Before I start defining the situation and asking a question, I'd like to make a few things clear:
- this is not a homework, merely a matter of personal interest and enthusiasm
- I am neither a physicist, nor a native English speaker, therefore I encourage anyone with proper knowledge to go ahead and edit anything in my post to correct/improve any terminology, thread title including
I'd like to start with an image I created for this purpose, I hope I didn't make any huge mistakes in it. So, here's the situation:
We have a biker weighing m1 riding a bicycle weighing m2, therefore the total weight m = m1 + m2.
The biker is driving at a constant speed of v and is about to jump down from a higher ground to a lower ground, while the total height he's jumping, the height difference of those two grounds is h.
According to how fast he's going and how high he's jumping from, he'll jump a distance d.
The question: What's the biker's weight at the moment of landing with which he's "pushing" the bicycle down? As I'm not a physicist and I'm not sure what's the proper term, I'll try to form it like this:
I know what the maximum weight the bicycle can hold is according to the manufacturer and I need to calculate whether I will exceed this limit when performing such a jump.
Things to consider: I believe there are many factors which can't be counted with properly, therefore
I think, and now correct me if I'm wrong:
- that we can assume that the forward speed when landing will be the same as the forward speed before jumping. Of course it won't, but I believe this speed will not drop so much, so possibly we can ignore it
- I believe there's a huge difference between locking one's joints and dropping with a rigid body versus using the proper technique, keeping joints free and flexing one's arms and legs to absorb the mass of landing. I don't know how it'd be possible to include this factor to the calculation, I'll let people decide.
Feel free to propose anything I forgot to include in my question, and again, if anyone who can edit posts can improve anything I wrote, please, don't hesitate and do so.
Best Answer
You are trying to figure out the force your bicycle can withstand, although you have posed the question so as to get that answer back as a "mass multiplier." The answer to your question really varies a ~LOT~ with technique. The best we can do here is give boundaries to the possible set of answers.
Assuming anything approaching good technique, the bike will land and start a rebound without bearing any appreciable weight from the rider. So mass of the bike ('m2') may be ignored.
Forward velocity 'v' and distance 'd' both have nothing to do with the problem.
What does matter is h, m1, hardness of the ground, and technique (to include rider's physical fitness and anthropomorphic measurments). Technique is the ability to decelerate the downward velocity of m1 to 0, while exerting the lowest peak downward force on the bike. Once you figure out what this peak force is, you have "effective" weight at landing.
Assume hard landing surface, so we can remove that from the equation. Assume a simultaneous two-wheel landing (I am not saying this is best, just makes calculation a bit easier). Assume that the tires plus your body mechanics give you 0.3m (just over 1 foot)of deceleration distance for m1. You did not give a weight, so to make math (and subsequent scaling) easy, I will assume 100 kg. y = 0.3m is height of m1 at touchdown and y=0 is height of m1 when your face and/or crotch smash into the bike frame.
F=ma. Or, more completely in this case: (The sum of forces) = ma. In the space of 0.3 m, m1 must come to rest (vertically). To do that you need to generate an 'a' which will result in dh/dt = 0. This in turn will require a net force upward. We need to find dh/dt at (wheels) touchdown; then calculate a; then derive a net force to accomplish that. Three problems. here we go...
1.) dh/dt at contact was given by dbrane (sign correction add for my frame of reference) --> -SQRT(2gh). So if h = 2m, then dh/dt = about -6.3 m/s at wheel contact.
2.) To bring dh/dt to 0 in the space of 0.3 m, we need to use the same form of the equation used in step one, but this time: -(dh/dt) = SQRT(2a(-dy)). For our 2m jump, this is -(-6.3 m/s) = 6.3 m/s = SQRT(0.6a). 'a' = about 66 m/(s^2), which is equivalent to about 6.7 gravities.
3.) We are back to F=ma, with a twist. Define the amount of force exerted by the rider = Fr. Define the amount of force exerted on the rider by gravity = Fg = (100Kg)(-9.8m/(s^2)) = -980N. Then (Fr + Fg) = (100kg)(66m/(s^2)) = 6600N. Fr = 6600N + 980N = 7580N.
The force exerted (assuming the rider can apply it smoothly) is about 1700 pounds force.
By doing a one wheel landing, you could simultaneously shorten the effective drop and lengthen the subsequent deceleration distance; both of which would significantly lower the required deceleration forces. My advice: Limit your jumps to under 2m vertical drop.