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Which Run? (Blue/Red/Black)



 
 
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  #21  
Old January 24th 05, 10:32 AM
jonnyboy
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why not get some lessons anyway? - you will always learn something.
Jonny.

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  #22  
Old January 24th 05, 11:04 AM
Ace
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On Mon, 24 Jan 2005 11:15:39 GMT, Ben Blaukopf
wrote:

Ace wrote:


I'd humbly suggest that if you were doing so in a high-speed situation
you would be losing a lot more than that. The faster you're going, the
(exponentially) greater the friction forces on the ski.


Yes, but it's not friction.

There are several things going on.

a) Friction. Completely independent of speed. Once you're slipping,
you're slipping, and it makes no difference how fast you're going.
b) Displacement of the snow ahead of the ski. Assuming the snow
displaces in the same way at high speed as at low speed, then the
force here is constant too.
c) Compressing the snow at the edge of the ski, causing the carve.
The snow will compress when carving, and by compressing it we
will slow down. The amount the snow compresses depends on how
fast we are going. and varies as the square of velocity.


You're right, these other factors are significant, but remember that
friction is proportional to the square of the speed, so the faster you
go, the more significant it bacomes.

--
Ace (brucedotrogers a.t rochedotcom)
Ski Club of Great Britain - http://www.skiclub.co.uk
All opinions expressed are personal and in no way represent those of the Ski Club.
  #23  
Old January 24th 05, 11:14 AM
Sammy
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I thought friction was proportional to the force normal (into) the
slope, which in turn is related to the change of speed (acceleration)
and not speed.

Sammy

  #24  
Old January 24th 05, 11:36 AM
Ben Blaukopf
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Ace wrote:
On Mon, 24 Jan 2005 11:15:39 GMT, Ben Blaukopf
wrote:


Ace wrote:



I'd humbly suggest that if you were doing so in a high-speed situation
you would be losing a lot more than that. The faster you're going, the
(exponentially) greater the friction forces on the ski.


Yes, but it's not friction.

There are several things going on.

a) Friction. Completely independent of speed. Once you're slipping,
you're slipping, and it makes no difference how fast you're going.
b) Displacement of the snow ahead of the ski. Assuming the snow
displaces in the same way at high speed as at low speed, then the
force here is constant too.
c) Compressing the snow at the edge of the ski, causing the carve.
The snow will compress when carving, and by compressing it we
will slow down. The amount the snow compresses depends on how
fast we are going. and varies as the square of velocity.



You're right, these other factors are significant, but remember that
friction is proportional to the square of the speed, so the faster you
go, the more significant it bacomes.


It's not. Friction is independent of speed. Do you mean air resistance?

Ben
  #25  
Old January 24th 05, 11:37 AM
Ben Blaukopf
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Ben Blaukopf wrote:
Ace wrote:

On Mon, 24 Jan 2005 11:15:39 GMT, Ben Blaukopf
wrote:


Ace wrote:




I'd humbly suggest that if you were doing so in a high-speed situation
you would be losing a lot more than that. The faster you're going, the
(exponentially) greater the friction forces on the ski.


Yes, but it's not friction.

There are several things going on.

a) Friction. Completely independent of speed. Once you're slipping,
you're slipping, and it makes no difference how fast you're going.
b) Displacement of the snow ahead of the ski. Assuming the snow
displaces in the same way at high speed as at low speed, then the
force here is constant too.
c) Compressing the snow at the edge of the ski, causing the carve.
The snow will compress when carving, and by compressing it we
will slow down. The amount the snow compresses depends on how
fast we are going. and varies as the square of velocity.




You're right, these other factors are significant, but remember that
friction is proportional to the square of the speed, so the faster you
go, the more significant it bacomes.


It's not. Friction is independent of speed. Do you mean air resistance?


Ah, I've missed a trick having looked at Sammy's answer. Apologies, I
misunderstood you. You're right.

Ben
  #26  
Old January 24th 05, 11:44 AM
Ace
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On Mon, 24 Jan 2005 12:37:57 GMT, Ben Blaukopf
wrote:

Ben Blaukopf wrote:
Ace wrote:


You're right, these other factors are significant, but remember that
friction is proportional to the square of the speed, so the faster you
go, the more significant it bacomes.


It's not. Friction is independent of speed. Do you mean air resistance?


Ah, I've missed a trick having looked at Sammy's answer. Apologies, I
misunderstood you. You're right.


Am I? I may have been thinking of drag, yes, but that's just another
form of friction, innit? I did try and answer Sammy's post but
realised that my physics wasn't really up to it. Can you elucidate
further?

--
Ace (brucedotrogers a.t rochedotcom)
Ski Club of Great Britain - http://www.skiclub.co.uk
All opinions expressed are personal and in no way represent those of the Ski Club.
  #27  
Old January 24th 05, 12:28 PM
Ben Blaukopf
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Ace wrote:
On Mon, 24 Jan 2005 12:37:57 GMT, Ben Blaukopf
wrote:


Ben Blaukopf wrote:

Ace wrote:



You're right, these other factors are significant, but remember that
friction is proportional to the square of the speed, so the faster you
go, the more significant it bacomes.


It's not. Friction is independent of speed. Do you mean air resistance?


Ah, I've missed a trick having looked at Sammy's answer. Apologies, I
misunderstood you. You're right.



Am I? I may have been thinking of drag, yes, but that's just another
form of friction, innit? I did try and answer Sammy's post but
realised that my physics wasn't really up to it. Can you elucidate
further?


OK, friction isn't the same as drag. You can have one without the other.
I was thinking of friction on the skis. I've got no idea whether air
resistance or friction is a more significant force here. Hmmm. On ice,
I guess it would be air resistance.

In order to work out friction you need to know the force exerted by the
snow on the skier, normal (perpendicular) to the surface. The frictional
force is then the coefficient of friction [1] multiplied by the normal
force.

The normal force has two components. One acts opposite to the weight
preventing the skier from simply falling straight through the snow, and
the second exists because we're moving in a curve (i.e. carving!). In
order to move in this curve, the snow must exert a force on us, equal to
mvv/r where v is our speed, and r is the radius of the carve, which we
can assume to be
constant (it isn't, of course but it doesn't really matter).

So the normal force equals mg i + mvv/r j (i and j being unit vectors
perpendicular and parallel to the slope). This force must act
perpendicular to the surface of our skis and the snow (provided we're
not slipping at all). So the frictional force is proportional to
sqrt(g^2 + v^4/r^2).

g is 10 (ish), let's say v is 13 (that'd be 30mph) and we're carving
with a radius of say 15m. It turns out that the weight and the "carving
force" are pretty much equal. The friction isn't proportional to the
square of the velocity, but the square of the velocity does figure in
the final equation.

F = um sqrt(g^2 + v^4/r^2)
u is coefficient of friction
m is mass in kilograms
g is gravitational force in N/kg
v is speed of skier in m/s
r is radius of carve in m

[1] The coefficient has two values, static and dynamic. As far as we're
concerned, we want the dynamic one, and it's constant.
  #28  
Old January 24th 05, 03:36 PM
John Ricketts
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[1] The coefficient has two values, static and dynamic. As far as we're
concerned, we want the dynamic one, and it's constant.


but IS IT?

What is the sliding mechanism? does the snow local to the underside of the
ski melt (similar to regelation under an ice skate)? If so then there would
be a viscous friction component (albeit small), which would indeed be speed
dependent and Ace would be right! ;-)



"Ben Blaukopf" wrote in message
...
Ace wrote:
On Mon, 24 Jan 2005 12:37:57 GMT, Ben Blaukopf
wrote:


Ben Blaukopf wrote:

Ace wrote:



You're right, these other factors are significant, but remember that
friction is proportional to the square of the speed, so the faster you
go, the more significant it bacomes.


It's not. Friction is independent of speed. Do you mean air resistance?


Ah, I've missed a trick having looked at Sammy's answer. Apologies, I
misunderstood you. You're right.



Am I? I may have been thinking of drag, yes, but that's just another
form of friction, innit? I did try and answer Sammy's post but
realised that my physics wasn't really up to it. Can you elucidate
further?


OK, friction isn't the same as drag. You can have one without the other.
I was thinking of friction on the skis. I've got no idea whether air
resistance or friction is a more significant force here. Hmmm. On ice,
I guess it would be air resistance.

In order to work out friction you need to know the force exerted by the
snow on the skier, normal (perpendicular) to the surface. The frictional
force is then the coefficient of friction [1] multiplied by the normal
force.

The normal force has two components. One acts opposite to the weight
preventing the skier from simply falling straight through the snow, and
the second exists because we're moving in a curve (i.e. carving!). In
order to move in this curve, the snow must exert a force on us, equal to
mvv/r where v is our speed, and r is the radius of the carve, which we can
assume to be
constant (it isn't, of course but it doesn't really matter).

So the normal force equals mg i + mvv/r j (i and j being unit vectors
perpendicular and parallel to the slope). This force must act
perpendicular to the surface of our skis and the snow (provided we're
not slipping at all). So the frictional force is proportional to
sqrt(g^2 + v^4/r^2).

g is 10 (ish), let's say v is 13 (that'd be 30mph) and we're carving
with a radius of say 15m. It turns out that the weight and the "carving
force" are pretty much equal. The friction isn't proportional to the
square of the velocity, but the square of the velocity does figure in
the final equation.

F = um sqrt(g^2 + v^4/r^2)
u is coefficient of friction
m is mass in kilograms
g is gravitational force in N/kg
v is speed of skier in m/s
r is radius of carve in m

[1] The coefficient has two values, static and dynamic. As far as we're
concerned, we want the dynamic one, and it's constant.



  #29  
Old January 24th 05, 11:00 PM
KMM
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"NIALLBRUCE" wrote in message
...

b) I wasn't asking about what each colour of run is like. I think that I've
been on every run in almost every condition! (I live and ski in Scotland)


Moreover, two of the instructors at the local 'dry' slope said that they
didn't know anything else to teach me.


Where are you in Scotland and what facilities have you used.What dry slope
do you spend time on? Sounds like you are being short changed here, there is
certainly room for some further skills to be taught.

However, I haven't had any instruction
on an icey run.
I might have missed an important aspect of skiing!? I tried to keep my

body
away from the hill so that my edges had more grip but this didn't seem to

help.
Is my technique wrong?? Any tips for ice??


Don't stand too tall. Try to stay relaxed, I use some lateral ankle flex
towards the hill to keep the edges biting on the icy surface. Why are you
trying to lean away from the hill for more grip, I don't see how that will
help you.


I watched the really good skiiers when I was on holiday so that I could

copy
their actions. I noticed that they were avoiding the regular christie

stops
(slow-downs in my case) by ignoring their speed (and thus reaching their
terminal velocity). I think that I had the ability to ski in this manner

but
other factors (outlined above) made this ambition unsafe. I returned to

the
blue slopes.


Its more than I can put down here but you don't need to ski as fast as
possible, shorter radius turns with good edge control is something you will
move on to in time with experience, its not the only way down as someone
else has posted, skilful skidding is another skill you may need to learn at
some point. You did the right thing, tried it, realised it was not quite
right and retreated to hopefully put in some more practice in a safer area.

Therefore, I can conclude that either:
1. My technique is wrong and I have to learn how to ski on ice /

hard-packed
snow


Yes.

2. The red runs, with the aforementioned snow conditions and crowds, were

not
suitable for those with little or no knowledge of the run (the majority)


Yes. On the other hand the only way to find out what skills you are still
lacking (find your limit, push your comfort zone) and to gain experience is
to try some harder runs. Not all reds will be like race tracks.

3. It is acceptable to ski slowly on red runs.


Yes, obvious safety point is not to spend time doing slow traverses when
others are bombing down the centre. If it is busy on the slope maybe stand
at the side for a few minutes until the crowds quieten down. The slopes are
for all. Some of the aforementioned fast skiers would have been fast because
they had no other skilful way of getting down the piste. Point and go
skiing. Skiing flat out on a busy piste is just asking for trouble.

What's the general opinion?
I'd really like to know the answer because if number 1 applies (and I

fully
accept that it might), then I really must get some lessons!!


You never stop learning. Its sounds to me that you have plenty of room for
improvement, that's not meant as a slight just that from reading your other
posts it appears you are still in the ascendancy of skiing skills/knowledge.
It takes time.

I think you have th right approach to the whole business, asking questions,
taking advice, think about the next step, trying to ski on varying pistes
and conditions.

K.


 




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