Forum home Road cycling forum The bottom bracket

How much kinetic energy does a cyclist have?

chronyxchronyx Posts: 455
edited February 2008 in The bottom bracket
Just a point I wondered on the way home, steaming along at 25mph (Wearing jeans and a big backpack!) with a great tailwind. How much energy would a 13 stone cyclist be carrying at that sort of speed?

25mph seems nothing in today's world where people can get in a car and be doing three figure speeds in less than ten seconds in some cases.

But when you think about it, a bicycles components must have to routinely deal with some pretty hefty (Technical term) forces, for the size of them. It's just that in comparison to other vehicles they seem trivial.

I dunno. Rambling now anyway, so over to you!
2007 Giant SCR2 - 'BFG'

Gone but not forgotten!:
2005 Specialized Hardrock Sport - 'Red Rocket'

Posts

  • Random VinceRandom Vince Posts: 11,374
    i've often wandered what kind of torque i'm producing in various situations
    My signature was stolen by a moose

    that will be all

    trying to get GT James banned since tuesday
  • KE=MVV/2
    Mass in Kg, V im m/s.
    Get your calculator out.
    Remember that you are an Englishman and thus have won first prize in the lottery of life.
  • chronyxchronyx Posts: 455
    I'll need my algebra for dummies first. Does that mean Mass x Velocity x Velocity again, or the sum of M x V x V again?
    2007 Giant SCR2 - 'BFG'

    Gone but not forgotten!:
    2005 Specialized Hardrock Sport - 'Red Rocket'
  • KE = 0.5*m*v*v

    where * is multiply

    Not sure how this figure will of much use . . . .
  • feelfeel Posts: 800
    i've often wandered what kind of torque i'm producing in various situations
    your max torque is probs when your pedal cranks are horizontal. Multiply your mass in kg by 10 (9.8 for accuracy) to turn it into a force in newtons and then multiply it by your crank length in metres, so 20cm becomes 0.2 M. This is not taking into account any turning force you are getting by pulling up on the other pedal, but i suspect this would probably decrease your force on the first pedal.

    with the kinetic energy calculation it is the mass of the rider plus luggage plus bike ie the total mass of every thing travelling at that velocity.
    We are born with the dead:
    See, they return, and bring us with them.
  • PositronPositron Posts: 191
    Oh about 5153 Joules.

    p. (yep a bloody physicist...)
    Never order anti-pasta to arrive at the same time as pasta.
  • Alex_Simmons/RSTAlex_Simmons/RST Posts: 4,161
    edited February 2008
    Positron wrote:
    Oh about 5153 Joules.

    p. (yep a bloody physicist...)
    plus weight of bike and bits, so more like 5,600 J I'd say.

    Instantaneous torque is applied in a pseudo-sinusoidal manner relative to crank position. From a maximal peak around the 3-9 o'clock crank position to minimal around the 6-12 o'clock crank position.

    Typically however it is measured as an average over the full crank revolution. This is how most leading power meters work, they measure the torque applied (either at the crank or at the rear hub) via strain gauges over a full revolution and multiply that by the pedal speed/cadence (or hub rpm) and a factor to account for crank length to determine the power output.

    Knowing torque on its own is pretty useless without knowing pedal speed (cadence). You could be applying a lot of torque but not be going anywhere. Likewise you can spin really fast but have torque so low that not much is happening. Bit like doing max cadence test on a trainer with no resistance. Tells you censored all.

    In layman's terms
    Power = Torque x Cadence (plus a factor for crank length)


  • LOL

    Its like the red arrows (way over my head)
    http://www.sketchymtb.co.uk/Blah.pl the new XC in Kent


    http://deadpool2e.pinkbike.com/channel/Afan-Vids/

    MOUNTAIN BIKING- The pastime of spending large sums of money you don't really have on something you don't really need.
  • Random VinceRandom Vince Posts: 11,374
    Positron wrote:
    Oh about 5153 Joules.

    p. (yep a bloody physicist...)
    plus weight of bike and bits, so more like 5,600 J I'd say.

    Instantaneous torque is applied in a pseudo-sinusoidal manner relative to crank position. From a maximal peak around the 3-9 o'clock crank position to minimal around the 6-12 o'clock crank position.

    Typically however it is measured as an average over the full crank revolution. This is how most leading power meters work, they measure the torque applied (either at the crank or at the rear hub) via strain gauges over a full revolution and multiply that by the pedal speed/cadence (or hub rpm) and a factor to account for crank length to determine the power output.

    Knowing torque on its own is pretty useless without knowing pedal speed (cadence). You could be applying a lot of torque but not be going anywhere. Likewise you can spin really fast but have torque so low that not much is happening. Bit like doing max cadence test on a trainer with no resistance. Tells you censored all.

    In layman's terms
    Power = Torque x Cadence (plus a factor for crank length)

    fair enough, was just wondering since i used to pull the back wheel of my mountain bike skewed from pedaling regardless of how tight i did the QR

    managed it with the road bike's bolt in rear wheel occasionally too... managed to get them both very tight (the mountain bike requires a tap with a hammer and a spanner as a lever to get it off)
    My signature was stolen by a moose

    that will be all

    trying to get GT James banned since tuesday
  • fair enough, was just wondering since i used to pull the back wheel of my mountain bike skewed from pedaling regardless of how tight i did the QR

    managed it with the road bike's bolt in rear wheel occasionally too... managed to get them both very tight (the mountain bike requires a tap with a hammer and a spanner as a lever to get it off)
    Not tight enough apparently ;)

    If Chris Hoy doesn't pull a wheel, then either you're stronger than him or your equipment isn't up to the job or is not tightened appropriately. Perhaps you have QR/bolts that are not suited to the job.
  • All very interesting, but nothing to do with kinetic energy, which is a function (as pointed out earlier) of mass and velocity. If two bikes/riders have identical mass and are travelling at the same speed, their kinetic energy will be the same even if one isn't pedalling at all (power output zero) and is being pulled along by holding on to a car...
  • ColinJColinJ Posts: 2,218
    fair enough, was just wondering since i used to pull the back wheel of my mountain bike skewed from pedaling regardless of how tight i did the QR

    managed it with the road bike's bolt in rear wheel occasionally too... managed to get them both very tight (the mountain bike requires a tap with a hammer and a spanner as a lever to get it off)
    Not tight enough apparently ;)

    If Chris Hoy doesn't pull a wheel, then either you're stronger than him or your equipment isn't up to the job or is not tightened appropriately. Perhaps you have QR/bolts that are not suited to the job.
    I was going to make that point!

    The loose/dodgy QR question has been covered many times before.

    I had a mate who was into body building who came out cycling with me a few times. He had enormously strong legs, but a ridiculous sense of gear selection. I saw him put so much force through the chain on hills that he destroyed the thing, but his wheel never came loose. I'd done up the QR and I didn't use a hammer to do it - just the press hard enough that the QR lever makes an imprint on the hand technique.

    I'm nearly 6' 2" and I climbed 25% gradients on my bike when I weighed 16+ stone. I had to use every last bit of my strength to do it but I never pulled a wheel loose.

    I know people who have pulled wheels out and in every case it was due to either using cheap nasty skewers or expensive stretchy Ti skewers.
  • know people who have pulled wheels out and in every case it was due to either using cheap nasty skewers or expensive stretchy Ti skewers

    or not riding a bike with vertical drop outs...
  • lateralus wrote:
    All very interesting, but nothing to do with kinetic energy, which is a function (as pointed out earlier) of mass and velocity. If two bikes/riders have identical mass and are travelling at the same speed, their kinetic energy will be the same even if one isn't pedalling at all (power output zero) and is being pulled along by holding on to a car...
    Oh that's certainly true but this is thread drift you know - when one topic loses momentum (& kinetic energy) and we have to accelerate onto another topic in another direction (requiring force/torque to be applied) :wink:
  • chronyxchronyx Posts: 455
    Oh that's certainly true but this is thread drift you know - when one topic loses momentum (& kinetic energy) and we have to accelerate onto another topic in another direction (requiring force/torque to be applied) :wink:

    :lol::lol::lol: Excellent
    2007 Giant SCR2 - 'BFG'

    Gone but not forgotten!:
    2005 Specialized Hardrock Sport - 'Red Rocket'
  • [Oh that's certainly true but this is thread drift you know - when one topic loses momentum (& kinetic energy) and we have to accelerate onto another topic in another direction (requiring force/torque to be applied) :wink:

    Touche. :)
  • ColinJColinJ Posts: 2,218
    know people who have pulled wheels out and in every case it was due to either using cheap nasty skewers or expensive stretchy Ti skewers

    or not riding a bike with vertical drop outs...
    Granted, vertical dropouts would also stop you pulling a wheel out, but they aren't strictly necessary! I'm big and strong and I don't have vertical dropouts on any of my 3 bikes and though I have broken chains and a crank in the past, I have never managed to pull a wheel out.

    The force required to pull a rear wheel out when it is held in place by a decent quality QR skewer done up properly is much greater than can be exerted through a bike chain.

    PS One other thing... the QR 'nut' has to be able to bite into the surface of the dropout. I've heard of people having problems with dropouts made of superhard 6/4 Ti, but there aren't too many of those about.
  • lateralus wrote:
    All very interesting, but nothing to do with kinetic energy, which is a function (as pointed out earlier) of mass and velocity. If two bikes/riders have identical mass and are travelling at the same speed, their kinetic energy will be the same even if one isn't pedalling at all (power output zero) and is being pulled along by holding on to a car...
    Oh that's certainly true but this is thread drift you know - when one topic loses momentum (& kinetic energy) and we have to accelerate onto another topic in another direction (requiring force/torque to be applied) :wink:

    The interesting thing is that the torque measured is not absolute but net torque as there is negative torque for the other leg moving from 6 o'clock to 12 o'clock.
  • blackhands wrote:
    The interesting thing is that the torque measured is not absolute but net torque as there is negative torque for the other leg moving from 6 o'clock to 12 o'clock.
    Perhaps you'd enjoy the discussions on the torque analysis using the feature available on some SRM power meters (it an add on to the CPU firmware). If you're not already a member, join the wattage forum and search the four recent threads on "Fun with SRM torque analysis". They discusses, inter alia, a method to isolate the individual leg torque application patterns.
    http://groups.google.com/group/wattage/

    But I have to say, this is hardly cake stop conversation for most cyclists. :wink:
Sign In or Register to comment.