Math Cycling Problem

Phil Holman said:

A rider is traveling at 37 ft/sec and starts to coast. The rider's
initial deceleration is 1 ft/sec^2 which varies with the square of the
velocity. How long will it take for the rider to decelerate from 37 to
29 ft/sec and how far will the rider have traveled?

Just plug it into Excel for a mathematically inelegant but adequate
solution. It's actually a silly problem because, rather than giving us
a Cd and a frontal area, they've used a kludge to tell us that 37 ft/sec
happens to give a retardation of 1 ft/sec^2 based on the rider's shape
and size.

"Zog The Undeniable" <[email hidden]> wrote in message
news:43c95880.0@entanet...

Quoted message said:

Phil Holman said:

A rider is traveling at 37 ft/sec and starts to coast. The rider's
initial deceleration is 1 ft/sec^2 which varies with the square of
the
velocity. How long will it take for the rider to decelerate from 37
to
29 ft/sec and how far will the rider have traveled?

Just plug it into Excel for a mathematically inelegant but adequate
solution. It's actually a silly problem because, rather than giving
us a Cd and a frontal area, they've used a kludge to tell us that 37
ft/sec happens to give a retardation of 1 ft/sec^2 based on the
rider's shape and size.

Inelegant is correct, the expectation is to use calculus. For your
suggestion you'd need the mass of bike plus rider as well as the Cd and
frontal area, and then the rolling resistance and slope to make it even
more realistic. Who is "they" by the way?

Phil H

On Sat, 14 Jan 2006 11:28:38 -0800, "Phil Holman"

piholmanc@yourservice said:

How long will it take for the rider to decelerate from 37 to
29 ft/sec

(37²/29)-37 seconds [~10.2 seconds]

Quoted message said:

and how far will the rider have traveled?

37²*(ln|37²/29|-ln|37|) feet [~333.5 feet]

"Greg Berchin" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

On Sat, 14 Jan 2006 11:28:38 -0800, "Phil Holman"

piholmanc@yourservice said:

How long will it take for the rider to decelerate from 37 to
29 ft/sec

(37²/29)-37 seconds [~10.2 seconds]

Quoted message said:

and how far will the rider have traveled?

37²*(ln|37²/29|-ln|37|) feet [~333.5 feet]

Excellent.

a = kv^2
-1 = k(37)^2, so k = -1/37^2
so dv/dt = (-1/37^2)v^2

Separate the variables and take the antiderivative of both sides

Int[-37^2 dv/v^2] = Int dt
37^2(1/v) = t + c
Using (0, 37), c = 37
So t = (37^2/v) - 37
When v = 29, t = (37^2/29) - 37
t = 10.21 seconds

v = 37^2/(t + 37)
s = 37^2 ln[t + 37] + c
Using (0, 0), c = 37^2 ln(37)
So s = 37^2 {ln[t + 37] - ln(37)}
s = 37^2 ln [(t + 37)/37]
when t = 10.21, s = 37^2 ln[(10.21 + 37)/37]
s = 333.52 ft

Phil H

On Sat, 14 Jan 2006 16:57:33 -0800, "Phil Holman"

piholmanc@yourservice said:

Excellent.

What did I win? A new Waterford would be nice.

Phil Holman said:

Quoted message said:

Quoted message said:

A rider is traveling at 37 ft/sec and starts to coast. The rider's
initial deceleration is 1 ft/sec^2 which varies with the square of the
velocity. How long will it take for the rider to decelerate from 37 to
29 ft/sec and how far will the rider have traveled?

Zog The Undeniable said:

Quoted message said:

Just plug it into Excel for a mathematically inelegant but adequate
solution.

Quoted message said:

It's not that hard to do with pencil and paper. It's a
shame for us colonials to see that UK standards
have gone so far downhill.

Just plug it into Mathematica, Maple or Matlab instead.

Donald Munro said:

Just plug it into Mathematica, Maple or Matlab instead.

% perl -e 'printf "%-8s %-8s %-8s\n","t","s","a";$dt=0.001;$v=37;$a=1;$vf=29;while(1){$t+=$dt;$vold=$v;$v-=$dt*($a=(($v-$a*$dt/2)/37)**2);$s+=($v+$vold)/2*$dt;if (($v<=>$vf)!=($vold<=>$vf)){printf "%-8g %-8g %-8g\n",$t-$dt*($v-$vf)/($v-$vold),$s,$a;exit}}'
t s a
10.2069 333.522 0.614327

Dan Connelly said:

s 333.522

I knew it was a track question.

--
E. Dronkert

"Dan Connelly" <d_j_c_o_n_n_e_l@i_e_e_e.o_r_g> wrote in message
news:[email hidden]...

Quoted message said:

Donald Munro said:

Just plug it into Mathematica, Maple or Matlab instead.

% perl -e 'printf "%-8s %-8s
%-8s\n","t","s","a";$dt=0.001;$v=37;$a=1;$vf=29;while(1){$t+=$dt;$vold=$v;$v-=$dt*($a=(($v-$a*$dt/2)/37)**2);$s+=($v+$vold)/2*$dt;if
(($v<=>$vf)!=($vold<=>$vf)){printf "%-8g %-8g
%-8g\n",$t-$dt*($v-$vf)/($v-$vold),$s,$a;exit}}'
t s a
10.2069 333.522 0.614327

I should have excluded Dan from even attempting the solution :-)

1*29^2/37^2 = .614317.............was that a typo?

Phil H

Donald Munro said:

Quoted message said:

Just plug it into Mathematica, Maple or Matlab instead.

Dan Connelly said:

% perl -e 'printf "%-8s %-8s %-8s\n","t","s","a";$dt=0.001;$v=37;$a=1;$vf=29;while(1){$t+=$dt;$vold=$v;$v-=$dt*($a=(($v-$a*$dt/2)/37)**2);$s+=($v+$vold)/2*$dt;if (($v<=>$vf)!=($vold<=>$vf)){printf "%-8g %-8g %-8g\n",$t-$dt*($v-$vf)/($v-$vold),$s,$a;exit}}'
t s a
10.2069 333.522 0.614327

Neanderthal.

Phil Holman said:

A rider is traveling at 37 ft/sec and starts to coast.

There's yer mistake right there! :-)

-Mark

In article <[email hidden]>,

Phil Holman' piholmanc@yourservice said:

"Greg Berchin" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

On Sat, 14 Jan 2006 11:28:38 -0800, "Phil Holman"

piholmanc@yourservice said:

How long will it take for the rider to decelerate from 37 to
29 ft/sec

(37²/29)-37 seconds [~10.2 seconds]

Quoted message said:

and how far will the rider have traveled?

37²*(ln|37²/29|-ln|37|) feet [~333.5 feet]

Excellent.

a = kv^2
-1 = k(37)^2, so k = -1/37^2
so dv/dt = (-1/37^2)v^2

Separate the variables and take the antiderivative of both sides

Int[-37^2 dv/v^2] = Int dt
37^2(1/v) = t + c
Using (0, 37), c = 37
So t = (37^2/v) - 37
When v = 29, t = (37^2/29) - 37
t = 10.21 seconds

v = 37^2/(t + 37)
s = 37^2 ln[t + 37] + c
Using (0, 0), c = 37^2 ln(37)
So s = 37^2 {ln[t + 37] - ln(37)}
s = 37^2 ln [(t + 37)/37]
when t = 10.21, s = 37^2 ln[(10.21 + 37)/37]
s = 333.52 ft

Phil H

This model leads only to solutions with an infinity in
finite time. In this case at t = -37. Do we not have a
model without infinities?

--
Michael Press

Michael Press said:

Quoted message said:

a = kv^2
-1 = k(37)^2, so k = -1/37^2
so dv/dt = (-1/37^2)v^2
Separate the variables and take the antiderivative of both sides

Quoted message said:

Quoted message said:

v = 37^2/(t + 37)
s = 37^2 ln[t + 37] + c

Quoted message said:

This model leads only to solutions with an infinity in
finite time. In this case at t = -37. Do we not have a
model without infinities?

Models are allowed to have infinities. It's only the real world
that doesn't have infinities (except in the case of things that are
actually infinite). In this case, assuming a = kv^2 is valid if the
only force is typical subsonic air drag. However, when you
extrapolate the equation backwards (negative time), the object's
velocity exceeds the speed of sound in air. At this point, long
before infinity, the model has ceased to be accurate;
supersonic and transonic drag behave differently. A bigger
issue is whether extrapolating backwards is meaningful at all.

It's a little clearer to write the equations symbolically.
a = kv^2 (k is negative)
v0 = 37 ft/sec, v1 = 29 ft/sec, t0 = 0.
dv/dt = kv^2, so dv / kv^2 = dt.
Integrating, t1-t0 = -1/k * (1/v1 - 1/v0), which gives t1 = 10.2 sec.

velocity: v(t) = v0 / (1 - v0*k*t).
distance: s(t) = -1/k * ln(1 - v0*k*t)

Extrapolating backwards to negative t, the velocity goes to
infinity at t = 1/(v0*k). This would mean that if the a = kv^2
force law always applied, an object at arbitrarily high velocity
could be decelerated in a finite time. There's two issues
here. One is that the force law may not always apply outside
the subsonic regime. The other is that you can only trigger
this mathematical singularity by starting with an object with
infinite velocity. If your initial conditions specify an object with
any finite velocity, its _future_ behavior governed by this
deceleration equation is always non-singular.

Because fluid drag is a dissipative process, the equations of
motion are not time reversible, so extrapolating backwards in
time is not guaranteed to give a sensible result.
(In the case of the bicycle rider, obviously some energy input
was required to accelerate to the initial speed, and while
pedaling the force law is different.)

In article
<[email hidden]>,

Quoted post said:

Michael Press said:

Quoted message said:

a = kv^2
-1 = k(37)^2, so k = -1/37^2
so dv/dt = (-1/37^2)v^2
Separate the variables and take the antiderivative of both sides

Quoted message said:

Quoted message said:

v = 37^2/(t + 37)
s = 37^2 ln[t + 37] + c

Quoted message said:

This model leads only to solutions with an infinity in
finite time. In this case at t = -37. Do we not have a
model without infinities?

Models are allowed to have infinities. It's only the real world
that doesn't have infinities (except in the case of things that are
actually infinite). In this case, assuming a = kv^2 is valid if the
only force is typical subsonic air drag. However, when you
extrapolate the equation backwards (negative time), the object's
velocity exceeds the speed of sound in air. At this point, long
before infinity, the model has ceased to be accurate;
supersonic and transonic drag behave differently. A bigger
issue is whether extrapolating backwards is meaningful at all.

It's a little clearer to write the equations symbolically.
a = kv^2 (k is negative)
v0 = 37 ft/sec, v1 = 29 ft/sec, t0 = 0.
dv/dt = kv^2, so dv / kv^2 = dt.
Integrating, t1-t0 = -1/k * (1/v1 - 1/v0), which gives t1 = 10.2 sec.

velocity: v(t) = v0 / (1 - v0*k*t).
distance: s(t) = -1/k * ln(1 - v0*k*t)

Extrapolating backwards to negative t, the velocity goes to
infinity at t = 1/(v0*k). This would mean that if the a = kv^2
force law always applied, an object at arbitrarily high velocity
could be decelerated in a finite time. There's two issues
here. One is that the force law may not always apply outside
the subsonic regime. The other is that you can only trigger
this mathematical singularity by starting with an object with
infinite velocity. If your initial conditions specify an object with
any finite velocity, its _future_ behavior governed by this
deceleration equation is always non-singular.

Because fluid drag is a dissipative process, the equations of
motion are not time reversible, so extrapolating backwards in
time is not guaranteed to give a sensible result.
(In the case of the bicycle rider, obviously some energy input
was required to accelerate to the initial speed, and while
pedaling the force law is different.)

Let me put it differently. How well does the
force ~ speed^2 model approximate the force of an
eccentric body in a turbulent air stream?

--
Michael Press