Percentage vs. linear weight...

A forum to discuss anything that has to do with pole vaulting that does not fit in the other forums.

Moderators: Russ, lonpvh

User avatar
Lax PV
PV Follower
Posts: 571
Joined: Fri Jul 25, 2003 10:02 pm
Expertise: Former HS and college vaulter, college and HS level coaching, CSCS certified
Lifetime Best: 475
Favorite Vaulter: Tarasov
Location: The Woodlands, TX
Contact:

Unread postby Lax PV » Wed Aug 22, 2007 1:26 pm

master wrote:
Lax PV wrote:its partially due to the idea that force is a parabolic curve... as force is continually added, the velocity continues to rise, therefore the position of the object will exponetially speed up.

huh????? :confused:

- master


poor wording... took a stab in the dark to explain it an did a poor job. either way, as the flex numbers get smaller and smaller, the amount of additional force it takes to bend the same gap in flex becomes greater.

User avatar
powerplant42
PV Rock Star
Posts: 2571
Joined: Sun Jul 01, 2007 10:58 am
Location: Italy

Unread postby powerplant42 » Wed Aug 22, 2007 3:15 pm

But that's not really what we're getting at here. Or is it? Would the flex numbers be a useful tool to figure this out?
"I run and jump, and then it's arrrrrgh!" -Bubka

User avatar
vault3rb0y
PV Rock Star
Posts: 2458
Joined: Wed Apr 26, 2006 6:59 pm
Expertise: College Coach, Former College Vaulter
Lifetime Best: 5.14m
Location: Still Searching
Contact:

Unread postby vault3rb0y » Wed Aug 22, 2007 11:32 pm

I think you need flex numbers to figure this out, and i think if i understand right, you are saying that if it takes 5 Ibs lets say to decrease the flex number 1cm, it will take MORE THAN 5 Ibs extra, at least, to go another 1 cm. You end up adding lets say 5 Ibs more, then have to add 7 Ibs more, then 11 Ibs more, etc. I dont think "velocity" of the weight added is the correct term, but anyway, the Ibs of weight needed to move from a flex number to 1cm lower increases at an exponential rate as you push farther and farther, until lets say you are at 2cm flex, and to move to that 1cm you are using 300 MORE pounds. Im taking physics and statistics this year, and already have background in high school calculus, I'll see if farther into this year a high school education will get me far enough to maybe express it better :P. But im almost 100% sure what you are saying, or what i think you are saying, can be expressed in a formula, ill take a shot. This is just for fun by the way...

Lets say we are solving for the amount of force in pounds it takes to move down 1cm flex from a certain point on a pole.

lets say for example we are using a 15' 19.0cm flex. We know for sure that with this pole, applying 35 Ibs (or is it 30, maybe 45? at any rate, the standard weight of a 15'...) of pressure will bend the pole 19.0cm. Since i jump on a line of poles similar to this, lets say that to move 1cm down from starting position requires 5Ibs more weight, and the amount of weight needed to move another 1cm downward increases by 2X as X=the weight needed to add for the previous cm. So 5 Ibs makes the pole an 18cm and 10 makes it a 17cm. Then 20 makes a 16cm., 40 makes a 15cm. Im sure theres a true formula for it, but im too stupid and lazy to think of it right now :p. I hope this is the direction we are going for....
The greater the challenge, the more glorious the triumph

User avatar
rainbowgirl28
I'm in Charge
Posts: 30435
Joined: Sat Aug 31, 2002 1:59 pm
Expertise: Former College Vaulter, I coach and officiate as life allows
Lifetime Best: 11'6"
Gender: Female
World Record Holder?: Renaud Lavillenie
Favorite Vaulter: Casey Carrigan
Location: A Temperate Island
Contact:

Unread postby rainbowgirl28 » Thu Aug 23, 2007 12:59 am

vault3rb0y wrote:lets say for example we are using a 15' 19.0cm flex. We know for sure that with this pole, applying 35 Ibs (or is it 30, maybe 45? at any rate, the standard weight of a 15'...) of pressure will bend the pole 19.0cm.


It's 50lbs

User avatar
rainbowgirl28
I'm in Charge
Posts: 30435
Joined: Sat Aug 31, 2002 1:59 pm
Expertise: Former College Vaulter, I coach and officiate as life allows
Lifetime Best: 11'6"
Gender: Female
World Record Holder?: Renaud Lavillenie
Favorite Vaulter: Casey Carrigan
Location: A Temperate Island
Contact:

Unread postby rainbowgirl28 » Thu Aug 23, 2007 1:10 am

vault3rb0y wrote:lets say for example we are using a 15' 19.0cm flex. We know for sure that with this pole, applying 35 Ibs (or is it 30, maybe 45? at any rate, the standard weight of a 15'...) of pressure will bend the pole 19.0cm. Since i jump on a line of poles similar to this, lets say that to move 1cm down from starting position requires 5Ibs more weight, and the amount of weight needed to move another 1cm downward increases by 2X as X=the weight needed to add for the previous cm. So 5 Ibs makes the pole an 18cm and 10 makes it a 17cm. Then 20 makes a 16cm., 40 makes a 15cm. Im sure theres a true formula for it, but im too stupid and lazy to think of it right now :p. I hope this is the direction we are going for....


I think you are getting it mixed up. If you hang a weight from a 15' pole and get a certain amount of deflection, adding more weight will make it deflect more, resulting in a higher number.


If you are trying to explain how flex numbers match up with relative stiffness, here is an example: (this is assuming whoever reads this understands flex numbers. If not you should use the search feature on this site)

Say you had an imaginary series of 13' poles that all feel exactly 5lbs stiffer. You start with a 13' 80 and end with a 13' 220

When you go from the 13'80 to the 13'85, there might be a 1.5 gap in flex numbers. (I don't know what the exact #s would actually be)

When you go from the 13'150 to the 13'155 there might be a 1.0 gap in flex numbers.

When you go from the 13'215 to the 13'220, there might be a 0.6 gap in the poles.


In general, for most poles that high school athletes would be on, a 1.0 gap in poles is going to feel like 5lbs.

User avatar
powerplant42
PV Rock Star
Posts: 2571
Joined: Sun Jul 01, 2007 10:58 am
Location: Italy

Unread postby powerplant42 » Thu Aug 23, 2007 9:43 am

I just figured out why it is in fact percentage. Vault3rboy has the right idea. A pole will be more difficult to bend as it becomes more and more bent. Right? Well, if you have a 13' pole at a low weight rating or high flex, then at a certain point in it's being bent it will become in essence a 13' pole of a higher weight rating/lower flex. This is slightly oversimplified i think. But you are right, there is a formula for it. e=mc2. That covers this situation, in the fact that the faster something goes, the more energy is needed to keep acceleration going. Subsitute this with bending a pole.
"I run and jump, and then it's arrrrrgh!" -Bubka

gtc
PV Whiz
Posts: 214
Joined: Fri Feb 04, 2005 2:41 pm

Unread postby gtc » Thu Aug 23, 2007 11:30 am

I think E=Mc2 Has to do with the relationship of the creating and destroying of Mass and Energy. You know nuclear power!

User avatar
Lax PV
PV Follower
Posts: 571
Joined: Fri Jul 25, 2003 10:02 pm
Expertise: Former HS and college vaulter, college and HS level coaching, CSCS certified
Lifetime Best: 475
Favorite Vaulter: Tarasov
Location: The Woodlands, TX
Contact:

Unread postby Lax PV » Thu Aug 23, 2007 12:28 pm

powerplant42 wrote:I just figured out why it is in fact percentage. Vault3rboy has the right idea. A pole will be more difficult to bend as it becomes more and more bent. Right? Well, if you have a 13' pole at a low weight rating or high flex, then at a certain point in it's being bent it will become in essence a 13' pole of a higher weight rating/lower flex. This is slightly oversimplified i think. But you are right, there is a formula for it. e=mc2. That covers this situation, in the fact that the faster something goes, the more energy is needed to keep acceleration going. Subsitute this with bending a pole.


Yeah... thats a bit different. Thats how the atomic bomb works. The equation that would best fit what you are decribing is similar though:

K = 1/2 mv^2

I am curious to ask, is the weight rating more based on the idea of human limitation? (i.e. its easier to increase in the 100m from 11.8 to 11.6 than it is from 9.9 to 9.7). A very basic question, (and I may be insulting my own intelligence here...) but how are the weights calculated? Are they arbitrary? Or does it model from the energy that a pole can potentially return.. or the threshold force before it breaks?

User avatar
rainbowgirl28
I'm in Charge
Posts: 30435
Joined: Sat Aug 31, 2002 1:59 pm
Expertise: Former College Vaulter, I coach and officiate as life allows
Lifetime Best: 11'6"
Gender: Female
World Record Holder?: Renaud Lavillenie
Favorite Vaulter: Casey Carrigan
Location: A Temperate Island
Contact:

Unread postby rainbowgirl28 » Thu Aug 23, 2007 1:02 pm

Lax PV wrote:but how are the weights calculated? Are they arbitrary?


Yes.

User avatar
powerplant42
PV Rock Star
Posts: 2571
Joined: Sun Jul 01, 2007 10:58 am
Location: Italy

Unread postby powerplant42 » Thu Aug 23, 2007 1:12 pm

Well, a really experienced vaulter could most likely shatter a pole at their weight rating or even slightly above.

To do with e=mc2... you need to look behind the formula, into what's inside nuclear physics. (I love this, we're talking about pole vaulting and nuclear physics all in the same thread!) Let's look at just the 'squared' part of the equation. Do any of you know what that really is getting at? A long time ago, there was a brilliant woman, I think she was French (I can't remember her name and I'm too lazy to go look it up...) who made an outrageous statement. She said that if one dropped a ball from height x, then the same ball from 2x, that the force of the ball when dropped from 2x would be the square of the force of the ball from height x. In other, more modern words, if you drove your car at 20 mph, then slammed on the brakes, you would leave a skid mark n ft long. Then, you drive your car at 40 mph and slam on the brakes again. How long will this skidmark be compared to the one at 20 mph? Not twice, but squared! (Don't go out and try this one folks! ;) )This works into e=mc2, and I'm sure there's a better formula to explain it, I just kind of threw it out there when I related the fact that more force is needed to make an object go faster to the 'squared' in e=mc2. So, let's now put this into pole vaulting: You bend your pole to 90 degrees, and that takes f amount of force. Now, to get the pole to bend to 45 degrees, you would need? Right, f squared amount of force! You see? And then you can tie this in with my other post about higher pole ratings and lower pole ratings.
I don't know. Maybe this is beside the point. I think that we just need an experiment: TO ANYONE WITH THE RIGHT EQUIPMENT: Try this. Take two poles of the same height and weight rating. Grab one of those large game scales (maybe this should go in the 'redneck pv' thread...) and set it to pull the pole with 50% it's weight. Repeat with the other pole. Now, pull with a set amount of force for each. For example, 100 pounds. Now measure the cord lengths of each pole.
WE ARE GETTING TO THE BOTTOM OF THIS! :D
"I run and jump, and then it's arrrrrgh!" -Bubka

User avatar
vault3rb0y
PV Rock Star
Posts: 2458
Joined: Wed Apr 26, 2006 6:59 pm
Expertise: College Coach, Former College Vaulter
Lifetime Best: 5.14m
Location: Still Searching
Contact:

Unread postby vault3rb0y » Thu Aug 23, 2007 9:07 pm

Phwew i definitely messed up on that equation last night, you are right becca i have it backwards!! Thats what i get for trying to act smart at 10:30 at night after the 3rd day of school :P. But im glad the general idea i was portraying got across. Pretty much, to know whether you are ready to jump from one pole to another has nothing to do with the weight rating, and not even as much to do with cm ratings, since going from 20cm to 19cm is easier than 10cm to 9cm. It sounds to me that it takes a trained mind, with a general idea of how hard it is to jump from one pole to another. The question that i find myself asking after this is, what would the cm flex of poles look like, according to athletes, who describe each bigger sized pole an equal jump from one to the other. It would only be able to come from the athletes, or by adding the same amount of energy EXTRA into each pole that you MOVE UP. flex would be a constant for lets say 10 poles, in which you try to achieve a 16cm flex. If the pole set were hypothetically "perfect" and you found an equal jump from one pole to the next, to the next, then it would take 5 Ibs more for each next pole to bring the pole to 16cm. Or 7Ibs, 10Ibs, whatever you consider your "perfect jump". Doing this would, or should, give you perfect pole jumps, but the flex numbers would start, at 16, then maybe 15.0, 14.3, 13.9, 13.6, 13.4, 13.2, 13.1, 13, etc.... proving that the weight you add makes less and less a difference to flex to more you add.
The greater the challenge, the more glorious the triumph

gtc
PV Whiz
Posts: 214
Joined: Fri Feb 04, 2005 2:41 pm

Unread postby gtc » Fri Aug 24, 2007 9:53 am

PP 42 The C in the formula stands for the speed of light. If you were going the speed of light squared and slammed on the brakes how long would your skid mark be?


Return to “Pole Vault - General”

Who is online

Users browsing this forum: No registered users and 39 guests