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Two kinds of Wedge Christie - Page 28

post #811 of 830
Quote:
Originally Posted by borntoski683 View Post


Perhaps this is the distinction I am trying to convey. when the axis of rotation is within the skier's CoM then an actual moment of inertia will be formed within the skier itself, which equates to the skier spinning. When the axis of rotation is outside the skier, then the skier no longer has their own moment of inertia that can perpetuate spinning, the skier becomes like a ball on the end of a string which is only rotating and turning by virtue of continuous external forces making it happen. By this analysis makes the skier just a lump of mass in a larger rotational system, that is rotating around some theoretical center point by virtue of the ski to snow external forces.

I don't know what language will explain this distinction, but there is a distinction and it matters a lot in skiing. Maintaining yourself as a ball on the end of a string rather then becoming yourself the center of the circle is the difference between turning and spinning. And since there is no string, it's manipulated in real time by our edges. Turn off the edges and a non spinning skier will continue straight. if the axis is within the skier and they are spinning then turn off the edges and the skier keeps spinning. That is the distinction. I am not sure what is the correct physics terminology at this point according to you PhDs to note this distinction.

 

Yep.... that's why I didn't want to touch the "pure carving" scenario..... the longer the turn radius, the more the skier becomes a lump of mass.

 

In addition, that's why I put on those disclaimers about edge bite and edge engagement, otherwise the skier will spin out. Along with the "handle bar", spinning out in a short turn is another hazard with the forward pressure. 

post #812 of 830
Quote:
Originally Posted by Ghost View Post
 

 

 

 

Some people deserve a refund from their schools, I however have gotten my money's worth from my education.

 

Sorry, but I will not stand by when someone claims I am "wrong" and is either not fully cognizant of high school physics  or attempts to mislead with half-truths.

 

Dude... chill. I see nothing wrong with either's point. Its just a question of which dominates, LM or AM. 

post #813 of 830
Further to what I was saying about where the axis of rotation is, let's talk about what it takes to get the axis of rotation within the CoM itself..... To start spinning.

An ice skater gets this by carving to the tightest radius possible until it's so tight that the axis is within their mass and linear inertia is converted to self sustained spinning. They are probably using a rotary push of some kind too, but what do I know. Not sure the Ph.D. word for that self sustained spinning.

In skiing we have a limit to how tight we can carve which really constrains how well we could do what an ice skater does that way. It's pretty rare to see a skier that has truly converted their linear momentum into spinning, except in Ariels.

The ironic thing is that when you oversteer or get ski tail washout or whatever, you do move the axis of rotation in closer but you also widen any previous carving turn shape you had previously pretty much to the point of moving straight actually. The edges become disengaged which drastically changes the external forces and your ability to tighten the radius any further and move the axis from the tips of the skis to within the body. You might get a little bit of rotational inertia but it's minimal. The string holding the ball is going to be cut as soon as the ski edge loses steering angle by washing out all the way to backwards.

We can do a 360 in the air which normally involves some kind of rotary push off, which CAN create conditions where the axis is within us and effective spinning happens. Rotary push off is generally frowned upon for ski turns. Likewise I doubt you will see too many arielists using tail washout to power 1040's.
Edited by borntoski683 - 11/26/15 at 7:47am
post #814 of 830
And further to what ghost eluded to, when you start using upper lower separation and counter rotation to twist the skis, you are not driving skier mass into rotational spinning one iota. The skis may twist but the skier does not. The twisted ski simply becomes a way to manipulate the ski/snow interaction, which may or may not tighten the radius depending on edging and other factors.
post #815 of 830
Quote:
Originally Posted by Ghost View Post

If a thing is rotating it has angular momentum.  If a thing has liner velocity it has linear momentum.  If it is both displacing and rotating it has both linear and angular momentum.  I hope you were not involved in any motor vehicular reconstructions.  Simple high school physics experiments show that conservation of energy must include both linear and angular velocity.    A car travel down a road at 70 mph in a normal linear fashion has less energy than a car spinning at one rotation per second while traveling down the road at 70 mph.  As Jamt correctly stated it is not that it is not there in a long or short curve, it is that it is minimal (for example car rotates from facing north to facing west in a long time so slow rotation).  

It is correct that the angular momentum conservation equation involves applied torque to calculate angular accelerations and the linear momentum equation involves forces to calculate accelerations and both equations are separate vector equations not to be added to a common sum.   However, the Energy equation must include all energy.

Some people deserve a refund from their schools, I however have gotten my money's worth from my education.

Sorry, but I will not stand by when someone claims I am "wrong" and is either not fully cognizant of high school physics  or attempts to mislead with half-truths.

A car driving around the outside edge of a circle only has angular momentum. Physics 101 look it up.
post #816 of 830
Quote:
Originally Posted by The Engineer View Post


A car driving around the outside edge of a circle only has angular momentum. Physics 101 look it up.

LM and AM are two different measures.

 

For a point mass:

 

angular momentum = radius x linear momentum = r x m x v

 

for the car:

 

Angular momentum around the circle center = r x m x v + I x w

 

where the first term corresponds to the rotation around the circle center and the second term corresponds to the angular momentum around the CoM.

 

What happens if all of a sudden the friction under the tires goes to zero?

post #817 of 830
Quote:
Originally Posted by Jamt View Post

LM and AM are two different measures.

For a point mass:

angular momentum = radius x linear momentum = r x m x v

for the car:

Angular momentum around the circle center = r x m x v + I x w

where the first term corresponds to the rotation around the circle center and the second term corresponds to the angular momentum around the CoM.

What happens if all of a sudden the friction under the tires goes to zero?

If you have nonzero linear momentum you will continually translate in space. When you go around a circle you go nowhere kind of like this forum. Like I said the instantaneous parameters can be helpful to determine rotational parameters. FYI things get interesting when you don't have a point mass. Some rotational and linear concepts converge for a point mass because the center of moment of inertia equals the center of mass which is not always true because the moment of inertia integrates over r squared whereas center of mass integrates over r. I have only been reading the posts that quote me, so I don't know where the conversation is, but jacks initial proposal has merit, and I will have to leave it with that.
Edited by The Engineer - 11/26/15 at 12:27pm
post #818 of 830
Quote:
Originally Posted by The Engineer View Post


 FYI things get interesting when you don't have a point mass. Some rotational and linear concepts converge for a point mass because the center of moment of inertia equals the center of mass which is not always true because the moment of inertia integrates over r squared whereas center of mass integrates over r. 

Yes, I used the car example above which is not a point mass:

 

L =  r x m x v + I x w

 

If the car is point mass the moment of inertia I=0, so L=r x m x v = r x (linear momentum), so in  non-strict way you could say that they converge.

 

I think one of the main problems with using total angular momentum in skiing is that it is always around a fixed point of rotation, but in skiing there are no fixed points.

even if we just use the part from CoM/CoMoI it is not fixed either since the body is changing shape continuously. 

Simple concepts, but how do you explain a cat landing on the paws with some simple equations?

post #819 of 830
Quote:
Originally Posted by Jamt View Post
 

 

I think one of the main problems with using total angular momentum in skiing is that it is always around a fixed point of rotation, but in skiing there are no fixed points.

even if we just use the part from CoM/CoMoI it is not fixed either since the body is changing shape continuously. 

 

 

 

 

Seems ice skaters do it at will..... using AM while the axis of rotation is moving along some path. What happens when she spins with her arms not tucked to her chest?

 

 

 

And btw...what's with this "total" angular momentum? I have acknowledged some LM, AM and other factors are involved with the  concept I brought up. 


Edited by jack97 - 11/26/15 at 4:04pm
post #820 of 830

Since this thread got railroaded.....  by looking at the trees and shrubs, the terrain looks flat to me but he is keeping pace with the car moving at 19 mph. He seems to be doing this by pushing or shifting his weight side to side using a low friction sled/board. Hmmm... seems like we talked about this.

 

 

 

 

I have to check out, go into stealth mode and fight the collective...... Have fun, it's the start of ski season!

post #821 of 830
Quote:
Originally Posted by jack97 View Post
 

Since this thread got railroaded.....  by looking at the trees and shrubs, the terrain looks flat to me but he is keeping pace with the car moving at 19 mph. He seems to be doing this by pushing or shifting his weight side to side using a low friction sled/board. Hmmm... seems like we talked about this.

 

 

 

Apparently skateboarders know things beyond the comprehension of physicists and expert skiers.

post #822 of 830
Quote:
Originally Posted by borntoski683 View Post

And further to what ghost eluded to, when you start using upper lower separation and counter rotation to twist the skis, you are not driving skier mass into rotational spinning one iota. The skis may twist but the skier does not. The twisted ski simply becomes a way to manipulate the ski/snow interaction, which may or may not tighten the radius depending on edging and other factors.

BTS, I have to check out too, but I want to give you one thing to consider.  I saw you questioning the ball on a string concept vs spinning around the COM.  Let's say you have a ball on a string and you spin it with a certain speed.  Now, let's add a second ball near the first and spin both around the same speed. Does the first ball now have any more energy than before? No. Does the second ball make any difference to the first ball's world?  No, excluding wind.  Now move the second ball to 45 degrees from the first. Has the first ball's life changed at all. It just keeps spinning the same as it had when it was by itself. The total energy with two balls is twice what it was with one.  Now, put the second ball at 180 degrees across the circle from the first. Does that effect the first ball's energy or momentum? Same ball spinning the same way, so no, but now with both balls with equal weight across the circle from each other you have an object spinning around it's own axis. This concept of equivalence is ubiquitous in engineering. From the ball's perspective, having another ball provide a force to keep it from flying off is equivalent to a rotating anchor.  It can't tell the difference. It's going around a circle. It has centrifugal force holding it, rotational energy and angular momentum the same with or without that second ball.  It doesn't matter whether that holding force is supplied by an equal and opposite mass across the circle, or an anchor, or gravity, or tires on the road, or ski edges dug in snow.  Reciprocally, for any shape rotating around the COM, you can look at just a tiny part of it near the outside and consider that tiny part is rotating around a circle not around it's own axis.  So, something rotating around it's axis is just a sum of many parts rotating not around their own axis.  Then you can sum up the energy of all the tiny parts to get the total energy which is actually how the equations of moment of inertia are derived. As far as angular momentum is concerned it's just how much material how far away spinning how fast.

post #823 of 830
Quote:
Originally Posted by jack97 View Post
 

 

 

 

Seems ice skaters do it at will..... using AM while the axis of rotation is moving along some path. What happens when she spins with her arms not tucked to her chest?

 

 

 

And btw...what's with this "total" angular momentum? I have acknowledged some LM, AM and other factors are involved with the  concept I brought up. 

When she is airborne there are no external forces and the axis is pretty close to the CoM  i.e. constant in the coordinate system moving with her. In skiing that is not true most of the time.

The total angular momentum is, again:

L =  r x m x v + I x w

The first part is due to the CoM passing the rotation axis at a discance of r, the second term is due to the rotation around the CoM.

Even if you are not rotating at all you have a large angular momentum if r is different from zero.

For example, say that you ski completely straight past a point 10 meters away (or that you counter so that you are not rotating around CoM), you weight 100 kg, and your speed is 10 m/s, then the angular momentum is:

10x100x10 = 10000, the point being, you don't have to rotate around the CoM to have a significant moment!

 

now if we add the term that is due to the rotation around the CoM. In a ski turn we could assume a rotational velocity of e.g. 1 rad/s. Further the moment of inertia around the longitudinal axis has been shown to be approximately I=1,5 kg m^2, so the second part is

I x w = 1,5

 

the total angular momentum is then 

 

10001,5

 

In other words for the total angular momentum the rotation around the CoM is negligible. 

 

Quote:
Originally Posted by jack97 View Post
 

Since this thread got railroaded.....  by looking at the trees and shrubs, the terrain looks flat to me but he is keeping pace with the car moving at 19 mph. He seems to be doing this by pushing or shifting his weight side to side using a low friction sled/board. Hmmm... seems like we talked about this.

 

 

 

 

I have to check out, go into stealth mode and fight the collective...... Have fun, it's the start of ski season!

 

Quote:
Originally Posted by MrGolfAnalogy View Post
 

 

 

Apparently skateboarders know things beyond the comprehension of physicists and expert skiers.

Well it is actually very easy from a physics point of view, it is just when you try to motivate everything with conservation of angular momentum that things get complicated.

We have seen repeatedly in the past that you can make the wrong conclusions, for example a lot of bump threads.

You can use energy or forces for an easier conclusion.

post #824 of 830
Quote:
Originally Posted by The Engineer View Post
 

BTS, I have to check out too, but I want to give you one thing to consider.  I saw you questioning the ball on a string concept vs spinning around the COM.  Let's say you have a ball on a string and you spin it with a certain speed.  Now, let's add a second ball near the first and spin both around the same speed. Does the first ball now have any more energy than before? No. Does the second ball make any difference to the first ball's world?  No, excluding wind.  Now move the second ball to 45 degrees from the first. Has the first ball's life changed at all. It just keeps spinning the same as it had when it was by itself. The total energy with two balls is twice what it was with one.  Now, put the second ball at 180 degrees across the circle from the first. Does that effect the first ball's energy or momentum? Same ball spinning the same way, so no, but now with both balls with equal weight across the circle from each other you have an object spinning around it's own axis. This concept of equivalence is ubiquitous in engineering. From the ball's perspective, having another ball provide a force to keep it from flying off is equivalent to a rotating anchor.  It can't tell the difference. It's going around a circle. It has centrifugal force holding it, rotational energy and angular momentum the same with or without that second ball.  It doesn't matter whether that holding force is supplied by an equal and opposite mass across the circle, or an anchor, or gravity, or tires on the road, or ski edges dug in snow.  Reciprocally, for any shape rotating around the COM, you can look at just a tiny part of it near the outside and consider that tiny part is rotating around a circle not around it's own axis.  So, something rotating around it's axis is just a sum of many parts rotating not around their own axis.  Then you can sum up the energy of all the tiny parts to get the total energy which is actually how the equations of moment of inertia are derived. As far as angular momentum is concerned it's just how much material how far away spinning how fast.

 

The points I have made have nothing at all to do with energy TE.  You have totally missed my point.  Furthermore I'm not interested in arguing about terminology with you science guys.  I have tried my best to explain the distinction I was talking about and that has obviously not been good enough because you still are missing what I was saying.

post #825 of 830
Quote:
Originally Posted by The Engineer View Post
 
Quote:
Originally Posted by Ghost View Post

If a thing is rotating it has angular momentum.  If a thing has liner velocity it has linear momentum.  If it is both displacing and rotating it has both linear and angular momentum.  I hope you were not involved in any motor vehicular reconstructions.  Simple high school physics experiments show that conservation of energy must include both linear and angular velocity.    A car travel down a road at 70 mph in a normal linear fashion has less energy than a car spinning at one rotation per second while traveling down the road at 70 mph.  As Jamt correctly stated it is not that it is not there in a long or short curve, it is that it is minimal (for example car rotates from facing north to facing west in a long time so slow rotation).  

It is correct that the angular momentum conservation equation involves applied torque to calculate angular accelerations and the linear momentum equation involves forces to calculate accelerations and both equations are separate vector equations not to be added to a common sum.   However, the Energy equation must include all energy.

Some people deserve a refund from their schools, I however have gotten my money's worth from my education.

Sorry, but I will not stand by when someone claims I am "wrong" and is either not fully cognizant of high school physics  or attempts to mislead with half-truths.

A car driving around the outside edge of a circle only has angular momentum. Physics 101 look it up.


The car's mass times the car's velocity = the car's momentum, p=mv

 

Now that we have shown who remembers their grade 11 physics and who does not, what were we talking about before this party got interupted?

 

Oh, yeah.  How does "rolling the knees" achieve very quick turns in mogul skiing.  What is the movement involved, how does it differ from twisting the ski and what is the mechanisms (i.e. torques and forces applied where and how) involved?

 

we CAN get this train back on the rails.

post #826 of 830
Quote:
Originally Posted by Ghost View Post
 


The car's mass times the car's velocity = the car's momentum, p=mv

 

Now that we have shown who remembers their grade 11 physics and who does not, what were we talking about before this party got interupted?

 

 

You are looking at this from 11th grade physics where it's over simplified.  In college level physics we realize that velocity and momentum are vectors not scalar quantities.  Linear momentum must have a velocity that doesn't change directions.  Since the driver around a circle keeps coming back to the same spot, the average linear momentum is zero.  The longer someone has linear momentum the farther that person must go.  No matter how long that person drives around a circle he keeps coming back to the same spot, so there must be no average linear momentum.  He does have instantaneous linear momentum for an infinitely small time before the direction changes which can be a useful concept to use in equations, but not to be used somehow to suggest that a skier going around a circle has negligible angular momentum.  If the spread of the mass is significant compared to the turning radius, there is significant rotational energy that must be dealt with different than linear kinetic energy.  A person's dimensions are significant compared to the turning radius, we feel the centrifugal forces we see the ski's mass spread along the radius when the ski rotates for a rotary turn, there is mass that can move to a shorter radius.  Look, something rotates the ski, whatever it is, if the mass that is rotating is closer to the center it will rotate faster even if the motivational force is also closer to the center.  I believe r squared if the force doesn't move and r if it does move.  A rotating object has energy 1/2 Iw^2, by your thinking we would have to add in 1/2 mv^2 for the simultaneous linear component, but that's already taken care of in the rotational equation, and the result boils down to something depending on the instantaneous linear momentum plus something else.  We only add in the linear component of energy if that circle is moving in space.

post #827 of 830

Are you guys still talking about wedge christies? Didn't read the whole thread but I'm 90% sure it's when you start the turn in a wedge, then later on go parallel. 

post #828 of 830
Quote:
Originally Posted by The Engineer View Post
 
Quote:
Originally Posted by Ghost View Post
 


The car's mass times the car's velocity = the car's momentum, p=mv

 

Now that we have shown who remembers their grade 11 physics and who does not, what were we talking about before this party got interupted?

 

 

You are looking at this from 11th grade physics where it's over simplified.  In college level physics we realize that velocity and momentum are vectors not scalar quantities.  Linear momentum must have a velocity that doesn't change directions.  Since the driver around a circle keeps coming back to the same spot, the average linear momentum is zero.  The longer someone has linear momentum the farther that person must go.  No matter how long that person drives around a circle he keeps coming back to the same spot, so there must be no average linear momentum.  He does have instantaneous linear momentum for an infinitely small time before the direction changes which can be a useful concept to use in equations, but not to be used somehow to suggest that a skier going around a circle has negligible angular momentum.  If the spread of the mass is significant compared to the turning radius, there is significant rotational energy that must be dealt with different than linear kinetic energy.  A person's dimensions are significant compared to the turning radius, we feel the centrifugal forces we see the ski's mass spread along the radius when the ski rotates for a rotary turn, there is mass that can move to a shorter radius.  Look, something rotates the ski, whatever it is, if the mass that is rotating is closer to the center it will rotate faster even if the motivational force is also closer to the center.  I believe r squared if the force doesn't move and r if it does move.  A rotating object has energy 1/2 Iw^2, by your thinking we would have to add in 1/2 mv^2 for the simultaneous linear component, but that's already taken care of in the rotational equation, and the result boils down to something depending on the instantaneous linear momentum plus something else.  We only add in the linear component of energy if that circle is moving in space.

 

Sorry for the poor state of education in your country.  Here in Canada we learn about force, velocity, acceleration momentum, angular velocity and angular momentum as vector quantities and about energy as a scalar quantity in high school.

 

While dealing with linear momentum (in high school) we learn that acceleration is ruled by the vector equation F=ma.  In fact if you were to turn to page 114 of "Fundamentals of Physics: A senior Course" (Martindale, David G.,  ISBN 0-669-95047-5), a high school physics text, you would find a section on uniform circular motion wherein a centripetal force applied perpendicular to an object's velocity (and momentum) accelerates the object around a circular path, altering the object's velocity (and hence its momentum) by changing the direction of said velocity in accordance with the vector equation given above.  Perusal of said text would show that Newton's relationship can also be expressed in a more general form as the rate of change in momentum of an object is equal to the net force applied to the object.  Even further perusal of this high school text and the more advanced text below (Halliday and Resnick) will provide examples of collision experiments where linear momentum and kinetic energy equations are used and explored. 

 

The high school teachings are in complete agreement with University physics texts such as the one found here http://www.amazon.ca/Fundamentals-Physics-David-Halliday/dp/111823071X/ref=dp_ob_image_bk (an earlier edition of which was used by my Alma Mater.

 

As it is apparent from your continued attempts to sidetrack the discussion and cloud the issue by casting aspersions on my understanding of physics (insinuating that I am not aware of the vector nature of velocity, momentum, force, acceleration, torque and angular momentum) and bloviating true but spurious random bits of physics, instead of responding constructively to my last post, I can only conclude that you are attempting a troll. 

 

As trolling is against the policies of this site, I will hear by cease and desist from feeding this troll.  I suggest that everyone else do the same.

 

Good day Sir.

 

 

post #829 of 830

Quote:

Originally Posted by Jim. View Post
 

Are you guys still talking about wedge christies? Didn't read the whole thread but I'm 90% sure it's when you start the turn in a wedge, then later on go parallel. 

We had moved on to mogul skiing, but that discussion seems to have dried up too.

 

EDIT: The Wedge Christie discussion was very enlightening and covered numerous different ways to accomplish a wedge christie fairly completely, including PSIA approved and non-psia approved, as well as not-quite-PSIA-sanctioned ways to pass the PSIA tests while maintaining various personal teaching philosophies.  I don't imagine there was much to more to add. 


Edited by Ghost - 11/27/15 at 6:20am
post #830 of 830
Quote:
Originally Posted by Ghost View Post
 

 

Sorry for the poor state of education in your country.  Here in Canada we learn about force, velocity, acceleration momentum, angular velocity and angular momentum as vector quantities and about energy as a scalar quantity in high school.

 

While dealing with linear momentum (in high school) we learn that acceleration is ruled by the vector equation F=ma.  In fact if you were to turn to page 114 of "Fundamentals of Physics: A senior Course" (Martindale, David G.,  ISBN 0-669-95047-5), a high school physics text, you would find a section on uniform circular motion wherein a centripetal force applied perpendicular to an object's velocity (and momentum) accelerates the object around a circular path, altering the object's velocity (and hence its momentum) by changing the direction of said velocity in accordance with the vector equation given above.  Perusal of said text would show that Newton's relationship can also be expressed in a more general form as the rate of change in momentum of an object is equal to the net force applied to the object.  Even further perusal of this high school text and the more advanced text below (Halliday and Resnick) will provide examples of collision experiments where linear momentum and kinetic energy equations are used and explored. 

 

The high school teachings are in complete agreement with University physics texts such as the one found here http://www.amazon.ca/Fundamentals-Physics-David-Halliday/dp/111823071X/ref=dp_ob_image_bk (an earlier edition of which was used by my Alma Mater.

 

As it is apparent from your continued attempts to sidetrack the discussion and cloud the issue by casting aspersions on my understanding of physics (insinuating that I am not aware of the vector nature of velocity, momentum, force, acceleration, torque and angular momentum) and bloviating true but spurious random bits of physics, instead of responding constructively to my last post, I can only conclude that you are attempting a troll. 

 

As trolling is against the policies of this site, I will hear by cease and desist from feeding this troll.  I suggest that everyone else do the same.

 

Good day Sir.

 

 

Linear momentum is not energy in any country.  It is a vector and needs to be treated accordingly.  The truth is not trolling, but insults to save face against overwhelming evidence and knowledge might be.


Edited by The Engineer - 11/30/15 at 6:11am
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