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# Dan Dipiro's Mogul Book - Page 26

Quote:
Originally Posted by borntoski683

Quote:
Originally Posted by Tog

I don't get this horizontal bit. Why would not the skier continue on a tangent to the curve? Which at the top of the pipe is near vertical.

Tog, thanks for the civil inquiry!

First, what causes the rider to follow the tangent of the curve?  The reaction force of the snow.  More pressure equals more reaction force, less pressure equals less reaction force.  So if you flex and reduce some of the pressure, you will be at least temporarily reducing the reaction force.  With less reaction force, the surface will be redirecting you upwards less then it would with more reaction force.   So if you flex your legs lower down on the wall, reducing pressure and reducing the redirection...then your CoM is getting closer to the wall, its not running in tangent anymore.  Its heading on a collision course for the wall.  As you continue to move on more of a horizontal path that does not match the wall, the more acute the angle of the wall becomes compared to the direction of travel by the CoM.  Eventually you will either be unable to flex your legs anymore and your body is going to compress against

I look at it like this as jamt mentioned in another thread. Hammer flies tangent to curve not directly out.

Reducing pressure is like lengthening the wire on the hammer ball. Or just lengthening your radius. I think. Pressure and standing height might complicate things beyond this?

ok, but if I understand you correctly, consider that at the transition, the tangent is not straight up in the air, its pointing in a direction that is towards the wall.  If you wait to flex until you are near the top, then yes...it won't really slow you down.

Kook described well in an earlier post how he flexes at the transitions...I believe that is roughly the area where the slope is curved and transitioning between vertical and horizantal.  So if you do the equivalent of releasing the hammer there, the tangent will not be straight up, it will be at an angle into the wall.

Just talking about point of exit at top of pipe and where the skier/rider will go.

Also, one doesn't necessarily go straight up on the wall. Point 2 is that if you do a trick up at the top you might push off. Now that may complicate things even more.

Consider in the example Kook gave a few posts again, if he flexes at the transition dropping in and flexes again at the next transition before going up, he feels he has killed his speed to almost nothing.  Might not even make it all the way up that wall.  I don't think there is any question about this technique in a half pipe having that effect.  The disagreement is about WHY it works and how it can all be applied to bump skiing.

Flexing at the transitions would be somewhat equivalent in your hammer throw analogy, to the rope slipping in your hands before the intended release, where is slips and then you grab it to stop it from slipping before finally letting go...something like that... hehe...

Quote:
Originally Posted by mdf

Even here, angular momentum is not a particularly fruitful way to do the calculations.  You have to have an initial movement to have something to be in phase with, sure.  But the simple energy balance is an easier way to look at it.  When you stand up at the bottom, you raise your center of mass relative to where it would be passively, adding energy into the system.  When you crouch at the top, the direction of crouching is not aligned with gravity, allowing you to reset your legs without as much loss of potential energy.

Can you do the calculation with angular momentum instead?  Probably, but why?

When you stand up, you're higher.  I don't see the news there.  If you push against centripetal force you can really transfer lots of energy to the rotation which is interesting and applicable.  It's also interesting to grow oscillations just by pumping.  In order to pump you have to squat at some point and you're either going to squat through gravity or centripetal force and give back energy you got from standing.  The perpendicular example compared to a vertical wall is interesting where you are suspended in a horizontal position, but you can't squat in that position being that there are no forces at that moment, and that makes the analysis complicated. You can squat before the perpendicular spot through centripetal force and gravity together, but can this give up less energy than you put in to stand?  I'm not sure.  We could put some more work into it, but I don't think it's worth it because it's not interesting enough to be pertinent to the discussion.  So, I would be curious to see if you have an example that proves oscillations can grow with a flat bottom (no extending on any curved surfaces), otherwise I'll have to consider this undetermined.

Quote:
Originally Posted by The Engineer

Quote:
Originally Posted by mdf

Even here, angular momentum is not a particularly fruitful way to do the calculations.  You have to have an initial movement to have something to be in phase with, sure.  But the simple energy balance is an easier way to look at it.  When you stand up at the bottom, you raise your center of mass relative to where it would be passively, adding energy into the system.  When you crouch at the top, the direction of crouching is not aligned with gravity, allowing you to reset your legs without as much loss of potential energy.

Can you do the calculation with angular momentum instead?  Probably, but why?

When you stand up, you're higher.  I don't see the news there.  If you push against centripetal force you can really transfer lots of energy to the rotation which is interesting and applicable.  It's also interesting to grow oscillations just by pumping.  In order to pump you have to squat at some point and you're either going to squat through gravity or centripetal force and give back energy you got from standing.  The perpendicular example compared to a vertical wall is interesting where you are suspended in a horizontal position, but you can't squat in that position being that there are no forces at that moment, and that makes the analysis complicated. You can squat before the perpendicular spot through centripetal force and gravity together, but can this give up less energy than you put in to stand?  I'm not sure.  We could put some more work into it, but I don't think it's worth it because it's not interesting enough to be pertinent to the discussion.  So, I would be curious to see if you have an example that proves oscillations can grow with a flat bottom (no extending on any curved surfaces), otherwise I'll have to consider this undetermined.

No, I am not claiming you can build oscillations without an external restoring force.  You have to have something cyclic to be in phase with.  All I am saying is that angular momentum adds little-or-nothing to the analysis.  The analysis is just so easy using energy balance.

Quote:
Originally Posted by mdf

No, I am not claiming you can build oscillations without an external restoring force.  You have to have something cyclic to be in phase with.  All I am saying is that angular momentum adds little-or-nothing to the analysis.  The analysis is just so easy using energy balance.

The oscillations can't start without an initial angular momentum, which makes me highly suspicious that you have to extend through centripetal force to grow the oscillations, which makes me think that extending on a flat bottom of the half pipe will not grow the oscillations.  Jamt, let me know if you can prove the oscillations will grow, but don't worry about it if it's too much trouble.

Do the physics people agree with this?:
You'll have to go to the page to see the equations.
Quote:

The snowboarder is able to increase his speed on the half-pipe with his feet remaining firmly on the board. This begs the question, what is the physics taking place that enables the snowboarder to increase his speed on the half-pipe?

To increase his speed, the snowboarder crouches down in the straight part of the half-pipe. Then when he enters the curved portion of the half-pipe he lifts his body and arms up, which results in him exiting the pipe at greater speed than he would otherwise.

The basic snowboarding physics behind this phenomenon can be understood by applying the principle of angular impulse and momentum.

The schematic in this analysis is given below.

"If you want to see a really interesting problem related to the physics of snowboarding, check out this analysis for determining the optimal trajectory for a maximum jump on a half-pipe. "

http://www.real-world-physics-problems.com/physics-of-snowboarding.html
Quote:
Originally Posted by Tog

Do the physics people agree with this?:
You'll have to go to the page to see the equations.
Quote:

The snowboarder is able to increase his speed on the half-pipe with his feet remaining firmly on the board. This begs the question, what is the physics taking place that enables the snowboarder to increase his speed on the half-pipe?

To increase his speed, the snowboarder crouches down in the straight part of the half-pipe. Then when he enters the curved portion of the half-pipe he lifts his body and arms up, which results in him exiting the pipe at greater speed than he would otherwise.

The basic snowboarding physics behind this phenomenon can be understood by applying the principle of angular impulse and momentum.

The schematic in this analysis is given below.

"If you want to see a really interesting problem related to the physics of snowboarding, check out this analysis for determining the optimal trajectory for a maximum jump on a half-pipe. "

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

No, this is very wrong.  Yes, conservation of angular momentum shows you speed up, but when you get short again you slow back down by the same amount.  (You can't just keep getting taller and taller.)

Don't believe me?  Look at the Bob Burquist vid from a few pages back.  He gets short at the top of the arc, not tall.

Quote:
Originally Posted by Tog

Do the physics people agree with this?:
You'll have to go to the page to see the equations.

When you extend while you have angular momentum you speed up, and when you crouch you slow down.  A spinning figure skater spins faster when the arms are brought closer to the body, COM closer to the center of the circle.

Quote:
Originally Posted by The Engineer

Quote:
Originally Posted by Tog

Do the physics people agree with this?:
You'll have to go to the page to see the equations.

When you extend while you have angular momentum you speed up, and when you crouch you slow down.  A spinning figure skater spins faster when the arms are brought closer to the body, COM closer to the center of the circle.

And then slows down again if they move their arms away from the body.  There is no net gain to build up cyclically.  Working against gravity is different, because you can get tall when you are aligned with gravity and short when you are not aligned with gravity.

From the US Patent Office

Quote:
Originally Posted by mdf

And then slows down again if they move their arms away from the body.  There is no net gain to build up cyclically.  Working against gravity is different, because you can get tall when you are aligned with gravity and short when you are not aligned with gravity.

When you are at the bottom of the half pipe you have lots of angular momentum, so raising your COM increases your speed, but then as you move up the wall gravity slows you down, so that you have less angular momentum, so that you can move your COM closer to the wall without changing your speed as much.  So, the oscillations grow by increasing your speed through the bottom by extending, but then not decreasing your speed as much on the sides by moving the COM closer to the wall.

Quote:
Originally Posted by The Engineer

The oscillations can't start without an initial angular momentum, which makes me highly suspicious that you have to extend through centripetal force to grow the oscillations, which makes me think that extending on a flat bottom of the half pipe will not grow the oscillations.  Jamt, let me know if you can prove the oscillations will grow, but don't worry about it if it's too much trouble.

Well I cheated and went to this simple physics calculator page page http://keisan.casio.com/exec/system/1225079475

If you have a flat bottom half pipe with 10 meters radius and you rise 1 meter at the bottom and retract your legs about 2,2 meters from the top of the ramp, you will land approximately 0,8 meters above the curved part of the ramp. You have gained m*g*0,8 of energy and your speed will be higher the next time you reach the bottom.

Note that this is probably the worst strategy because you will impact the wall at a 90 degree angle and loose some energy in that impact. If you retract later you will land at a better angle.

With the equations you see on the above page you could make an analytical solution, but that would just be too much fun.

In comparison the centripetal force in a perfect half pipe without a flat bottom will give you a centripetal force which is twice as big as the g force at the bottom.

In other words the gain you can get from a curved ramp is three times larger if you count both centripetal and gravity.

For any one still fuzzy on parametric resonance.  I think this guy does a good job of explaining it.

Quote:
Originally Posted by mdf

...

Don't believe me?  Look at the Bob Burquist vid from a few pages back.  He gets short at the top of the arc, not tall....

He gets short to pump subsequently, it's not what you think.

Quote:

Thanks @CTKook

Pump the transition for speed = extension?

More or less, yes. A particular type and timing of extension, which works for pumping, and also there is factoring in other contributors to pumping, but more or less that is correct.

Edited by CTKook - 5/6/15 at 3:16pm

I would like to make a heartfelt apology heard by the forum, that I did not treat Borntoski with respect.  He has such a different approach to some things in life compared to me that it completely caught me off guard, and I did not know how to behave properly.  I hope that you will accept my apology.

Quote:
Originally Posted by Jamt

Well I cheated and went to this simple physics calculator page page http://keisan.casio.com/exec/system/1225079475

If you have a flat bottom half pipe with 10 meters radius and you rise 1 meter at the bottom and retract your legs about 2,2 meters from the top of the ramp, you will land approximately 0,8 meters above the curved part of the ramp. You have gained m*g*0,8 of energy and your speed will be higher the next time you reach the bottom.

Note that this is probably the worst strategy because you will impact the wall at a 90 degree angle and loose some energy in that impact. If you retract later you will land at a better angle.

With the equations you see on the above page you could make an analytical solution, but that would just be too much fun.

In comparison the centripetal force in a perfect half pipe without a flat bottom will give you a centripetal force which is twice as big as the g force at the bottom.

In other words the gain you can get from a curved ramp is three times larger if you count both centripetal and gravity.

I've been working very hard to make detailed arguments that riding up a bump curved or not is counter productive to speed control.  I've been working very hard to demonstrate that CTkooks parametric resonance method for speed control has limited use based on the difficulty in having angular momentum without curved surfaces.  Unfortunately, if it's true that you can amplify your height by extending on a flat spot before going into a quarter pipe, it would also be true that you could amplify the decrease in height in the reverse method and give slowing by CTkooks ideas without as much stringent requirement on the surface.  Still though, I just don't believe it's true, because I see this violating conservation of energy.

If you are standing on top of a lossless halfpipe, you have energy mgh.  You can drop down into the halfpipe and emerge at the other side with exactly the same energy and height given by mgh.  Now if you pull apart the half pipe to give a flat spot in the middle you can still drop into the spread halfpipe and emerge on the other side with exactly the same height no matter how long the flat section.  If during the flat section you double your amount of energy, you must end up with twice the energy, so you will have twice the height at your maximum point.  When you stand during the flat spot you increase your energy by mg(delta)h.  Therefore your final energy must also be increased by mg(delta)h by conservation of energy.  The end height delta must be the same as the flat height delta.  I don't believe you can get any height amplifications merely by pushing against gravity, this violates conservation of energy.  I am very certain of this, so please give lots and lots of consideration before negating.

Quote:
Originally Posted by The Engineer

...

I don't believe you can get any height amplifications merely by pushing against gravity, this violates conservation of energy. ...

This is true.  Extending solely on flat bottom only would increase wind resistance, aka God don't care how tall you are.,

This however does not negate my point re speed control.  The God of bumps don't care how tall you are, but does reward work.  Way of the world.

Aka you can help keep speed by floating over the lip, but only because cognitively and athletic floating over the lip helps ensure a good pump.

Quote:
Originally Posted by mdf

From the US Patent Office

I'm guessing nobody recognizes what this is?

^^^

English majors don't recognize crap like that. However, we can read the clues and we can Google.

Quote:
Originally Posted by mdf

I'm guessing nobody recognizes what this is?

isn't this akin to fission in a mason jar???

Quote:
Originally Posted by mdf

Dude.  A tramp ain't a flat surface, nor a rigid one.  Try boosting on granite or limestone, and compare to a tramp.  There are many objective differences.

is that a perpetual energy machine?

lol, I'm still trying to understand if absorption slows you down or speeds you up.

Professor, anyone?

Nail

Quote:
Originally Posted by mdf

Quote:
Originally Posted by mdf

From the US Patent Office

I'm guessing nobody recognizes what this is?
Kind of the opposite of a steam locomotive? Instead of linear to rotary it's rotary to linear.

.Although we are now in driftum absurdum relativium, this thread could be called "What Dan Didn't Tell You About Skiing Moguls".
Quote:
Originally Posted by The Engineer

The end height delta must be the same as the flat height delta.

I don't believe you can get any height amplifications merely by pushing against gravity, this violates conservation of energy.

I don't see how you can first say that the height has increased with delta and then in the next sentence say that you cannot get any height amplifications.

A better approach to get amplification with a flat bottom would be to double pump. It is not possible without it.

It is gravity that enables pumping and it can be utilized in many ways. Without gravity you would have true conservation (would also need a full pipe)

Another way to look at things is the forces involved. The vertical forces pushing on the rider will determine how high he gets. It should be intuitively obvious that the forces are larger in the case when you extend at the bottom and flex towards the top than if you don't.

I have tried to explain in various ways how it works and I have even calculated an example. The next step would be to build this into a physics simulator, but that just feel like too much work at the moment.

Be my guest to prove that it is not possible. To prove that you need to show that the energy returned by flexing the legs close to the top is exactly equal to mg(delta h)

New idea. What about a ramp with a flat part towards the top?

Is this possible? Will the speed be higher when we reach C for the second time?

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