Ken Roberts - - Skating

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simple model of propulsive force in skating leg-push

08dec

what's here

  • simplest model

  • components of force at ground contact

  • sideways weight shift

  • future propulsive work

  • vectors defined in xyz coordinate frame

  • dynamic equation of motion

 

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simplest model

I'll start with the simplest model of propulsive force and work

dW  =  (fG n) (n drB)

where

dW  is the increment of propulsive Work. ("work" is a carefully defined concept of physics, objectively measurable, with measurement units of Joules)

fG  is the force applied at the point of ground contact rG.

rG  is the position of effective contact point for transfer of force between the pushing foot and the ground surface.

   is scalar product or "dot" product.

n  is the unit vector along the ground surface, perpendicular to the gliding direction of the foot.

rB  is the current position of the center of mass of the skater's Body.

drB  is the incremental change in position of skater's Body center of mass.

drB  =  v dt,   where

v  is the skater's current velocity. The direction of v changes continually during the stroke-cycle of skating. Usually the current direction of the skater's center-of-mass is not the same as the "aiming" direction  of the skater's foot the gliding along the ground surface.

concepts so far:

  • propulsive Work in skating is performed only in the course of the skater's body moving some distance in some direction.

  • propulsive Work is done by overcoming resistive forces which oppose the motion in that direction.

  • the skating foot can only transmit possibly propulsive forces in the direction of unit vector n. There are always other forces at the point of ground contact rG, but they are either cannot be propulsive (e.g. a force component aimed down into the ground surface), or they cannot be transmitted to the ground (e.g. a force component in the same direction as the gliding of the foot).

  • the direction which the skating foot can transmit propulsive force to the ground is not the same as the direction the skater's body is moving (v) - (unlike with running and walking), so only a portion of force fG transmitted to the ground is effective currently to add to propulsive work. The proportion is determined by angle between the vectors n and v.

components of force at ground contact

Getting into some more useful detail, the force applied at the point of ground contact can be thought of as having three components:

fG  =  fext  +  faim  +  fswe

where

fext  is the "extension" force roughly along the direct line from the skater's hip or center-of-mass rB to the foot-to-ground contact point rG. The articulations of kinesiology which apply this "extension" force include hip extension, knee extension, and ankle plantar flexion.

faim  is the force applied in the "aiming" direction of the skater's foot gliding along the surface of the ground. In normal skating situations, no significant portion of  is transmitted to the ground, so I will ignore it.

fswe  is the "sweeping" force, applied by moving the foot sideways out from underneath its hip, in the direction perpendicular to both fext and faim. This kind of force is often ignored, but the human body has muscles which can apply this force, and winning skaters exploit it effectively. The articulations of kinesiology which apply this "sweep" force include hip abduction, medial hip rotation (if the knee is partly bent), and ankle pronation.

substituting these components into the propulsive force and work model:

dW  =  (fext n) (n v dt)  +  (fswe n) (n v dt)

and the magnitude of the component of fG in the direction of n is

fn  =  fext sin λ  +  fswe cos λ

and get the result

dW  =   fn sin (α η) v dt

and the propulsive component of fG is then

fprop  =  fext sin λ sin (α η)  +  fswe cos λ sin (α η)

where

fn  is the component of fG in the direction of n, and its magnitude | fn | = fn.

fprop  is the propulsive force magnitude -- the portion of force applied to the ground which is effective applied in the direction of current motion v.

v  is the magnitude of the current velocity vector v, the speed of the skater's body center-of-mass in the current direction of motion:  v = | v |.

x  is the direction unit vector of the overall "average" direction of forward motion of the skater's body over the whole stroke-cycle of skate-pushes with both legs - (unlike the direction of v which is the current instantaneous direction of motion, and varies throughout the stroke-cycle).

y  is the direction unit vector along the surface of the ground and perpendicular to x the direction of the overall "average" direction of forward motion of the skater's body over the whole stroke-cycle of skate-pushes with both legs. So y stays constant throughout the whole stroke-cycle of  of skate-pushes with both legs.

z  is the direction unit vector perpendicular to the surface of the ground, up and away from the ground. Note that if the ground surface is sloping, then z is not vertical with respect to gravity.

λ  is the angle of the skater's leg away from vertical, or more precisely the angle of the skater's extension leg-push direction away from vertical. So if the skater's foot is roughly underneath the skater's hip, then λ = 0.

α  is the angle between: (1) the "aiming" direction of the skater's foot gliding along the surface of the ground; and (2) the direction vector x which is the overall "average" direction of forward motion of the skater's body over the whole stroke-cycle of skate-pushes with both legs.

η  is angle between the v and x -- between the overall average direction of forward motion and the current instantaneous direction of motion.

(α η)  is the angle between: (1) the "aiming" direction of the skater's foot gliding along the surface of the ground; and (2) current instantaneous direction of motion v.

sideways weight shift

But it's more helpful for understanding skating propulsion to refine this further and add two further concepts:

ay  is sideways acceleration -- the sideways component of acceleration of the skater's body center-of-mass in the y direction.

fQext is the force which the "extension" muscles could sustain repeatably without any help from sideways acceleration ay. The magnitude of this sustainable force is fQext.

Actually skating without help of sideways acceleration requires moving the shoulders opposite to the direction of the hips: The hip moves away from the pushing direction of the foot, but the shoulders move toward the pushing direction of the currently pushing foot.

fQswe is the force which the "sweep" muscles could sustain repeatably without any help from sideways acceleration ay. The magnitude of this sustainable force is fQswe.

m  is the mass of the skater's body.

This allows us to write:

fn  =  fQext sin λ  +  fQswe cos λ  +  ky m ay cos α

where

ky < 1  is the effectiveness of transmitting sideways kinetic energy of (mostly) the upper body into force the contact with the surface of the ground. It must be less than 1 because this force must be transmitted thru the muscles of the leg, so it interferes with the force which those muscles could otherwise generate without that extra load. The contribution of the ay force is less than what it would have been if it had been transmitted just on its own, but even with the offset from interference with other loads, the total result is higher power output, so sideways weight shift ay makes a net positive contribution.

concepts

  • If you believe that sideways weight shift is important for skating (and it really is), then you should believe that the ay term is important to recognize explicitly.

  • In the earlier formulas for fn and fprop, the ay force was already there, but "buried" in the fext and fswe force components. The  force is transmitted through the "extension" (fext) and "sweep" (fswe) muscles, so in an important sense it is part of their force. But it's a level of force which they could not sustain at that rate of delivering work.

  • What the  force does is enable the same muscles to operate at a different section of their force-speed-power curve, a section which is more favorable for delivering and sustaining higher force at a higher power rate.

  • Actually there should also be other similar terms for ax and az. But for many styles of skating technique those quantities self-cancel over the whole stroke-cycle -- including the technique styles used by most of the fastest skaters in real race situations. For technique styles where they're not self-cancelling they're usually trickier to analyze than ay.

  • ay is roughly proportional to the maximum sideways speed of the skater's body mass. The reason is that sideways speed has to go to 0 roughly in the middle of the leg-push, because it the body mass has to be stopped before it can be started and accelerated toward the opposite side for the next leg-push toward the side. So the higher the maximum sideways speed, the quicker it has to be de-accelerated and stopped, then quicker re-accelerated to attain that same maximum speed in the opposite sideways direction.

Therefore a key determinant of the propulsive contribution if sideways weight shift is the quickness of the sideways move, not the size of the side-to-side motion.

  • the proportional contributions of sweep force fQswe and sideways weight shift ay tend to be larger than we might expect, because they're contribution to fn is based on the cosine function (instead of the sine function for extension force)

propulsive portion of force currently being transmitted to the ground is:

fprop  =  fQext sin λ sin (α η)

 +  fQswe cos λ sin (α η)

 +  ky m ay cos α sin (α η)

future propulsive work

Force applied thru the foot also makes a contribution to propulsive work in the next leg-push, by building sideways kinetic energy.

KEy  =  m (max vy)2

where

KEy  is the sideways kinetic energy

max vy  is the maximum sideways speed of the skier's body mass (averaged over all its parts).

The portion of current force which goes into this future propulsive energy and work is:

kKE cos α fn

The portion which is successfully transmitted into propulsive force in the next leg-push must by multiplied by ky cos α.  And if we take kKE = ky  (which is not unreasonable), we get this as the as the portion of fn which is effective for future propulsive work:

ffut-prop  =  ky2 cos2 α fn

adding together the current and future contribution to propulsion:

fprop  +  ffut-prop  =  [sin (α η)  +  ky2 cos2 α] fn

concepts

  • so the skating leg-push is much more "effecient" for propulsion than might be guessed -- provided that sideways weight transfer is exploited strongly and skillfully.

vectors defined in xyz coordinate frame

v  =  (cos η, sin η, 0) v

drB  =  (cos η, sin η, 0) drB

"aiming" direction of foot gliding  =  (cos α, sin α, 0)

faim  =  (cos α, sin α, 0) faim

fext  =  (sin α sin λ, cos α sin λ, cos λ) fext

fswe  =  (sin α cos λ, cos α cos λ, sin λ) fswe

n  =  (sin α, cos α, 0)

fn  =  (sin α, cos α, 0) fn

g  =  (sin γ, 0, cos γ) g

(so if γ = 0, then g = gz)

component of gravitational force in the direction of v

m g v / v  =  mg sin γ cos η 

effective slope angle in the direction of v

angle  =  cos−1[(g × y) v / (g v)]  =  cos−1(cos γ cos η)

(note that overall slope angle  is defined in the direction of x)

where

g × y  =  (cos γ, 0, sin γ) g

(vector product of g with y, or "cross" product)

dynamic equation

Here's the basic dynamic equation for acceleration of the skater's body mass:

m a  =  m g  +  fG  +  fresist

where

g  is the acceleration vector of gravity, which on flat ground is in the z direction:  g  =  gz.

fresist  is the sum of the resistive forces, usually sliding friction, air and wind resistance, and (if carefully defined) hill slope.

From the perspective of propulsive work and power, we're interested mainly in the components of force in the direction of current forward motion   dr  =  v dt.

m a dr  =  m g dr  +  fn dr  +  fresist dr

calculating in more detail

m g dr  =  mg sin γ cos η dr

where

γ  is the angle of the slope of the hill in the direction of overall forward travel thru the whole stroke-cycle with pushes by both legs.

next

fn dr  =  sin (α η) fn dr

and

fFriction dr  =  kF (fG z) cos α dr

fWind dr  =  kW cbody | w v |2 cos ω dr

where

kF  is the coefficient of sliding friction (which could vary during the stroke cycle as the pressure distribution along the foot changes -- and the coefficient of friction definitely varies with different surface conditions)

w  is the wind velocity and direction.

ω  is the relative angle between the wind direction and the current direction of forward motion v.

cbody  is the cross-sectional area of the skater's body in its current configuration.

kW  is the aerodynamic factor for the skater's body configuration.

Expressing everything as incremental Work we get (roughly)

dW  =  m a dr  =  {

    fQext sin λ sin (α η)  . . . . . . . [extension muscle push]

fQswe cos λ sin (α η)  . . . . . . [sweep muscle push]

ky m ay cos α sin (α η)  . . . . [sideways weight shift]

  kF (fG z) cos α  . . . . . . . . . . [sliding friction]

  kW cbody | w v |2 cos ω  . . . . [air + wind resistance]

  m g sin γ cos η  . . . . . . . . . . . [gravity on hill slope]

} dr

concept

  • at steady state, total external Work done on the skater's body mass must integrate to 0 over the entire stroke-cycle pushing with both legs, and the first three terms for propulsive Work must integrate to the same as the last three terms for resistive Work.

where does the power get "wasted"?

first written: April  2008

I keep reading posts on various forums about "efficiency", which makes me think about how to analyze it.

Here's a try:

aerobic power is generated mainly from carbohydrate fuel and oxygen.

about 75% of the energy in the chemical bonds in the carbohydrate fuel is lost in this process -- the resulting work of the muscle fibers contracting is only 25% of the energy available. (Also carbon dioxide is released).

Therefore if "not wasting energy" is the goal, just stay home and watch TV. Active life is inefficient.

  • A significant percentage of the carbohydrate energy goes to non-muscular processes -- especially for operating the brain.

  • Another portion goes to muscles which can never deliver propulsive work -- especially those for operating the heart and lungs.

What's "left over" after that is available for power from "skeletal" or "peripheral" muscles. Here's some of the things it gets used for:

  • posture: maintaining the basic configuration of the bones and joints, mostly against gravity.

Unlike some machines, we don't just "lock" our bones and joints into a desired configuration -- just holding a stable "isometric" configuration requires muscular effort. It burns calories and uses up some oxygen.

Different motion patterns for propulsion might require more or less energy to be used to maintain the "posture" configurations which are appropriate for that motion. Typically motions in which the posture is more "bent over" require more energy to hold that posture stable.

Some postural configurations which require more energy to sustain, also offer more effective leverage or "gearing" of key muscles or a higher percentage of direct transmission of muscular moves into propulsive work -- (so there are trade-offs)

Some people use more muscular tension than necessary to maintain posture.

Some energy to maintain postural configuration is required for the desire propulsive motion, other energy might be avoidable.

  • balance: small adjustments to keep from falling over.

People with well-practiced specific balance use smaller more accurate corrective moves, which take less work.

Also sometimes a person with better specific balance can sustain a configuration which offers more effective leverage or "gearing" of key muscles or a higher percentage of direct transmission of muscular moves into propulsive work.

Some energy to maintain balance is required for the desire propulsive motion, other energy might be avoidable.

  • recovery moves: needed to complete the stroke-cycle, to bring body parts into a configuration where they can make the same push again.

Some recovery moves themselves add positive propulsive work. Often this depends on timing and other tricky points, since often the attempted propulsive work in a recovery move is self-cancelling. For example, recovering the leg forward is partly propulsive in running, but not in normal walking (with continual ground contact).

Some propulsive muscle moves are "self-recovering". For example pelvis - lower spine rotation in walking and running. If pushing with the Right leg, the rotation of the pelvis about the spinal axis to bring the non-pushing Left hip forward to add propulsive work to the push thru the Right leg also finishes with the Right hip backward. Which is exactly the configuration needed to make a propulsive pelvis - lower spine rotation move for the push of the Left leg.

  • static friction to enable transmitting propulsive work to ground: In some situations, downward force in addition to body weight is needed to have sufficient static friction to transmit a forward-backward force to the ground.

An example might be classic striding in cross-country skiing, especially when climbing a steep hill. Extra downward pushing force is needed at the same time forward-backward force is being applied -- in order to prevent the ski from losing grip and slipping back. This extra force causes the mass of the upper body to move upward in reaction, and can sometimes it has so much upward momentum that both feet are lifted off the ground simultaneously after the leg-push finishes -- so the next leg-push cannot start until time has elapsed for the mass of the upper body to fall down again.

In most situations for most propulsive motions, body weight and the propulsive forces themselves are sufficient to provide sufficient static friction.

Some performers might apply more downward force than necessary, while other more skillful performers might take it closer to the "edge" of only the minimum required.

  • transmit work thru body parts: Sometimes some joints and muscles are held static ("isometric") to transmit the propulsive work from another muscle move to the ground. Or transmit it to the mass of the remainder of the body.

Human muscles and joints are useful in this static transmission role. Especially smaller muscles can sometimes make a larger contribution to overall power by statically transmitting the work from larger muscles, rather than attempting to add active work of their own. This function of "isometric" static transmission does not do propulsive work, but it does consume fuel and oxygen.

  • wasted muscle moves that are unrelated to any propulsive goal.

I guess this is what lots of people mean by "wasted energy" or "inefficient" motion. But it's only one way in which the power generated from fuel and oxygen is not available to add to propulsion work.

Also moves with lots of people think are just "wasted" actually do make a contribution to propulsion (thru a less obvious exploitation of the physics of propulsion), or are necessary to support propulsive contribution of other muscles (in one of the ways described above).

What remains after those uses can be applied to propulsive work. But it's not so simple . . .

mainly because the moves are not in exactly the right direction for propulsion. The ideal propulsive force pushes straight backward against the ground and straight forward against the mass of the body.

But most actual propulsive moves by the human body push diagonally. Either a combination of backward and downward against the ground (as in striding or pole-pushing). Or a combination of backward and sideways against the ground (as in skating) -- well actually the skating leg-push is a diagonal combination of three directions: backward and sideways and downward against the ground.

Only the directional "component" of force which is in the forward-backward direction is directly + immediately added to propulsive work.

Other directional components (downward and sideways) of the pushing force can also contribute to propulsion, but only indirectly, deferred to a later phase of the stroke-cycle, or to the next stroke-cycle.

These deferred forces may result in additional losses of power:

  • they need to be transmitted (again) thru body parts.

If some of these muscles are used in static "isometric" mode, that's a loss, as described above under "transmit".

  • after the force has been transmitted, it's still not in exactly in the ideal  foreward-backward direction. So again only a portion of the force is applied directly and immediately to propulsive work. The remainder must be transmitted again, with further losses as described here.

Again and again and again.

  • may result in time delays in the stroke-cycle.

Since one of the three main drivers of propulsive power is stroke-cycle time, delay means loss of power.

  • may result in other moves being pushed into a less favorable power point on the "muscle force intensity versus muscle speed". curve.

Or some other moves could be pushed into a more favorable power point on their curve.

So even the muscle moves and forces which are not "wasted" still have power losses.

Why does power matter so much?

Because the goal of human propulsion is to move some distance in some amount of time -- so some positive speed is required. The rate of speed is determined by this equation:

[Resistance Force] * [Foreward Speed]  =  [Direct Propulsive Power from muscles]

where [Resistance Force] is the sum of the forces opposing forward motion: like air resistance, or friction against the ground, or if climbing up a hill then gravity is included. Especially for air resistance, the intensity of resistance depends on the overall body posture and to some extent which muscle moves are being used - (so a change to postural configuration which takes less energy to maintain might also increase air resistance -- there are trade-offs.)

Actually it's more complicated, since each of those quantities varies at different times in the stroke-cycle, so the equation has to be sort of "integrated" over the whole stroke-cycle.

It gets more complicated, because

[Resistance Force] depends on [Foreward Speed]

because air resistance gets disproportionately larger at higher speeds.

[Direct Propulsive Power from muscles] depends on [Foreward Speed]

because muscles produce different power levels depending on the speed, sometimes higher, sometimes lower -- but above some speed (different for each muscle move) the power capability only declines with speed.

Yet another complication is that how much power each muscles can deliver depends on the duration of time it needs to sustain that power level.

At higher speeds, as the performer tries to raise the speed rise even higher, [Resistance Force] tends to only rise and [Direct Propulsive Power] tends only to fall. So there's an upper limit.

But despite all that complexity, more Power generally means higher speed -- in getting from one place to another.

 

Chasing efficiency is a complicated game.

Trying to use "efficiency" to increase speed is a very complicated game.

more . . .

see also

 

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