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# Why are men racers faster than women? And factors influencing speed - Page 3

 M= 50 kg g 9.8 m/s/s q= 35 degrees mu 0.05 ro 0.986 cd 0.9 A 0.8 m*(dV/dt) = m*g*sinQ - mu*m*g*cosQ - 0.5*ro*Cd*A*V^2 (dV/dt) = g*sinQ - mu*g*cosQ - 0.5*ro*Cd*A*V^2/m V gsinQ-mugcosQ -0.5*ro*CD*A*V^2/m Air Friction/(driving force-ski friction) a mph 1 260.9832 -0.35496 -0.001360087 5.212565 2.236936 2 260.9832 -1.41984 -0.00544035 5.191268 4.473873 3 260.9832 -3.19464 -0.012240787 5.155772 6.710809 4 260.9832 -5.67936 -0.021761398 5.106077 8.947745 5 260.9832 -8.874 -0.034002185 5.042185 11.18468 10 260.9832 -35.496 -0.13600874 4.509745 22.36936 20 260.9832 -141.984 -0.544034958 2.379985 44.73873 30 260.9832 -319.464 -1.224078657 -1.16962 67.10809 35 260.9832 -434.826 -1.66610706 -3.47686 78.29277

Quote:
Originally Posted by Ghost

 M= 80 kg g 9.8 m/s/s q= 35 degrees mu 0.05 ro 0.986 cd 0.9 A 0.8 m*(dV/dt) = m*g*sinQ - mu*m*g*cosQ - 0.5*ro*Cd*A*V^2 (dV/dt) = g*sinQ - mu*g*cosQ - 0.5*ro*Cd*A*V^2/m V gsinQ-mugcosQ -0.5*ro*CD*A*V^2/m Air Friction/(driving force-ski friction) a mph 1 417.5732 -0.35496 -0.000850055 5.215228 2.236936 2 417.5732 -1.41984 -0.003400218 5.201917 4.473873 3 417.5732 -3.19464 -0.007650492 5.179732 6.710809 4 417.5732 -5.67936 -0.013600874 5.148673 8.947745 5 417.5732 -8.874 -0.021251366 5.10874 11.18468 10 417.5732 -35.496 -0.085005462 4.775965 22.36936 20 417.5732 -141.984 -0.340021849 3.444865 44.73873 30 417.5732 -319.464 -0.76504916 1.226365 67.10809 35 417.5732 -434.826 -1.041316913 -0.21566 78.29277

Pretty much what I said earlier.  More mass is a big advatage at GS and higher speeds, not so much at SL speeds.

Edited by Ghost - 3/24/11 at 3:35pm

Ghost

Add the Snow friction and increase the frontal area for the heavier person then what is the result.

Quote:
Originally Posted by RossiGuy

I'm not a physics guru and it's been a long time since high school physics...

My 70 lb son just did the "slush cup" (pond skim) last weekend. He started at the same place as his buddy that probably weighs 90 lbs. Both pointed them straight down the hill and had great sit back technique when they hit the water - tips up just out of it.

Guess which one ran out of speed and sank at the end?

The weight advantage only applies on a slope where with gravity creates/amplifies acceleration, and on a very slick surface.  At some point skimming across the flat water extra weight becomes a huge disadvantage.  Did it that happen before the end of the pool?  Who knows.. but I bet the heavier kid was going a little faster assuming same wax and other variables close.

Did they both have the same size skis?

Snow friction is included (coefficient of friction = 0.05), but I see I've underestimated air density and overestimated projected area.

 M= 60 kg g 9.8 m/s/s q= 30 degrees mu 0.05 ro 1.5 cd 0.9 A 0.5 m*(dV/dt) = m*g*sinQ - mu*m*g*cosQ - 0.5*ro*Cd*A*V^2 (dV/dt) = g*sinQ - mu*g*cosQ - 0.5*ro*Cd*A*V^2/m V gsinQ-mugcosQ -0.5*ro*CD*A*V^2/m Air Friction/(driving force-ski friction) a mph 1 268.5389 -0.3375 -0.001256801 4.470023 2.236936 2 268.5389 -1.35 -0.005027206 4.453148 4.473873 3 268.5389 -3.0375 -0.011311212 4.425023 6.710809 4 268.5389 -5.4 -0.020108822 4.385648 8.947745 5 268.5389 -8.4375 -0.031420034 4.335023 11.18468 10 268.5389 -33.75 -0.125680138 3.913148 22.36936 20 268.5389 -135 -0.50272055 2.225648 44.73873 25 268.5389 -210.938 -0.78550086 0.960023 55.92341 30 268.5389 -303.75 -1.131121238 -0.58685 67.10809 35 268.5389 -413.438 -1.539581685 -2.41498 78.29277
 M= 90 kg g 9.8 m/s/s q= 30 degrees mu 0.05 ro 1.5 cd 0.9 A 0.5 m*(dV/dt) = m*g*sinQ - mu*m*g*cosQ - 0.5*ro*Cd*A*V^2 (dV/dt) = g*sinQ - mu*g*cosQ - 0.5*ro*Cd*A*V^2/m V gsinQ-mugcosQ -0.5*ro*CD*A*V^2/m Air Friction/(driving force-ski friction) a mph 1 402.8083 -0.3375 -0.000837868 4.471898 2.236936 2 402.8083 -1.35 -0.00335147 4.460648 4.473873 3 402.8083 -3.0375 -0.007540808 4.441898 6.710809 4 402.8083 -5.4 -0.013405881 4.415648 8.947745 5 402.8083 -8.4375 -0.02094669 4.381898 11.18468 10 402.8083 -33.75 -0.083786758 4.100648 22.36936 20 402.8083 -135 -0.335147033 2.975648 44.73873 25 402.8083 -210.938 -0.52366724 2.131898 55.92341 30 402.8083 -303.75 -0.754080825 1.100648 67.10809 35 402.8083 -413.438 -1.02638779 -0.1181 78.29277

So, could you interpret these charts for the benefit of the rest of us?

Three fourths of the audience is running for the exits.

Here, come on back....

http://www.lindseyvonn.com/photos

GAAAAAH!!  (sound of panting and pounding ski boots)

Gravity forces (and ski friction) are bigger on the more massive skier, allowing him to better counter the air resistance force which is about the same on both skiers.  Lighter skiers need to make up for it with better technique.

Inertia

Quote:
Originally Posted by Ghost

Gravity forces (and ski friction) are bigger on the more massive skier, allowing him to better counter the air resistance force which is about the same on both skiers.  Lighter skiers need to make up for it with better technique.

If a feather and a rock fall at the same rate, how are gravity forces bigger for the greater mass?

Quote:
Originally Posted by Ghost

 M= 80 kg g 9.8 m/s/s q= 35 degrees mu 0.05 ro 0.986 cd 0.9 A 0.8 m*(dV/dt) = m*g*sinQ - mu*m*g*cosQ - 0.5*ro*Cd*A*V^2 (dV/dt) = g*sinQ - mu*g*cosQ - 0.5*ro*Cd*A*V^2/m V gsinQ-mugcosQ -0.5*ro*CD*A*V^2/m Air Friction/(driving force-ski friction) a mph 1 417.5732 -0.35496 -0.000850055 5.215228 2.236936 2 417.5732 -1.41984 -0.003400218 5.201917 4.473873 3 417.5732 -3.19464 -0.007650492 5.179732 6.710809 4 417.5732 -5.67936 -0.013600874 5.148673 8.947745 5 417.5732 -8.874 -0.021251366 5.10874 11.18468 10 417.5732 -35.496 -0.085005462 4.775965 22.36936 20 417.5732 -141.984 -0.340021849 3.444865 44.73873 30 417.5732 -319.464 -0.76504916 1.226365 67.10809 35 417.5732 -434.826 -1.041316913 -0.21566 78.29277

Pretty much what I said earlier.  More mass is a big advantage at GS and higher speeds, not so much at SL speeds.

Thanks for the detailed analysis, Ghost.  Forgive me spelling it out....just trying to understand this, tell me if I've got it right....so m in the equations is M the mass?  And dV/dt is the acceleration, positive being downhill.  And in the equation for dV/dt, m only appears in one place, and that's in the air friction term - 0.5*ro*Cd*A*V^2/m.  So since the constants and V are all positive, this term always reduces the downhill acceleration (makes sense), and the magnitude of the retardation gets less as m increases.  And the amount the retardation gets less with an increase in m depends on the magnitude of - 0.5*ro*Cd*A*V^2, which you give in the table for various speeds V.  So for SL speeds (say 22 mph) we get -35 and for DH speeds (say 78 mph) we get -434.

To take it a step further, an increase in mass from, say, m=80 kg to 100 kg reduces the deceleration due to air friction at SL speeds from 35 to 35*80/100=28, a difference of 7, but at DH speeds it reduces it from 434 to 434*80/100=347, a difference of 87.

Wow, you're right, a 20kg increase in mass makes much more difference at DH speeds.

.

I am trying to understand this too.

Why wouldnt you simply take the driving force (gravity) subtract the resisting forces....to get a net driving force.  Divide that by the mass to work out an accelaration figure?

Sure you would need to do this for varying speeds, but of course the ratio of driving to wind increases for a heavier skier...but you need more force to get the heavier skier to go at the same rate....

Further, just trying to understand your chart there, if we look at say your last line there...417.5732 is that driving force?  and 434.826 is resisting force from air?  If that is correct, how can the resisting force exceed the driving force?

Quote:

Originally Posted by Tog

If a feather and a rock fall at the same rate, how are gravity forces bigger for the greater mass?

I'll try to explain this without equations.

If you drop a feather and a rock from the same height, it's obvious that they don't fall at the same rate.  The rock falls much faster.  The difference is due to air resistance.  If you dropped them both in a vacuum, which eliminates the air resistance, they would indeed fall at the same rate.  So gravity pulls with a greater force on the heavier rock, but also the rock, having more mass, takes more force to get it up to a particular speed.  This effect exactly counters the increased pull of gravity on the rock, so the rock and the feather fall at the same rate in a vacuum.  But in air, the increased pull of gravity on the rock makes a difference, since it helps the rock overcome the air friction much more than the lesser pull of gravity on the feather.  So the rock falls faster.

Hope that helps.

Quote:
Originally Posted by Skidude72

I am trying to understand this too.

Why wouldnt you simply take the driving force (gravity) subtract the resisting forces....to get a net driving force.  Divide that by the mass to work out an acceleration figure?

Actually, I think that's what he did in the first equation:  mass x acceleration = driving force - resisting force.  The driving force is proportional to mass (since it comes from gravity), but the resisting force is independent of mass.  So when he divides through by mass to get the 2nd equation, mass only remains in the denominator of the air friction term.

Quote:
Originally Posted by renenkel

Actually, I think that's what he did in the first equation:  mass x acceleration = driving force - resisting force.  The driving force is proportional to mass (since it comes from gravity), but the resisting force is independent of mass.  So when he divides through by mass to get the 2nd equation, mass only remains in the denominator of the air friction term.

Yeah, I got that.

I just dont get how this can be accurate...if it was then ski racing, and particulary speed skiing should be split into weight classess.  But clearly it isnt.

Logically I understand it...but I am shocked that the difference is that great for skiers at ski speeds.  My gut tells me the difference should be almost negligable.  I get the feather vs. brick...but that is when weight vs. surface area ratios are orders of magnitude different...with people the difference is alot less in a relative sense.

Quote:
Originally Posted by Ghost

I think dominant factor is number 2.  The ratio of gravity to other forces is greater for heavier skiers.  The amount of energy available is greater for heavier skiers. You have to read those physics books past the 1st order theory.  I have.

==

if that were the case then the heavier skier would always win.  ,  men are stronger.  its just a fact.

It's interesting physics discussion, but it has very little to do with skiing... at least when it comes to original question in this thread ;) Let's say that with SL, weight, drag factor, airfoil shape etc. have very little to do with final results. Agreed? So if it doesn't matter if someone is 50kg and someone is 90kg, then it really comes down to just more power in general, how could you otherwise explain 7-8sec difference between men and women on normal SL course (5-6sec if course is easy and flat). I have seen way too many trainings when men and women team shared same course, and on full course (around 50sec), this is pretty much standard difference... no matter what racer, no matter what nation.

So if we concentrate just on DH or SG, we can go into weight and gravity, and size and aerodynamics, but with SL all these things don't matter, yet there's still huge difference. So on the end there's really just pure power left on stake, and that's what makes difference.

Quote:
Originally Posted by Skidude72

I am trying to understand this too.

Why wouldnt you simply take the driving force (gravity) subtract the resisting forces....to get a net driving force.  Divide that by the mass to work out an accelaration figure?

Sure you would need to do this for varying speeds, but of course the ratio of driving to wind increases for a heavier skier...but you need more force to get the heavier skier to go at the same rate....

Further, just trying to understand your chart there, if we look at say your last line there...417.5732 is that driving force?  and 434.826 is resisting force from air?  If that is correct, how can the resisting force exceed the driving force?

Just for you Skidude

 g 9.8 m/s/s q= 30 degrees mu 0.05 ro 1.5 cd 0.9 A 0.5 m*(dV/dt) = m*g*sinQ - mu*m*g*cosQ - 0.5*ro*Cd*A*V^2 (dV/dt) = g*sinQ - mu*g*cosQ - 0.5*ro*Cd*A*V^2/m M= 90 kg V gsinQ-mugcosQ -0.5*ro*CD*A*V^2/m Total Force (Newtons) a =F/m  in m/(s^2) mph 1 402.8083 -0.3375 402.4707797 4.471897552 2.236936 2 402.8083 -1.35 401.4582797 4.460647552 4.473873 3 402.8083 -3.0375 399.7707797 4.441897552 6.710809 4 402.8083 -5.4 397.4082797 4.415647552 8.947745 5 402.8083 -8.4375 394.3707797 4.381897552 11.18468 10 402.8083 -33.75 369.0582797 4.100647552 22.36936 20 402.8083 -135 267.8082797 2.975647552 44.73873 25 402.8083 -210.9375 191.8707797 2.131897552 55.92341 30 402.8083 -303.75 99.05827969 1.100647552 67.10809 35 402.8083 -413.4375 -10.62922031 -0.118102448 78.29277 M= 50 kg V gsinQ-mugcosQ -0.5*ro*CD*A*V^2/m Total Driving  (Newtons) a m/(s^2) mph 1 223.7824 -0.3375 223.4448776 4.468897552 2.236936 2 223.7824 -1.35 222.4323776 4.448647552 4.473873 3 223.7824 -3.0375 220.7448776 4.414897552 6.710809 4 223.7824 -5.4 218.3823776 4.367647552 8.947745 5 223.7824 -8.4375 215.3448776 4.306897552 11.18468 10 223.7824 -33.75 190.0323776 3.800647552 22.36936 20 223.7824 -135 88.78237761 1.775647552 44.73873 25 223.7824 -210.9375 12.84487761 0.256897552 55.92341 30 223.7824 -303.75 -79.96762239 -1.599352448 67.10809 35 223.7824 -413.4375 -189.6551224 -3.793102448 78.29277

When the drag force exceeds driving force, the skier slows down (as in when you come out of your tuck on your way to the speed trap at the Jay-peak citizens race, because some moron is standing in the middle of the course looking up hill to see if anyone else is coming down).

Think of it this way, the more massive skier is less affected by the wind, which is a significant factor for GS, SG, and DH.

For SL, as I said before, strength is a determining factor.  I will add that muscle leverage is also a factor.

The variation between individuals will allow some women to beat some men, but on average, men will always be stronger, bigger and faster than women.

(edit: at least until the next evolutionary change)

Quote:
Originally Posted by Skidude72

I just dont get how this can be accurate...if it was then ski racing, and particulary speed skiing should be split into weight classess.  But clearly it isnt.

Back in the good old days I fought in full contact karate with no weight classes.   The heavier guys did have an advantage, and some (especialy some of the lighter guys) argued that it should be split into weight classes, but it clearly wasn't.  Maybe some day speed events will be split into weight classes, until that day they will continue to do things to take away that advantage, like lay out sg courses and call it DH or put SL turns on a SG (no symbol for hyperbole).

Well based on this, it seems your equation is  correct.

http://www.physics.ox.ac.uk/schools/inset/physics%20of%20skiing.htm

and this site seems to agree with your conclusion:

http://www.real-world-physics-problems.com/physics-of-skiing.html

Still not sure on the magnitude of the difference...but the theory at least seems sound.  Would be interesting to do some empirical tests.  I like the answer thou...I REALLY am the better skier, my competition is just fatter!

Quote:
Originally Posted by crgildart

The weight advantage only applies on a slope where with gravity creates/amplifies acceleration, and on a very slick surface.  At some point skimming across the flat water extra weight becomes a huge disadvantage.  Did it that happen before the end of the pool?  Who knows.. but I bet the heavier kid was going a little faster assuming same wax and other variables close.

Did they both have the same size skis?

The bigger kid was going faster and just made it across. My son ran out of speed and sank about five feet from the end. I don't know about the skis but I would imagine the bigger kid's skis are bigger than my son's which would help. I'll have to see if he's there this weekend to see how big his skis are.

There were quite a few kids smaller than my son and none of them made it. They just didn't have enough speed to get across. Or their tips caught and there was a big old splash...

At the top of my game as a senior in high school I weighed under 100 pounds.  I couldn't beat any of my friends that skied as often as I did when racing full tuck though my wind resistance was far less and wax/base contact was even or superior,   I could usually beat most of them as the number of turns on the race course  increased.  See Pinewood Derby references if you don't understand Ghost's math proof-basically the same theory and variables but substitute axle and hub  lube and prep for snow friction .

But, before folks start whining about losing because they are skinny I must point out that at least at the level of NASTAR (somewhere between SL and GS usually) the weight disadvantage wasn't insurmountable.  Even weighing about 50% less than most others in my age category I was able to earn gold pretty easily (they didn't have platnum back then).  But, all other things equal, weight does matter without a doubt, especially on straight a ways..

Quote:
Originally Posted by Tog

If a feather and a rock fall at the same rate, how are gravity forces bigger for the greater mass?

It's the acceleration due to gravity that is constant rather than its force. That's why all objects will fall at the same speed if they are travelling straight down in a vacuum, even though more massive objects have more force acting on them due to gravity than those with less mass.

Edited by CerebralVortex - 3/25/11 at 10:11am

Skidude -- Really nice explanations in those links, especially the 2nd one.  Thanks for finding/posting them!

Wow,

so we needed a physics formula to explain how being bigger and stronger with more muscle mass will make you better in athletic competitions???

Hey can someone run the numbers on why I can't beat Shaq one on one in Basketball?

Quote:
Logically I understand it...but I am shocked that the difference is that great for skiers at ski speeds.  My gut tells me the difference should be almost negligable.  I get the feather vs. brick...but that is when weight vs. surface area ratios are orders of magnitude different...with people the difference is alot less in a relative sense.

The difference in top speed by adding a few tens of kilos of weight to a skier isn't orders of magnitude different either.  In Ghost's simulated example, nearly doubling the weight (50kg to 90kg) took top speed from ~60mph to ~70mph.  Acceleration at the 22MPH mark was only ~7% faster for the heavier skier.  In an SL race that would be swamped by skill and strength.  In an SG/DH, the difference would be more pronounced, to the point where the lighter skier would be at a significant disadvantage.  At 44MPH, the heavier skier had ~67% faster acceleration.

Basically what's happening in real life is that when a person's weight/size goes up, their cross-sectional 'surface area' goes up proportionally less than their weight -- especially in a tuck.  So 'bigger' people have better top speed and acceleration, everything else being equal.

Quote:
Originally Posted by CerebralVortex

It's the acceleration due to gravity that is constant rather than its force. That's why all objects will fall at the same speed if they are travelling straight down in a vacuum, even though more massive objects have more force acting on them due to gravity than those with less mass.

Very nice explanation. You've earned your user name...

Also the proof in the soap box derby or whatever they're callled, pretty much does it. So, you want to make the cars out of depleted uranium for max speed?

btw...

Quote:
Originally Posted by Tog

Also the proof in the soap box derby or whatever they're callled, pretty much does it. So, you want to make the cars out of depleted uranium for max speed?

For speed skiing like the guy in the aero suit that Richie-Rich posted, where they only go in a straight line, I guess they should make the skis out of depleted uranium.  That way, the extra weight would make you go faster without your muscles having to hold it up.  For a course with turns though, heavy skis would have agility disadvantages?  Would it severely affect the ability of a DH racer to ski well if he had really heavy skis?  For doing smoothly carved turns, maybe not so much?  But sometimes they have to hop up and suddenly apply the skis at a different angle, then it would be a problem....

If you really wanted to answer this, you could go back to the WC and take the weight of the racers and calculate the number of wins (and number of races, assign a value), and the weight.

Go back to the beginning to increase the sample size and predict validity

Ghost

Drag on a skier in a tuck at 70 Mph is like 18 lbf.

Somewhere your drag calculation is off by a factor of 5.

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