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Question re archive thread about sine and force and if this applies to bicycling?

  1. Oct 15, 2008 #1
    Hi, I'm new to the forum so I hope this is the right place to post this, I actually joined to ask this question. I found something in the archive through a google search. The thread can be found at: https://www.physicsforums.com/archive/index.php/t-117265.html

    Some of the things in there were hard for me to understand, like how you get the 1-cos(t) from sin(t) , because we haven't covered integration in math yet, so any simple explanation of that would be appreciated.

    However, (also considering the graph provided by Integral @ http://home.comcast.net/~Integral50/Math/sinforce.jpg [Broken] ) my main question is: do you think that this applies to bicycling? In other words, considering how bicycle pedals work, CAN the forward force on the bicycle be represented by a function like f(t)=sin(t), where t=time. Obviously if that is true then it follows that the other 2 curves on the graph are true for bicycles aswell. The most important thing is wether or not the force is represented by a sine curve. Also, should the sine curve be moved up so that the trough only touches the x-axis without intersecting it because (and correct me if I'm wrong) but I don't think there is a time when the forward force on the bicycle is less than 0, i.e. force in backward direction.

    I think the question is straightforward but is there anything that I need to explain more?

    Thanks in advance
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 15, 2008 #2
    i dont think you can write the forward force of a bicycle as an equation of sin....im pretty sure the equation would be a little more complicated dealing with gears and what not....i could be wrong...thats just my thought....
    Last edited: Oct 16, 2008
  4. Oct 16, 2008 #3
  5. Oct 16, 2008 #4
    ok that makes sense, thanks.

    oh, right I guess I forgot to mention that its a simple system. so lots of stuff is ignored because you can actualy control them. so differences in friction from the road, wind resistance, using different gears, difference in the incline og the hill, cadence of the pedalling, most extra variable stuff can actually be ignored, because it's kept constant. so in other words the system we're looking at is a much simplified one, like just the pedals and the bike. So the downward motion of the feet on the pedals goes to forward motion in the bike. If you split up the rotation into 4 parts then there is the power stroke from your left leg, then a break where you don't push down with either leg, then the powerstroke from your right leg and then another break. Do you think that this will translate to a sine curve? ,so that the crests of the curve are the power strokes and the troughs are the breaks... does this make sense?
  6. Oct 16, 2008 #5


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    Note that a rider can push forward, pull backwards, and pull upwards on the pedals, if using toe clips or clip in type pedals, so those "breaks" aren't zero force breaks.
  7. Oct 20, 2008 #6
    Thanks for the suggestion. Riders using platform pedals, not clipless or toe clips. Out of interest, what curve would you get for a rider wearing toe clips / clipless shoes?
  8. Oct 21, 2008 #7
    I am guessing that the curve looks more like a unit step function (Step curve). Forgoing the toe clip discussion, I would think the force would be a constant on-off type force. Similar to pushing your finger on a table top for 2 seconds and releasing the pressure.
  9. Oct 21, 2008 #8
    Good point.
    I'm not sure but i think there is a peak point in the force from the pedal stroke ...but I may be wrong. If you can explain your reasoning to me in a way that I can understand then your theory can replace mine :)
  10. Oct 21, 2008 #9
    Here is another option. Measure the force indirectly using a DIY Force Sensitive Resister attached to a multimeter. "Instructables" has instructions for making a simple FSR. Place the resistor on the pedal and note how the voltage varies with time as you push against the pedal with your foot. The resistor may not respond fast enough to give an instatenous measure but might give you a rough idea if the force is sinusoidal.
  11. Oct 21, 2008 #10
    Rather than making a Force Sensitive Resistor (FSR) you can purchase them at fairly low cost. Trossen Robotics sells a 1.5 inch Interlink FSR for $8.75. Also MIT has a good introduction to FSR, google "Force Sensitive Resistor introduction". The multimeter will measure resistance rather than voltage.
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