Math Cycling Problem

[Phil Holman]

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

 

[Zog The Undeniable]

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.

 

[[email protected]]

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

 

[Phil Holman]

"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

 

[Donald Munro]

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.

 

[Dan Connelly]

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

 

[Ewoud Dronkert]

Dan Connelly wrote:
> s 333.522

I knew it was a track question.

--
E. Dronkert

 

[Phil Holman]

"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]

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.

 

[Mark Janeba]

Phil Holman wrote:
> A rider is traveling at 37 ft/sec and starts to coast.

There's yer mistake right there!

-Mark

 

[Phil Holman]

"Donald Munro" <[email protected]> wrote in message
news[email protected]...
> 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.
>
Here's the calculus solution posted to rbt. Probably has origins even
further to the left on the evolutionary scale.

"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