what's here

3 contact points formulas : (based on torques, not body position)

2 contact points formulas : forward

(special case of rear skate lifted completely off the ground)

  • main formula

  • implications

2 contact points formulas : rear-ward

(special case of braking wheel off the ground due to high muscular torques)

3 contact points formulas : (based on body position)

  • main formula

  • implications

see also


3 contact points formulas (based on torques)

In normal heel-brake stopping, there are three simultaneous contact points between the skater and the ground:

  • the front skate's rear wheel (identified by subscript F in these formulas)

  • the pad of the heel-brake (identified by subscript B in these formulas)

  • the rear skate's wheels (identified by subscript R in these formulas)

Total decelerating force is the sum of the braking force from the heel-brake, the rolling or sliding resistance of the rear wheel of the front skate and the wheels of the rear skate, and air resistance of the cross-section of the skater's body. Our focus here is on the contribution of the heel-brake.

For a different analysis focusing on the contribution of body position, see the other 3 contacts formula based on body position further below.

My formula for the braking deceleration force from the heel-brake is:

braking force from heel-brake  =  μB wB

where

μB  is the coefficient of friction of the brake pad against the ground.

wB  is the weight-bearing force between the ground and the brake pad.

wB  =  { τA + (xA - xF) (m g - wR) + τC - k1 + k2 } / { (xB - xF) - k3 }

where

/  is the division operator

τA  is the shin-muscle torque thru the ankle joint (method A)

(xA - xF) (mg - wR)  is the torque from the skater's total body weight relative to the axis of ankle joint. Sometimes it may be convenient to identify this quantity as τB (method B).

τC  is the torque applied thru force against the inside back of the boot cuff (method C) -- torque whose "original" source is body weight, but must be transmitted to the boot by (isometric) muscular force.

xA  is the horizontal position of the axis of the ankle joint when the skate is tilted back in the heel-braking position, measured backward from the ground contact point of the rear wheel of the front skate.

xF  is the horizontal position of the ground contact point of the rear wheel of the front skate.

m  is the total mass of the skater's body (including clothing and attached equipment).

g  is the constant downward acceleration of gravity at the surface of Earth.

mg = the force of the skater's body weight

wR  is the weight-bearing force between the ground and the rear skate, which it is assumed the skater can control by shifting the position of the mass of the upper body.

k1  is a small positive value: the torque from the weight of the front braking foot relative to the axis of the ankle joint.

k2  is typically a very very small positive value (unless the rear skate is used for simultaneous T-stop braking, in which case k2 is very small positive), arising from the inertial force in reaction to the deceleration of the mass of the front braking foot.

xB  is the horizontal position of the heel-brake pad against the ground, measured backward from the ground contact point of the rear wheel of the front skate.

k3  is a very small positive value arising from the inertial force in reaction to the deceleration of the mass of the front braking foot.

simplified formula

If the wheels of the rear skate are rolling and not skidding (as they would in a simultaneous T-stop), then their friction is very small compared with the heel-brake pad, and it is helpful to approximate by setting  μR = 0, and since we assume that the front skate's rear wheel is rolling, we can set  μF = 0.  And since k1, k2, k3 are small quantities and not subject to much active control by the skater, it could be helpful to ignore them, which yields this simpler formula:

wB  =  { τA + (xA - xF) (mg - wR) + τC } / { (xB - xF) }

constraints

The main 3-point formula above is valid only if these inequalities hold:

wF  >  0

where wF  is the weight-bearing force between the ground and the front skate's rear wheel. If the front skate's rear-wheel comes up off the ground, then the skater has shifted into the situation covered by the 2 contact points formulas.

wB  >  0

because if there is not positive weight-bearing force through the brake-pad, then we are not really doing any heel-brake stopping.

wR  ≥  0

because the ground contact through the rear skate cannot be used to help pull the skater's mass down toward the ground -- it can only push up against the weight of the skater.

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implications

Here's some key implications of the main formula:

  • The closer the brake pad ground-contact point xB is the front skate's rear wheel xF, the stronger the braking force, because of the (xB - xF) term in the denominator.

  • If the ankle joint is in front of the front skate's rear wheel, then xA - xF < 0, so the gravitational force of body weight mg is working against the force of braking. In that case it makes sense to increase weight-support by the rear skate wR to reduce this negative impact.

  • If the ankle joint is behind the front skate's rear wheel, then xA - xF > 0, so the gravitational force of body weight mg is working to help the force of braking, and so method B is working. In that case it makes sense to decrease weight-support by the rear skate wR, to maximize the contribution of body weight to downward force on the brake-pad.

  • If the ankle joint is even with the front skate's rear wheel, then xA - xF = 0, so there is no contribution from method B -- so any significant braking force will have to come from the muscular torques τA and τC.

  • less weight on the rear skate wR results in higher braking force (other things being the same).  This formula doesn't say how to reduce that weight -- see the other 3-contact formula which deals with body position instead.

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details of small terms

These are the detailed formulas for the small-value terms in the main formula:

k1  =  mQ g (xA - xQ)

k2  =  (zA - zQ) (mQ / m) [ μF m g + (μR  -  μF) wR ]

k3  =  (zA - zQ) (mQ / m) (μB - μF)

where

mQ  is the mass of the front braking foot (including clothing and attached equipment).

xQ  is the horizontal position of the center-of-mass of the front braking foot, when the skate is tilted back in the heel-braking position.

zQ  is the vertical position of the center-of-mass of the front braking foot, when the skate is tilted back in the heel-braking position.

zA  is the vertical position of the axis of the ankle joint when the skate is tilted back in the heel-braking position, measured upward from the ground.

μF  is the coefficient of friction of the front skate's rear wheel rolling on the ground, which is normally very small.

μB  is the coefficient of friction of the brake pad against the ground.

μR  is the coefficient of friction of the rear skate wheels against the ground. Which is normally very small -- unless the rear skate is being used for simultaneous T-stop braking.

derivation notes

Some key assumptions for deriving the main formula above:

  • All torques and forces from the skater's body of outside the front braking foot must be transmitted to the ground either through the rear skate wheels or the ankle joint of the front foot. (This assumption makes the ankle joint a key focus of force/torque analysis.)

  • The skater has full freedom to move the center-of-mass of the body outside the front braking foot to whatever position is needed to provide the desired rear-skate weight support wR and balance the shin-muscle torque through the ankle joint τA and deliver the desired boot-cuff torque τC.

(I think this is reasonable description of a major objective of the skater's neuromuscular control center naturally operates during braking -- and it avoids the complexity of needing to worry about the exact position of the skater's upper body. This assumption is not used for the 2-contact-points formulas.)

  • Air resistance forces have a uniform effect on all parts of the skater's body, including the front braking foot. (This assumption permits us to ignore air resistance in the details of the formulas)

Then some key formulas used as a basis for deriving the main wB formula above are:

skater's total body weight is fully supported by the three ground contact points:

mg  =  wF + wB  + wR

In the equilibrium situation, braking deceleration of the front braking foot must be the same as the braking deceleration of the skater's total body mass (since the foot has a stable connection to the rest of the body), and the braking force on the total body is the sum of the braking forces applied at the three ground contact points. So if bQ is the inertial force of deceleration on the mass of the skater's braking foot mQ, then we have:

bQ / mQ  =  (μFwF + μBwB  + μRwR) / m

and the torque equilibrium equation for the skater's front brake foot, measured relative to the axis of the ankle joint is:

τA + τC + bQ (zA - zQ)  =  wF (xF - xA) + wB (xB - xA) + mQ g (xA - xQ)

From those three formulas, the main wB formula can be derived by straightforward algebraic manipulation.

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2 contact points formulas: front skate + brake

This formula is for the special case where the rear skate is lifted completely off the ground, so the skater's body-weight is supported by only two ground-contact points:

  • the wheels of the front skate (identified by subscript F in these formulas)

  • the heel-brake pad on the front skate (identified by subscript B in these formulas)

In this special case, the distribution of weight-support forces is determined by the horizontal and vertical position of the skater's body, relative to the two ground-contact points. Muscular torques matter only through the way that they influence the position of the center-of-mass of the skater's body.

This formula is also useful for analyzing at least on non-heel-brake stopping method: the normal (rear-ward) T-stop.

My formula for the braking deceleration force from the heel-brake is:

braking force from heel-brake  =  μB wB

where

μB  is the coefficient of friction of the brake pad against the ground.

wB  is the weight-bearing force between the ground and the brake pad.

(Since the rear skate is off the ground, there is no braking force from the rear wheels, so μR is irrelevant.)

wB  =  m g { xS - xF } / { (xB + μB zS) - xF }

where

/  is the division operator

μR  is the coefficient of friction of the rear skate wheels against the ground. Which is normally very small -- unless the rear skate is being used for simultaneous T-stop braking.

xB  is the horizontal position of the heel-brake pad against the ground (measured in the direction opposite to the skater's forward motion).

xF  is the horizontal position of the effective center of ground contact of the rear wheel of the front skate (measured in the direction opposite to the skater's forward motion).

xS  is the horizontal position of the center-of-mass of the skater's body (measured in the direction opposite to the skater's forward motion).

zS  is the vertical position of the center-of-mass of the skater's body (measured upward from the ground).

m  is the total mass of the skater's body (including clothing and attached equipment).

g  is the constant downward acceleration of gravity at the surface of Earth.

 

also

wF  =  m g { (xB + μB zS) - xS } / { (xB + μB zS) - xF }

where

wF  is the weight-bearing force between the ground and the rear wheel of the front skate.

implications

This formula suggests that in order to increase braking force, (other things being the same) the skater's center of mass should be:

  • farther back behind (higher xS), and

  • closer to the ground (lower zS).

since wF  ≥  0, and wB  ≥  0, and the denominator is positive we get the constraint that:

xF  ≤  xS  ≤  xB + μB zS

The right-hand higher inequality means that if the skater's center of mass is too far behind the brake pad, the skater falls over backward.

The nice thing about using these two points of contact is that it's difficult to fall forward, since the left-hand lower inequality means that only you can't allow the center-of-mass to get forward of the front skates rear wheel, or the skater falls forward. The reason it's the skater has a wide range to shift body weight forward is because the center-of-mass moves further forward, it tends to take weight off the brake pad, which reduces the stopping force. So this "forward" 2-point position tends to be self-correcting. Putting both constraints together gives

xF  ≤  xS  ≤  xB + μB zS

 

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2 contact points formulas: brake pad + rear skate

This formula is for the special case where the torque applied thru the ankle joint is so strong that it lifts the front skate's rear wheel off the ground, so the skater's body-weight is supported by only two ground-contact points:

  • the heel-brake pad on the front skate (identified by subscript B in these formulas)

  • the wheels of the rear skate (identified by subscript R in these formulas)

Actually this rarely happens in heel-brake stopping by most skaters -- but is relevant to the analysis of some non-heel-brake stops, such as the "forward T-stop".

In this special case, the distribution of weight-support forces is determined by the horizontal and vertical position of the skater's body, relative to the two ground-contact points. Muscular torques matter only through the way that they influence the position of the center-of-mass of the skater's body.

Total decelerating force is the sum of the braking force from the heel-brake, the rolling or sliding resistance of the wheels of the rear skate, and air resistance of the cross-section of the skater's body. Our focus here is on the contribution of the heel-brake.

My formula for the braking deceleration force from the heel-brake is:

braking force from heel-brake  =  μB wB

wB  =  m g { (xR + μR zS) - xS } / { (xR + μR zS) - (xB + μB zS) }

where

/  is the division operator

μR  is the coefficient of friction of the rear skate wheels against the ground. Which is normally very small -- unless the rear skate is being used for simultaneous T-stop braking.

xR  is the horizontal position of the effective center of ground contact of the wheels of the rear skate (measured in the direction opposite to the skater's forward motion).

We can also calculate

wR  is the weight-bearing force between the ground and the wheels of the rear skate.

wR  =  m g { xS - (xB + μB zS) } / { (xR + μR zS) - (xB + μB zS) }

constraints

The main 2-point formula above is valid only if these inequalities hold:

wB  >  0

because if there is not positive weight-bearing force through the brake-pad, then we are not really doing any heel-brake stopping.

wR  ≥  0

because the ground contact through the rear skate cannot be used to help pull the skater's mass down toward the ground -- it can only push up against the weight of the skater. Of course if  wR = 0  then the entire weight of the skater is being supported by only one point of contact, the heel-brake pad -- which would require amazing balance control.

These two constraints imply that the horizontal position of the skater's center of mass xS must be in between the two limits (xB + μB zS) and (xR + μR zS), without saying which of those two limits must be larger. If xS stays outside those limits then the skater falls over forward or backward.

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implications

stability of denominator

Consider the denominator of the main  formulas. Let 

D  =  (xR + μR zS) - (xB + μB zS)

If the skater allows D = 0, and if xS is "in between" the two limits, then xS must be equal to both of the limits, and the formula is underdetermined. But the limit converges.

The problem is that when D is small, then small variations in xS (or zS) can result in large changes in braking force (due to substantial shifts in weight from one ground contact point to the other) -- so the balance is tricky because there are significany unexpected changes in the skaters speed. And a small variation in xS (or zS) could move the center of mass outside the two limits -- so the skater starts to fall over backward or forward. (Actually falling forward puts the skater into the different forward 2-contact mode, which has very different characteristics for the situation of the center-of-mass moving forward).

While this could be avoided by keeping D < 0, if μB > μR, (friction of brake pad is much larger than friction of rear wheels -- which is typically the case), I'm thinking it makes more sense for the skater to chose a strategy of keeping D > 0 (substantially greater than zero),  because this fits with having a longer "wheel-base" distance (xR - xB) between the two ground-contact points, as is allowed by this inequality:

(xR - xB) > (μB - μR) zS

Allowing the choice of longer "wheel-base" distances seems like it should be more stable for controlling balance when uneven pavement conditions are encountered.

So for the remainder of this analysis of implications, I will assume that D > 0, that is, the rear skate is substantially out behind the brake pad.

maximum braking with rear skate rolling

In the normal situation (where there is no simultaneous T-stopping by the rear skate), we may assume that  μR = 0, and the braking force is maximized with nearly all of the skater's body weight supported by the brake pad, so in the limit of control,

braking force    μB m g

which is obtained as the skater's center of mass approaches some position on a diagonal line which passes through the front skate's brake pad with a slope of 1 / μB  that satisfies this equation:

(xS - xB) / zS  =  μB

so if the effective coefficient of friction μB for the brake pad is around 0.5, then the skater's center of mass must be around half as far back behind the brake pad as it is off the ground. The higher the coefficient of friction, the farther back must go the skater's center-of-mass. And with μR = 0 and D > 0, the formula for wB implies that the position of the rear skate must be back still farther -- behind the skater's center-of-mass.

maximum braking with rear skate skidding in simultaneous T-stop

When the rear skate is skidding instead of rolling, we must take μR > 0, and try to maximize the total braking force from both ground-contact points:

B  =  μBwB  + μRwR

Taking the convenience of setting xB = 0, and substituting for wB and wR yields a total braking force

B  =  m g { μBxR - (μB - μR) xS } / { xR - (μB - μR) zS }

If we assume that  μB > μR > 0 (the plausible situation where the friction of the heel-brake is greater than the friction of rear T-stopping), then differentiating this with respect to xR and setting to zero and solving finds a local maximum value -- which is obtained when the skater's center of mass is positioned on a diagonal line which passes through the front skate's brake pad with a slope of 1 / μB  that satisfies this equation:

xS / zS  =  μB

which is the same as the earlier result for the rear skate rolling, but with the convenient choice of setting  xB = 0. It means that full body weight is supported by the brake pad, and virtually none on the rear wheel.

If we instead assume that the rear T-stop braking has more friction than the heel-brake:   μR > μB > 0, then that  xS / zS  ratio for the center-of-mass position gives a local minimum braking force, which is not what we're looking for. Instead we find the maximum by putting full body weight on the rear skate: by setting wB = 0, which is obtained when the skater's center of mass is positioned on a diagonal line which passes through the rear skate ground contact point with a slope of 1 / μR  that satisfies this equation:

xS  =  xR + μR zS

which implies that  xS > xR , which means that the skater's center-of-mass must be behind the rear skate. Which makes sense in the physics, since all the weight is on the rear skate.

overall braking maximization strategy

So the theoretical strategy for maximizing braking force in the 2-contact-points situation is to position the skater's center-of-mass so that virtually all the skater's body-weight is supported through the ground-contact point which has the higher coefficient of friction. I say "virtually all", because I would expect that in practice  μB > μR, (even if simultaneous rear-skate T-brake stopping is used, since the skate wheels being dragged behind in normal-T-stop mode do not have such a high coefficient of friction) -- so in practice the strategy is to shift virtually all the weight on the front skate's brake pad.

But this places the skater in a position which is on the edge of being out of balance, about to fall forward with the slightest disturbance. So for stable control of balance, it is useful to retain some significant weight-support through the rear skate.

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derivation notes

Some key assumptions for deriving the main 2-contact-points formula above:

  • Air resistance forces have a uniform effect on all parts of the skater's body, including the front braking foot. (This assumption permits us to ignore air resistance in the details of the formulas)

Then some key formulas used as a basis for deriving the main wB formula above are:

The skater's total body weight is fully supported by the two ground contact points:

m g  =  wB  + wR

The inertial deceleration force on the skater's body mass (ignoring air resistance) is

μBwB  + μRwR

The torque equilibrium equation for the skater, measured relative to the heel-brake pad is

m g xS  =  wR xR + (μB wB + μR wR) zS

From those formulas, the main wB and wR formulas can be derived by straightforward algebraic manipulation.

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3 contact points formulas (body position)

In normal heel-brake stopping, there are three simultaneous contact points between the skater and the ground:

F = rear wheel of the front braking skate

B = brake pad

R = rear skate wheels effective center of ground contact.

The purpose of these formulas is to analyze the effect of the skater's body position on heel-braking. But with three points of ground contact, knowing body position is not enough to determine the force at each contact point. So we need to make . . .

another assumption, that the skater's muscles are controlled to maintain a roughly linear relationship between the force thru the braking skate's wheel and the force thru the brake pad:

wF  =  k wB + h

where k and h are constants such that  wF ≥ 0  for  0 wB ≤ mg.

For simplicity we assume that there is no braking by the wheels of the rear skate:

μR  =  0

which leads to

wB  =  { m g (xR - xS)  -  h (xR - xF) } / { (xR - xB)  +  k (xR - xF)  -  μB zS }

implications

The formula gives these implications for body position:

  • forward: more braking force when the skater's body center of mass is more forward, i.e. smaller .

  • higher: more braking force when the skater's body center of mass is higher off the ground, i.e. larger zS.

derivation

The formula for equilibrium of moments (or torques) about an axis thru the skater's center of mass is:

μB wB zS  +  wR (xR - xS)  =  wF (xS - xF)  +  wB (xS - xB)

then substituting  wR = (mg - h) - wB(1 + k)  and solving for wB gives the result.

Note that this formula is equivalent to the 2-point formulas for wB derived earlier:

rear-ward (braking skate rear wheel up off ground) by setting  k = h = 0  and μR = 0.

forward (rear skate wheels all off ground) by setting  k = -1  and  h = mg.

The equivalence to the other 3-contact point formula is tricky to make because that formula "buries" the body position information in the torques and wR term, while this formula "buries" the torques in the h and k constants.

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