Math Cycling Problem



P

Phil Holman

Guest
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?

Phil H
 
Phil Holman wrote:

> 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 wrote:
> Phil Holman wrote:
>
> > 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 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.

> 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.


If one were measuring speed distance and time, as in a
coast down test, deceleration (I prefer that to retardation
because retard means something else in RBR) is the
directly derived parameter. Cd and A are a further inference
from that. And you also have to know the air density.

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

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


(37²/29)-37 seconds [~10.2 seconds]

>and how far will the rider have traveled?


37²*(ln|37²/29|-ln|37|) feet [~333.5 feet]
 
"Greg Berchin" <[email protected]> wrote in message
news:[email protected]...
> On Sat, 14 Jan 2006 11:28:38 -0800, "Phil Holman"
> <piholmanc@yourservice> wrote:
>
>>How long will it take for the rider to decelerate from 37 to
>>29 ft/sec

>
> (37²/29)-37 seconds [~10.2 seconds]
>
>>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> wrote:

>Excellent.


What did I win? A new Waterford would be nice.
 
Phil Holman wrote:
>> > 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 wrote:
>> Just plug it into Excel for a mathematically inelegant but adequate
>> solution.


[email protected] wrote:
> 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 wrote:

> 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 wrote:
> 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 protected]...
> Donald Munro wrote:
>
>> 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 wrote:
>> Just plug it into Mathematica, Maple or Matlab instead.


Dan Connelly wrote:
> % 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 wrote:
> A rider is traveling at 37 ft/sec and starts to coast.


There's yer mistake right there! :)

-Mark
 
In article <[email protected]>,
"Phil Holman" <piholmanc@yourservice> wrote:

> "Greg Berchin" <[email protected]> wrote in message
> news:[email protected]...
> > On Sat, 14 Jan 2006 11:28:38 -0800, "Phil Holman"
> > <piholmanc@yourservice> wrote:
> >
> >>How long will it take for the rider to decelerate from 37 to
> >>29 ft/sec

> >
> > (37²/29)-37 seconds [~10.2 seconds]
> >
> >>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 wrote:

> > 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


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


> 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 protected]>,
"[email protected]" <[email protected]> wrote:

> Michael Press wrote:
>
> > > 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

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

>
> > 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
 

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