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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:
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:
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:
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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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. 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
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. back to Top |
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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. back to Top |
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