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Up and Down Acceleration
- Thread starter Mike King
- Start date
Thank you.
Fortunately, progressive flex boots provide a nice cue for this.
yes, actually
Thread Starter
TS
Wow. Just Wow. That is a completely new way of thinking about flexion in the finish of the turn that I had never considered. Can I triple like this?
But that is in the horizontal plane, i.e. perpendicular what you brought up in this thread...
It is almost the opposite of what I think is fun when skiing, I want a lot of redirection and a lot of differences in force magnitude. I can also enjoy turns with consistent forces, but then they better be brushed in order to not lose speed control.
About that averaging....
First a little ground work:
1) When you just stand still on the ground, there is a gravity force pulling you down, and a normal force from the snow pushing you up. As you are not accelerating (your velocity is constant at 0 m/s), those two forces add up to 0 force.
2) If you stand there and flex and extend your legs, you go up and down. Gravity is always pulling you down with a force of your mass (m) times acceleration due to gravity (g), but you change how much you push the Earth down with your legs, and thereby change how much the Earth pushes you back up. When you push the Earth down with your legs, the earth pushes you back up just as hard. In effect you are pushing yourself up.
3) When you push yourself up harder than gravity is pulling you down, you have a net (adding both forces together) up-force. and up-acceleration. When you push yourself up less hard than gravity is pulling you down you have a net down force and down acceleration. The direction of the acceleration only depends on the force, how hard you push; it doesn't matter if your going down or up at the time of the push.
Now for the averaging:
1) For the sake of keeping things simple, let's say you start at a given height from the ground with no vertical velocity. If during your run you push more than gravity pulls you down, you will end up with a positive (which we will define as away from the ground) velocity. If you during your run, you push less than gravity pulls you will end up with a negative velocity (which we define as towards the ground).
2) Since you end up at the same height from the ground (compare any slight difference in your height with the vertical you've just descended to see how true that is), your average velocity on your run (either average over distance or average over time) will be total displacement (a fancy word for distance with direction) travelled /total time taken = 0.
Therefore, during your run, you have pushed yourself up just as much as gravity has pulled you down.
You can do that by pushing with the constant force of gravity for the whole run, OR THE BETTER CHOICE by pushing more at some times and less at other time. Choosing when to push harder and when to "be lighter" allows you to have more force maintaining your grip in the belly of the turn where it's needed, and less in transition where less is needed.
There is only so much pushing up (vertical force) to go around; it has to average out to balance gravity in the end. Don't spend all your vertical force at the wrong time in the wrong place.
First a little ground work:
1) When you just stand still on the ground, there is a gravity force pulling you down, and a normal force from the snow pushing you up. As you are not accelerating (your velocity is constant at 0 m/s), those two forces add up to 0 force.
2) If you stand there and flex and extend your legs, you go up and down. Gravity is always pulling you down with a force of your mass (m) times acceleration due to gravity (g), but you change how much you push the Earth down with your legs, and thereby change how much the Earth pushes you back up. When you push the Earth down with your legs, the earth pushes you back up just as hard. In effect you are pushing yourself up.
3) When you push yourself up harder than gravity is pulling you down, you have a net (adding both forces together) up-force. and up-acceleration. When you push yourself up less hard than gravity is pulling you down you have a net down force and down acceleration. The direction of the acceleration only depends on the force, how hard you push; it doesn't matter if your going down or up at the time of the push.
Now for the averaging:
1) For the sake of keeping things simple, let's say you start at a given height from the ground with no vertical velocity. If during your run you push more than gravity pulls you down, you will end up with a positive (which we will define as away from the ground) velocity. If you during your run, you push less than gravity pulls you will end up with a negative velocity (which we define as towards the ground).
2) Since you end up at the same height from the ground (compare any slight difference in your height with the vertical you've just descended to see how true that is), your average velocity on your run (either average over distance or average over time) will be total displacement (a fancy word for distance with direction) travelled /total time taken = 0.
Therefore, during your run, you have pushed yourself up just as much as gravity has pulled you down.
You can do that by pushing with the constant force of gravity for the whole run, OR THE BETTER CHOICE by pushing more at some times and less at other time. Choosing when to push harder and when to "be lighter" allows you to have more force maintaining your grip in the belly of the turn where it's needed, and less in transition where less is needed.
There is only so much pushing up (vertical force) to go around; it has to average out to balance gravity in the end. Don't spend all your vertical force at the wrong time in the wrong place.
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Thread Starter
TS
Ok. Just had a tech talk with @tomgellie and I think I get it -- I need to resist the force and take the impulse across the hill.
Thread Starter
TS
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I get it now. Very cool.
I finally just read through this. It makes sense, and can be remembered. You have explained what and why with clarity. Thanks.
Visual representation:
At this point there's basically no GRF as there's no C and G being applied to the snow.
Apex and after - plenty of C and G. The bigger the better in terms of creating GRF to redirect skier across the hill. More G (from arresting the vertically down movement) allows more C whilst maintaining platform angle.
So a question: how long does this effect last? SL turns are all over in around 0.8 sec. What about medium and wider radius turns? @Jamt ?
At this point there's basically no GRF as there's no C and G being applied to the snow.
Apex and after - plenty of C and G. The bigger the better in terms of creating GRF to redirect skier across the hill. More G (from arresting the vertically down movement) allows more C whilst maintaining platform angle.
So a question: how long does this effect last? SL turns are all over in around 0.8 sec. What about medium and wider radius turns? @Jamt ?
It doesn't last very long, and the more force you exert the shorter it lasts.
The usual statement F=ma or F=M(dv/dt), or breaking it down into exceedingly small intervals F=M delta V/ (delta t) can be taken as F(delta t) = delta V
Force times time gives you your velocity change. More force less time, or less force more time.
Example:
If we want to talk about pushing down with two gs (one to counter gravity and one to accelerate you up, effectively doubling your traction), then acceleration is 1 g based on net force.
a is acceleration
V is velocity Vo is starting velecity
D is distance Do is starting distance, set the start point at 0.
doing the math starting from V= 0.
a=g= (32 ft/s)/s most places we ski
V=Vo+at
d = Do +1/2(at)t
Lets say you have about a foot to move up and down,
1 =(1/2)(32 ft/s/s)( t)(t)
t = 0.25 seconds to move up a foot, then you probably want to move back down.
Use
d=(1/2)(a)t^2 or t = the square root of (2d) to find the timing for more or less force (more or less acceleration) and more or less range.
The extra grip is only temporary.
I'll let someone else check for errors in my math; I have to exercise now
. The point is the same, despite any possible errors above: more force over less time or less force over more time; spend your force wisely.
The usual statement F=ma or F=M(dv/dt), or breaking it down into exceedingly small intervals F=M delta V/ (delta t) can be taken as F(delta t) = delta V
Force times time gives you your velocity change. More force less time, or less force more time.
Example:
If we want to talk about pushing down with two gs (one to counter gravity and one to accelerate you up, effectively doubling your traction), then acceleration is 1 g based on net force.
a is acceleration
V is velocity Vo is starting velecity
D is distance Do is starting distance, set the start point at 0.
doing the math starting from V= 0.
a=g= (32 ft/s)/s most places we ski
V=Vo+at
d = Do +1/2(at)t
Lets say you have about a foot to move up and down,
1 =(1/2)(32 ft/s/s)( t)(t)
t = 0.25 seconds to move up a foot, then you probably want to move back down.
Use
d=(1/2)(a)t^2 or t = the square root of (2d) to find the timing for more or less force (more or less acceleration) and more or less range.
The extra grip is only temporary.
I'll let someone else check for errors in my math; I have to exercise now
. The point is the same, despite any possible errors above: more force over less time or less force over more time; spend your force wisely.
Nice vid, but I think if they pressured the ski ruffly they'd lose their line
Loved the banana line though. I had a race coach in the 90's that talked about taking the banana shape through the turn. Think like a Banana. Ski like a Banana. Be the Banana. we use to say.
good times.
Correct, but to get it into perspective we can consider the entire turn.It doesn't last very long, and the more force you exert the shorter it lasts.
The usual statement F=ma or F=M(dv/dt), or breaking it down into exceedingly small intervals F=M delta V/ (delta t) can be taken as F(delta t) = delta V
Force times time gives you your velocity change. More force less time, or less force more time.
Example:
If we want to talk about pushing down with two gs (one to counter gravity and one to accelerate you up, effectively doubling your traction), then acceleration is 1 g based on net force.
a is acceleration
V is velocity Vo is starting velecity
D is distance Do is starting distance, set the start point at 0.
doing the math starting from V= 0.
a=g= (32 ft/s)/s most places we ski
V=Vo+at
d = Do +1/2(at)t
Lets say you have about a foot to move up and down,
1 =(1/2)(32 ft/s/s)( t)(t)
t = 0.25 seconds to move up a foot, then you probably want to move back down.
Use
d=(1/2)(a)t^2 or t = the square root of (2d) to find the timing for more or less force (more or less acceleration) and more or less range.
The extra grip is only temporary.
I'll let someone else check for errors in my math; I have to exercise now
When you have accelerated up for 0.25 seconds you will have some up speed. If you remove the pressure completely it will take you another 0.25 seconds until the speed is zero again.
Then it will take another 0.25 seconds to again reach the same speed but on the way down. If you again set the pressure so you have 1g up acceleration it will take another 0.25 seconds until you have zero speed and you are back where you started.
So, in total we rise and fall 2 feet, and it will take 1 second for an entire turn.
Of course it is impossible to set and remove pressure instantaneously, so in one second you will rise and fall less than that.
The following pictures show CoM height and pressure for typical 10 and 13 meter SL turns.
The time axis is in percentage, but a turn was slightly less than a second.
The up-down is about a foot and the "2g up force phase" is just some tenths of a second. about half a second with 1g or more.
Finally, the calculations above assumed a flat ground. The steeper it is the longer it will take to fall a certain distance towards the ground (If it is 90 degrees you will never land). That explains some of the difference in the empirical data and the simplified model above. This also means that when it is steep you need more distinct pressure management, and of course this is also when it is more difficult to get early pressure....
Attachments
I was just going to add that this morning for more perspective, sort of. For simplicity consider up acceleration of 1 g starting half way down for a foot (0.25 seconds) and continues on the way up one foot for 0.25 s, then you let gravity do its thing - down acceleration for 0.25 seconds still on the way up for another foot, and continues for the first foot of the way down.
Once you can picture that, make it a little more real. It's not instantly twice your weight half the time and instantly weightless the other half (nobody has that big of a jerk in them); you change it smoothly, like the graphs from @Jamt , and take slope into account, but that will complicate the math (changes in distance, direction of gravity versus snow surface and vector mechanics). Once you get it, it's easy to translate to slope by feel without doing the math.
Also, you decide where to put the most force and what shape you want that force graph to have.
BTW, thanks for checking my math (I only chose 1 foot to make the math easier).
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Hmmm, I've been thinking about the bold part for the last couple of days and I still cannot wrap my head around it.
I try to be light as much as possible, but once I engage the edges I try to be heavy and really feel I am standing on the outside for better engagement and balance. You see lots of athletes lift the inside leg at the top of the turn as they engage for example. But that is way before my/their CoM is close to the snow. I want/try to engage without having too much edge angle already. If I ski a gs course on a hard surface and stay light till my CoM is close to the ground, let's call it toppling so everyone understands, I usually wash out.
I understand that turn forces are still low at the top of the turn where engagement has just started, compared to the belly of the turn with max. edge angle. But they are still bigger than your weight as far as I understand. If you are engaged and there are turn forces and you add that up, shouldn't that be more than what you weigh? Cause merely standing on a ski , is 1g grf, and as far as I understand turn forces will add to that 1g, no matter how little the turn forces are. So following that logic, I was always under the impression that you need way bigger turn forces than your weight to get your CoM close to the snow. It is easier to get your CoM close to the snow when going 50 km/h with a 12 meter radius, compared to going 45 km/h with 15 meter radius. The latter will have less grf/turn forces and you will be lighter. Hence it will be harder to get your CoM close to the ground. That is at least how I understand it.
Race coaches tell you to edge early, but try to delay edge angle as much as possible and be patient. If I do that in my perception I get more force in the belly of the turn and I feel more grip that way. Which would be my description of the concept 'stay light for as long as possible.' But I do not necessarily feel extremely light (like floating in a transition where it feels more like -1g) before I start to let my CoM drop.
What am I not understanding/doing wrong?
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I think it's more the vertical snow reaction force has to be less than your weight until you need the traction, much less, but not zero.
I think you are understanding and doing things correctly. It is more a matter of what "close to the snow means". You ski with very high edge angles, so for you close to the snow is more or less hip to the snow. Most people never come close to that. Perhaps a better term should be "close enough to the snow" or something like that.
It is more of a mental concept, you should feel light long enough to get high edge angles, but not so long so you actually are close to the snow before you engage at all. It is more gradual than that. With high angles in the turn you probable engage the edges when they have an angle of 45-60 degrees.
Most non-racers have fear issues which will make them engage long before they risk landing on the snow with the hip.
All coaches say that you should have early pressure, but if you try to do that too early you defeat the purpose. For example if you engage with a lot of force when the inclination angle is 45 degrees the turning force and the up force will have the same magnitude, so by definition if you turn a lot then you will also push yourself up a lot, and that will not last for very long. You will not reach high edge angles and you may even wash out before the turn is finished. Alternatively you have to be static for a while to get enough turn, which is even worse (not meaning static balance here
)At 60 degrees the turn force will be almost twice as large as the up force so the extra 15 degrees makes a huge differences in how much turn vs up forces you get.
^ Toppling. Isn't this what Tom Gellie means by learning to and getting comfortable with falling. Not falling into a crash but falling to incline further before the skis catch us and we restore lateral balance.
Thread Starter
TS
@Jamt and @François Pugh thanks for all of this. One question that comes to mind has to do with the endless debate in the other thread: Would the extra movement from and up and over transition result in greater down force than the compact transition?
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