow mass (in slugs) is mass (grains)/(7000*32.174)

7000*32.174 = 225218

The K.E. = 1/2mv^2, one-half m v squared.

So the total equation for K.E. is: mass (grains) *v^2 / (2*225218)

2*225218 = 450436 I think this is a better number based on gravitational acceleration. The number 450240 comes from using 32.16fps^2 for acceleration of gravity.

I hope I explained this clearly enough, I sort of confused myslef for a minute while writing it.

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Re: WTT3 bow photos and FD curves (images on page 3 added May 9)

Postby Kirkll » Mon May 18, 2009 8:58 pm

Cupcake, that was clear as mud.

speed, times speed, times weight, divided by 450240 = KE that's about as simple as a guy can put it. if H6 = weight, then it makes sense.

Thanks wtk
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Re: WTT3 bow photos and FD curves (images on page 3 added May 9)

Postby jwillis » Tue May 19, 2009 7:57 am

cupcake wrote:...I will share the origin of the 450240 divisor...

Thanks for sharing this. I couldn't find it anywhere online. Jim
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Re: WTT3 bow photos and FD curves (images on page 3 added May 9)

Postby Kirkll » Sun May 24, 2009 12:22 pm

Hey Badger,

How long has the divisor 450240 been an accepted standard for calculating KE? I've been using it for 20 years. :? :? :?

Do you think Cup cakes opinion that we been using the wrong number all these years has any weight?
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