Math Cycling Problem

Discussion in 'Road Cycling' started by Phil Holman, Jan 14, 2006.

  1. Phil Holman

    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
     
    Tags:


  2. 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.
     
  3. 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
     
  4. Phil Holman

    Phil Holman Guest

    "Zog The Undeniable" <[email protected]> wrote in message
    news:[email protected]
    > 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
     
  5. Donald Munro

    Donald Munro Guest

    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.
     
  6. Dan Connelly

    Dan Connelly Guest

    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
     
  7. Dan Connelly wrote:
    > s 333.522


    I knew it was a track question.

    --
    E. Dronkert
     
  8. Phil Holman

    Phil Holman Guest

    "Dan Connelly" <[email protected]_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
     
  9. Donald Munro

    Donald Munro Guest

    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.
     
  10. Mark Janeba

    Mark Janeba Guest

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


    There's yer mistake right there! :)

    -Mark
     
  11. Phil Holman

    Phil Holman Guest

    "Donald Munro" <[email protected]> wrote in message
    news:p[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"
    > <[email protected]> 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
     
Loading...
Loading...