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# 4 Questions - Page 6

Quote:
Originally Posted by Rick

Michael, what you're talking about is exactly what my skid angle refers to.  Now you're seeing why I adopted a different term.  Steering angle is harbored with definitions that just don't work for what I'm trying to describe.

LeMaster talks not only about steering angles under foot, but also about steering angles within various parts of the ski, including both tail and forebody.  Those were the other reference points I spoke of earlier.  So yes, by his description a carving ski has a built in steering angle.  It's not zero.  Skid angle would be 0.

You're right, the length of the ski alters the steering angle, even when the shape of the turn is exactly the same.  Quite confusing indeed. Ski length has no effect on skid angle when carving.  With skid angle carving is 0, period.

Interesting too, he has a term he uses for the steering angle created solely via the sidecut of the ski.  He calls it the "local steering angle". It harbors the same peculiarity you just recognized.  A longer ski with the exact same sidecut radius will have a larger local steering angle. Two skis that will produce the same turn shape, but have different local steering angles.

Well since my earlier questions are too difficult how about answering these:

How does your "skid angle" concept help us understand skiing?
Does it allow us to measure somthing?

(I think it wont allow any meaningful measure, because in reality the ski is almost always bent somewhat, and that bend has a huge effect...so "skid angle" cannot dictate turn shape, or anything else really...except maybe for amount of breaking being applied, but I again I cant really see the value in measuring it, nor how you would do it while standing on the hill doing D&C, or even with video do to the various angles and distortions etc.)

Does it provide insight into how skis work?
Does it provide insight into the tradeoffs between various ski techniques?
Does it explain forces we feel in a turn?

The steering angle concept does all of that......."skid angle"....seems to be lacking still.  But I would be pleased if you could prove me wrong.
Michael,
I don't have any books to refer to either, but I think we are going to have to delve a little deeper into the idea behind the concept.

It seems that the term steering angle was invented to explain how a ski redirected our path of travel. In a skidding ski, a force is applied at an angle of 90 degrees to that edge (with some approximations neglecting along-edge friction).  Basically he is saying we need only consider the normal force on the ski.  Next he must, it seems, consider the force arises because of our actions to change our momentum, and consider that the normal force arises in some fashion as a reaction force to our "inertial force" (term used with distaste, but it was popular some years ago force and inertia having the same units), so there must needs be a directional difference between this normal force and our existing direction of travel.

An edge-locked carving ski is a little different, in that the edge is perfectly capable of applying a force without being across our direction of motion.  (because as skidude pointed out our intended path does cross the ski).  The existing solution was to pick an arbitrary point, the ski tip, that was easy to use and held the same general principal of more steering angle = more turning force.

At first glance, it would seem the solution is just to use the centrifugal force as the inertial force giving us the normal force and it should all work the same, since the edge is perpendicular to the inertial force.  Since most of the inertia is coming from us and not the ski, we can use the direction of the centrifugal force as being straight sideways at the boot centre, and the angle we are interested in is the angle of the ski compared to this direction.  We could then average (integrate) the angle over the ski assigning positive values for both tip and tail deviations from the sideways direction.

Let's see how that would work out. The steering angle would do it's job of assigning how much the edge resists our inertia wonderfully, but the magnitude of the force would not increase with a tighter turn using a steering angle so defined.  We don't get more force for a tighter turn! This increase would need to be represented somehow.  We need to include mV^2/R.  Most folks wouldn't go for that level of complexity.  It's good to know what's going on, but not all that helpful to the garden variety skier.  Maybe we better stick with tip direction.

Upon rereading my post it occurred to me that skidding is not necessary for the original concept (referred to as Leamaster's) to work. It could be that the existing concept totally ignored centrifugal force (which is the only inertial force needing consideration in an edge-locked carving ski), and the MV^2/R needs to be added to the non-carving ski too.
Edited by Ghost - 11/8/09 at 4:01am
You cannot turn with a zero degree steering angle.  To suggest otherwise is to comploetely misunderstand the concept.  That is part of the definition -- the skis are pointing in the SAME direction that the CM is moving =>  steering angle = 0.

Lemaster uses the direction that the ski underfoot is pointing.
Quote:
Originally Posted by BigE

You cannot turn with a zero degree steering angle.  To suggest otherwise is to comploetely misunderstand the concept.  That is part of the definition -- the skis are pointing in the SAME direction that the CM is moving =>  steering angle = 0.

Lemaster uses the direction that the ski underfoot is pointing.

I think everybody agrees with this.
EDIT:  "Lemaster uses the direction that the ski underfoot is pointing."  and compares it to the direction of travel to get steering angle for a non-edge-locked turn.
Edited by Ghost - 11/8/09 at 10:52am
Quote:
Originally Posted by Skidude72

Well since my earlier questions are too difficult how about answering these:

How does your "skid angle" concept help us understand skiing?
Does it allow us to measure somthing?

(I think it wont allow any meaningful measure, because in reality the ski is almost always bent somewhat, and that bend has a huge effect...so "skid angle" cannot dictate turn shape, or anything else really...except maybe for amount of breaking being applied, but I again I cant really see the value in measuring it, nor how you would do it while standing on the hill doing D&C, or even with video do to the various angles and distortions etc.)

Does it provide insight into how skis work?
Does it provide insight into the tradeoffs between various ski techniques?
Does it explain forces we feel in a turn?

The steering angle concept does all of that......."skid angle"....seems to be lacking still.  But I would be pleased if you could prove me wrong.

Damn, now some of those are actually pretty good questions.  Some actually have the potential of leading to better understanding.

If they'd been asked by someone else I'd have taken the time to try to formulate clear answers.  Unfortunately your behavior and demeanor in this thread leaves me suspect that you would just use the answers as fodder for more endless debate that's not really aimed at coming to understand.

Your question above about skid angle in which you ask, "does it allow us to measure something?" is a good example of what brings me to that conclusion. It's very siimple what it measures, it measures one specific thing, and I've explained what that is over and over in this 6 page thread.  There's no way if you've actually read even a portion of what I've wrote with a genuine interest in coming to understand what I'm sayiing that this would still be a question for you, so I can only conclude your reason for asking is born of some other motivation.  I don't have the time or interest in playing the games associated with whatever that motivation is.  Sorry

Folks, the above only applies to skidude.  If you see something in his questions that you would be interested in sincerely discussing with me I'll be happy to entertain your questions and explain the best I can.
Quote:
Originally Posted by BigE

You cannot turn with a zero degree steering angle.  To suggest otherwise is to comploetely misunderstand the concept.  That is part of the definition -- the skis are pointing in the SAME direction that the CM is moving =>  steering angle = 0.

Lemaster uses the direction that the ski underfoot is pointing.

BigE
In an edge-locked carved 8-m radius turn, my skis underfoot are pointed straight ahead in my direction of motion, I am turning. What would you say is my "steering angle", and how would you measure it?
Now see, I guess I misunderstood LeMaster, maybe I gotta read it again?  I thought that the ski tip being bent created a significant steering angle.  Measuring under the foot doesn't make any sense for steering angle.  The front half of the ski is bent inwards, creating significant steering angle(s) and that redirects the ski in a carved turn.
Ghost,

You're skis are pointing "straight ahead", but you are turning. How?

Clearly, your reference is not the inertial path of the CM. That's what steering angle is measured against.  It's a fairly straightforward calculation....
Edited by BigE - 11/8/09 at 12:38pm
Quote:
Originally Posted by borntoski683

Now see, I guess I misunderstood LeMaster, maybe I gotta read it again?  I thought that the ski tip being bent created a significant steering angle.  Measuring under the foot doesn't make any sense for steering angle.  The front half of the ski is bent inwards, creating significant steering angle(s) and that redirects the ski in a carved turn.

LeMaster writes that there is insufficient power in the ski for the tip to define the path. The turning is defined by the steering angle of the ski underfoot.
Yes, correct, that is where the ski is able to effectively redirect the CoM.  The tip and tail are what influence how the ski itself is redirected on the snow.
Just reading some LeMaster now, what the heck... p23.
Quote:
"Skis have an hourglass shape, called sidecut.  its interesting effect is that the ski's steering angle varies continuously along its entire length.  Its greatest at the tip an decreases steadily toward the tail.  Because the tip always has a greater steering angle than the rest of the ski, the ski will turn as it moves forward.  The ski turns itself"
That passage refers to the steering angle of the ski. around pages 42-46 is some other stuff.
Ok, I just read up some LeMaster to get his viewpoint a little more crystalized in my mind.  I will try to summarize for those that either don't have the book or want to know.  By no means am I implying that this is the last word on the subject and I'm still wanting to hear more about what Rick has in mind if we can get past the the bashing that has been happening.

So here is my take from LeMaster on steering angle (p22-26, 42-46)

The steering angle is the angle between the edge of the ski at any point of the ski and the direction the CoM momentum is currently pointed.

A ski with a sidecut is such that during a turn, at the tip there is more steering angle than further back.  This effect is enhanced even more when the ski is bent.

Because there is more pressure under the front of the ski then at the back, the ski will tend to pivot itself around the center.

He didn't put it the following way, but I will liken it unto an airplane wing.  A wing rises in the air because there is more air pressure under the wing than over it, because of the shape of the wing.

Essentially it comes down to the ski having a self steering effect due to the progressive steering angle over the length of the ski.

Of course he also goes into more detail about skidding and carving and how steering angle can be used in a variety of situations to carve or skid.  "Too much" steering angle results in more skidding, and less actual radius reducing turning.

Now back to Rick, I believe he is attempting to distinguish a difference between (a) the effective radius-reducing effects of steering angle.....and (b) more steering angle which creates skidding.   Isolating the actual steering effects of steering angle from the skidding effects of more steering angle...that is something I am very much interested in discussing.  LeMaster kind of leave it as an art, Rick is trying to break it down a little bit more into science.  I'm game.

At least that's what I thought, but Rick does go on to say that any radius turn can be skied with any amount of skidding, which is where he loses me, particularly if you include "at any speed".  Need to hear a lot more details about exactly how that will be accomplished.
Quote:
Originally Posted by BigE

Ghost,

You're skis are pointing "straight ahead", but you are turning. How?

Clearly, your reference is not the inertial path of the CM. That's what steering angle is measured against.  It's a fairly straightforward calculation....

Consider the following facts about an 8-m radius edge-locked carved turn at the instant the skier is in the fall line heading straight down the fall line.
Skis are pointed straight down the fall line directly underfoot.  Ski tips and tails are not.
The skier's momentum is pointed straight down the fall line.
The skiers instantaneous velocity is straight down the fall line.
The steering angle directly underfoot is zero.
The steering angle defined above (angle between edge and direction of CoM momentum) varies from zero underfoot to local maximi at the tip and tails.
The underfoot section of ski is probably providing most of the turning force to accelerate the skier into the turn despite the local steering angle being zero at that point.

It would appear that Lamaster didn't even consider centrifugal force, but we can.
Ghost, you have errors in your scenario:

Quote:
Originally Posted by Ghost
Consider the following facts about an 8-m radius edge-locked carved turn at the instant the skier is in the fall line heading straight down the fall line.

check

Quote:

The skier's momentum is pointed straight down the fall line.

nope.  At that point of the turn the skiers momentum force vector will be pointing to the outside.

Quote:

The steering angle directly underfoot is zero.

Nope, because of the previous statement, there will be a steering angle.

Quote:

The steering angle defined above (angle between edge and direction of CoM momentum) varies from zero underfoot to local maximi at the tip and tails.

not quite.  The steering angle is maximum at the tip, minimum at the tail.  Underfoot is somewhere in between
Ghost,

Let's make a few assumptions:

The arc is round and can be described by y(t)^2 = x(t)^2 -64.

y'(t1) is a tangent. It is the inertial path of the CM at time t1.
y'(t2) is the inertial path of the CM at time t2..
y'(t3) is the inertial path of the CM at time t3.

Since tha arc is perfectly round, the angle between the tangents y'(t1) drawn at
(x(t1),y(t1))  and the tangent y'(t2) drawn at (x(t2),y(t2)) should be the same as the angle between the two tangents y'(t2) drawn at (x(t2),y(t2)) and y'(t3) drawn at drawn at (x(t3),y(t3)), provided that t2-t1 = t3-t2.

Moreover,  dividing the Angle betweens between tangents y'(t1) and y'(t2) by t2-t1 should be the same as dividing the angle between tangents y'(t3) and y'(t2) by t2-t1.

This indicates that the rate of change of steering angle is zero, therefore the instantaneous steering angle at any point on the arc will be a constant.  It must be non-zero, otherwise there would be no difference between the tangent lines y'(t1) and y'(t2).

It's pretty tedious to chew through the math....but I hope that helps.
It is true that the centrifugal force is directed to the outside of the curve, but this force only exists in a frame of reference that is rotating.  The direction that an object traveling in a circular path is traveling in at any given moment is tangential to the circle.  If you swing a weight on the end of a thread, and cut the thread with a razor blade the weight will travel away in the direction of the tangent.   (it would only appear to move outward to someone who continued to go around the path the weight was on before the thread got cut)
I can't draw the fancy pics, but maybe this will help.
http://www.physicsclassroom.com/mmedia/circmot/ucm.cfm
and this
http://www.physicsclassroom.com/Class/circles/u6l1a.cfm  (scroll down)

If you were to suddenly stop turning at the fall line you would continue to head straight down the hill.  I know because I've done it many times..
Ghost if your momentum where traveling straight down the hill at that instant, which it is not, then you would not be turning, you would be straight running down the hill.  Having a steering angle is required to make the skis turn.  No steering angle, no turn.

When you start out straight running there is no steering angle obviously because you are flat on the snow.  However if you tip your skis while straight running then you get a very small bit of steering angle built into the sidecut of the tip of the ski which LeMaster refers to as local steering angle.  That little bit of steering angle in the tip of the ski starts the ski turning, which then creates a bit more steering angle and so on.

There is not an imaginary groove in the snow that the ski falls into and rides it.  The steering angle is in fact what creates the turning and your ski will leave a groove behind.

Without steering angle, there is no turn.
BigE, the steering angle is defined for a particular instant in time  as the angle between the edge direction and the direction the CoM is pointed in at that instant: zero under foot, more at the tip, more at the tail.
Quote:
Originally Posted by Ghost

If you were to suddenly stop turning at the fall line you would continue to head straight down the hill.  I know because I've done it many times..

define "suddenly stop turning".  If you mean, low side....then I beg to differ.  You will slide off the side of the trail.

If you intentionally end your turn and go down the hill, that involves a whole lot of movements to specifically reduce the steering angle as you approach the fall line, so that you smoothly end up straight running.
-
That's because you usually low-side before you reach that point, or perhaps your in the back seat and your skis have gone around the fall line but you'rw still travelling across it.
As soon as you stop applying any force to the skier, the skier will move in a straight line.  If you stop applying force when the skier is travelling straight down hill, the skier will travel straight down hill.  At the fall line your force is pushing you straight across the fall line to make you turn, just like the thread and the weight pulling to the centre.  When you cut the thread the weight doesn't go straight out from the centre; it just stops turning.
Quote:
Originally Posted by borntoski683

And also, related to your example about swinging a weight on the end of a string and cutting the string at just the right instant.  Its the same, but your facts are not quite there.  The ball's direction is tangental to the arc, but not in a direction that is 90 degrees to the string.
the string is radial, the ball travels in a tangential direction.  When the string is cut, the the two directions are 90 degrees from each other.
In any case, we aren't attached to a string.  We are using steering angle in the ski to turn.  think it through more mate.
Did you get the check your understanding question  right?
http://www.physicsclassroom.com/Class/circles/u6l1a.cfm
It doesn't matter if we use a string, tube, ski, or tires.  If an object is going around a circle at a uniform speed, once you stop forcing an object to go around a circle, it travels in a direction tangential to the point in that circle where the forces stop accelerating the object.

We know the steering angle is constant, and there are 180 degrees in a semi-circle.  The perimeter of the 8m semi-circle is 8pi meters.  If the skier is travelling at a linear rate of about 10m/sec,  he'd loop the circle in .8pi seconds.  That's close enough to 2.5 seconds.

So 180 degrees/2.5 seconds gives 72 degrees/second, .72 degrees in a hundredth of a second -- the skier would travel .1 meters or about 4 inches in that time.

So providing he's only going 10m/sec, I'll go with a steering angle of about .72 degrees.
Well at least your trying to get speed into it.   The centripetal force required to accelerate the skier in an arc is the product of the mass, the velocity squared, and 1 over the radius.  Doesn't change the steering angle, though as the steering angle as defined is independent of velocity, or time.  It is defined at any given instant, and it seems to only be defined by Lemaster based on linaerCoM momentum.
Quote:
Originally Posted by Ghost

Did you get the check your understanding question  right?
http://www.physicsclassroom.com/Class/circles/u6l1a.cfm
It doesn't matter if we use a string, tube, ski, or tires.  If an object is going around a circle at a uniform speed, once you stop forcing an object to go around a circle, it travels in a direction tangential to the point in that circle where the forces stop accelerating the object.

In terms of skiing..  How exactly do you remove the steering angle suddenly at the fall line as you suggest.  Be more specific.

A string and ball does not apply because we do not have a string attached to us holding us on an arcing path.

If you start getting into more physics book arguments about radius and tangents, I will just ago away and the discussion will end with me because I'm not a physics professor, I don't wish to be one and you're missing the point while trying to bully your own idea with fancy equations.

The steering angle in the ski is what causes the radius to happen.  How can you eliminate the steering angle and where exactly would you need to eliminate the steering angle so that you end up straight running down the fall line?

You see in a normal turn, at the fall line, in order to keep turning, there fundamentally HAS to be a steering angle there or you would not continue to turn.  That implies that in a normal turn, at the fall line, the skis are straight but the force vector is out, not down.  They are not lined up the same direction.   There is a "STEERING ANGLE".  The steering angle is kind of like a delay.

In order to be cut loose from the turn and end up skiing staight down, you need to have the steering angle removed at the fallline.  How might you do that?  Well you could turn your skis towards the outside, theoretically so that steering angle is removed.  But you wouldn't be skiing straight down anymore.  You could try to do it ahead of time, which is what I believe you are doing when you perform this ski move, which means that as you approach the fall line you do things that will reduce the steering angle so that by the time you reach the apex, there is no more steering angle and you're skiing straight down.
Ghost:

From lemaster:

" In a pure carved turn, the ski cuts a groove in the snow, which it then slides along.... At every moment, the groove has an infinitesimal steering angle under the skiers foot and so the turn is extremely efficient."

My approximation agrees completely with lemaster -- the smaller the duration is that I examine the changes in tangents, the smaller the steeering angle will be.

Also from lemaster:

"Get your skis waxed and look at the bases after a half day of skiing.b  The wax will be worn off under the foot, but not under the tip/tail.

This is direct evidence that the big forces in skiing happen directly under your feet.  These are the forces that make you, the skier, turn.  The middle of the ski does the work of changing your direction of travel largely due to its steering angle there, as shown in figuer 3.6a ( figure shows skiers direction of travel as different from the dirction the midbody of the ski points.)

The tip and tail are simply not stiff enough to push very hard on you, and therfore don't have enough force to make you turn.

Compared to the force a skier applies to a ski, that is to say his weight and centrifugal force, the ski is notvery siff.  It typically takes only 10 to 15 punds to flatten a ski on the snow.  From this point on, as you add more weight and centrifugal force, the pressyure builds under the middle of the ski, bot not much elsewhere.

what this tells us is thathe tip abnd tail of the ski do little to make the skier turn.  What do they do, then? THey make the ski itself turn."

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