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How to Measure Force Exerted?

post #1 of 9
Thread Starter 

Reading all these reviews, I notice a distinct difference in impressions on how a ski feels in terms of stiffness when it's a lighter vs heavier skier.  And that got me thinking.......how to measure the force exerted on a ski relative to body weight?  That would help explain why such differences in opinions of skis and how stiff they are, if it's a noodle, etc.

 

I understand F=mA, but how do we determine A?  If Skier X is going at Y speed, that doesn't provide for Acceleration since that's defined as change over time.  So then it's only in the turns apparently when "actual" force is exerted and when the stiffness is truly noticed (not on the flats obviously, since flex/stiffness is felt in response to force applied, primarily in the turns).  So, how is that determined to solve the equation??

 

My curiosity is this: Say I weigh 80kg and am on Ski Z but someone else is 100kg on Ski Z, going at the same speed, with the same exact turns, what is the difference in force being against on the ski?  The REASON for this is it should give some indication of the difference being felt in skis that are stiff for some but not others.  I had a pair of skis that felt good when I was 30lbs heavier.  But since I've lost the weight they feel like planks and I get seriously bounced around at speed.  I would love to know the actual difference in force being applied on that ski since I'm 30lbs lighter now.  And this would also then give some indication why a ski gets excellent reviews for a 170lb guy but called a noodle for a 205lb guy (for example).  I truly feel this is the missing link in ski reviews, as a heavier person simply cannot relate to a review written by a lightweight, especially considering the forces being applied are magnified at higher speeds.  So that 30lbs difference would seem even MORE significant at, say 50km/h carving down a run.  See my point?

 

Maybe this is all foolishness, but I would appreciate any input/help.  I'm not a mathematician; I'm a concept guy.  I have the crazy thought, but not the knowledge how to solve the question LOL.

post #2 of 9

What type of answer are you looking for here?  As you mentioned, the mass and speed of a skier come into play here.  Speed isn't constant throughout a ski turn.  You would also have to account for the ever-changing-angles of the ski edges into the snow.

 

Force is measured in newtons, which isn't a unit most people have any good intuitive understanding of.

 

Maybe I'm too much of a math geek, but to me the "answer" to this is in one hell of a complicated integral.  Sadly, I used to think figuring that stuff out was interesting.  Thankfully, I don't remember enough to do it anymore.

post #3 of 9

If I understand you right, you want to compare two skiers that are skiing the same path but have different weights.  In that case acceleration is constant.  There fore will be directly proportional to the mass of the skier.  100/80 = 1.25.  Therefore, the force will be 25% greater for the 2nd skier.  Real world data might show some errors in the assumptions but in a simple hypothetical scenario, the math is simple and the hypothetical scenario should give you a general idea of what to expect.

post #4 of 9
Thread Starter 

Know when people are told to wear seatbeats because in an accident the "stopping force" is the same as X times your body weight?  Or fighter pilots are told they're pulling X G's, which is equivalent to Y times their body weight?  Or something like that.

 

I was hoping to "simplify" the same thing for skiing and weight.  Like......"If you weigh 30lbs less, at 50km/h it's like 150lbs difference!"  or something like that.  A vast oversimplification, I understand.  But my point is that it's hard to gauge what effect body weight has on a ski.  I was hoping that might help put things into perspective.  So when someone who weighs 170lbs says a ski is great but a 200lbs person says it's a noodle, a layperson could identify with something simple by saying, "Well that 30lbs feels more like 150lbs when moving at 50km/h, so that's a huge difference in how stiff a ski feels".


Maybe it's a complex way to address a simple question, but I was hoping there was some "simple-ish" way to quantify such an issue........??

post #5 of 9

athanvg has given you the simple answer: all other things being equal the force depends directly on the weight.  The force needed to turn you if you weigh twice as much because you have twice as much mass is twice as much force.

 

But there seems to be more on your mind, and you just didn't ask the right question.

 

If you lost weight and want to feel a relationship between the ski bending into a curve effortlessly as you push on it the way you used to, you will need to go around that same curve faster, or you will need to go around a tighter curve at the same speed.

 

The relationship, which will hold as long as you have enough weight to provide enough traction for the ski not to slip out on you, is

F=M x A

= M x (V x V / R )

where M is mass, A is acceleration, V is velocity, R is turn radius and F is the force.

 

If you weigh 1/4 as much you will need to ski 2 times as fast or ski a turn with halve the radius to get the same force exerted through your legs.  Skiing faster is the simplest comparison, and is straightforward.  (bold corrections edited in)

 

Changing the radius is more complicated because it's harder to bend the ski into a tighter turn.  You may reach a point where the force needed to bend the ski is higher than the force needed to accelerate you around that size of a turn.  If you try to push hard enough to, say get your old SGs into a 16 m turn at 10 mph, you will just push yourself right into the inside of the turn and fall down without even bending the ski much.  If you had more mass, the amount of push needed to throw you off balance into the inside of any given turn at any given speed is greater, and therefore, not having enough mass/force to bend the ski for that turn at that speed while in balance is less of a problem.


Edited by Ghost - 4/10/13 at 6:04pm
post #6 of 9

Somebody actually understands simple physics. Well said; bravo, Ghost.

 

Quote:
Originally Posted by Ghost View Post

 

If you weigh half as much you will need to ski 4 times as fast or ski a turn with halve the radius to get the same force exerted through your legs.  Skiing faster is the simplest comparison, and is straightforward.  

 

 

Something fishy about that 4 though.  

post #7 of 9
Quote:
Originally Posted by Spooky View Post

Somebody actually understands simple physics. Well said; bravo, Ghost.

 

Quote:
Originally Posted by Ghost View Post

 

If you weigh half as much you will need to ski 4 times as fast or ski a turn with halve the radius to get the same force exerted through your legs.  Skiing faster is the simplest comparison, and is straightforward.  

 

 

Something fishy about that 4 though.  


Oops, typing too fast, got it bassackwards.  F = m V^2/R;  Twice as fast is the same as 4 times the mass.  Only need to ski at square root of two times as fast to equal twice the mass.

post #8 of 9
Quote:
Originally Posted by Ghost View Post

Quote:
Originally Posted by Spooky View Post

Somebody actually understands simple physics. Well said; bravo, Ghost.

 

Quote:
Originally Posted by Ghost View Post

 

If you weigh half as much you will need to ski 4 times as fast or ski a turn with halve the radius to get the same force exerted through your legs.  Skiing faster is the simplest comparison, and is straightforward.  

 

 

Something fishy about that 4 though.  


Oops, typing too fast, got it bassackwards.  F = m V^2/R;  Twice as fast is the same as 4 times the mass.  Only need to ski at square root of two times as fast to equal twice the mass.

That would be my typical blown final exam question.

post #9 of 9
Quote:
Originally Posted by Ghost View Post

F = m V^2/R; Twice as fast is the same as 4 times the mass. Only need to ski at square root of two times as fast to equal twice the mass.

 

 

I forgot to carry the one

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