what's here
3 contact points
formulas (normal situation)
2 contact points
formulas (special case of high muscular torques)
see also
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. 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 muscular torque applied thru the boot cuff (method C)
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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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.
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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.)
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. back to Top |
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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:
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
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
= m g { xR - xS
+
μR
zS } / { (xR
- xB) - (μB
-
μR)
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.
xB is the
horizontal position of the heel-brake pad against the ground (measured
in the direction opposite to the skater's forward motion).
xR is the
horizontal position of the effective center of ground contact of the wheels of the
rear skate (measured backward from the heel-brake pad xB).
xS is the
horizontal position of the center-of-mass of the skater's body (measured
backward from the heel-brake pad xB).
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.
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
- xB) - (μB
-
μR)
zS }
simplified formulas
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.
It is also convenient to set xB = 0,
which yields these simpler formulas:
wB = m g (xR - xS)
/ (xR -
μB
zS)
wR
= mg (xS -
μB
zS) / (xR
- 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 an amazing
balance control.
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stability of denominator
Consider the denominator of the main
formulas. Let
D = (xR
- xB) - (μB
-
μR)
zS If the skater allows
D = 0, then the balance control goes
unstable. While this could be avoided by keeping
D < 0, I think it makes more sense
for the skater to chose a strategy of keeping
D > 0, 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.
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
get to around twice 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 important to retain some
significant weight-support through the rear skate. back to Top |
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derivation notes
Some key assumptions for deriving the main
2-contact-points formula
above:
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
(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. back to Top |
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