# What's my weight when jumping on a bicycle from a higher to a lower ground?

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.

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A physician is the person who tells you to say "aaaahhhh" while they look at your tonsils. Physicists are the people who study physics. :) – Colin K May 11 '11 at 14:51
hahahah yeah thanks, I fixed it, lol – Richard Rodriguez May 11 '11 at 14:52
On a more serious note: I understand what you are asking, but I think the term "weight" may be misleading. The weight of the bike and biker are the same at all times in the situation you describe, and are equal to their mass times the acceleration due to gravity. You are asking about the load on the bike. The second issue here is that this is actually a very difficult question to answer precisely. As you allude to when you reference "rigid body versus using the proper technique," the amount of force is dependent on the time over which it is distributed. This is very hard to know accurately. – Colin K May 11 '11 at 14:54
can you please edit my post, Colin K, to improve anything you proposed? – Richard Rodriguez May 11 '11 at 14:55
To expand on Colin's remarks a little, the answer is fundamentally wrapped up in the degree to which the ground-tires-[shocks-]frame-rider system is not a rigid body. – dmckee May 11 '11 at 14:59

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.

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Could you please join all those formulas to one formula? So that I can count various values by changing weight of the driver and height of the jump? I can't really work it out of those many formulas, sorry :( I know this stuff is very clear and easy to you guys, but I just ... can't :/ – Richard Rodriguez May 14 '11 at 1:17

Well, this will only be a partial answer, because why the physics is simple, this is actually a relatively hard question to answer.

As has been alluded to, the force will depend heavily on the amount of time over which the impact occurs. This, in turn, depends on the rigidity of the objects involved. Let me give some background here. (I'm going to be very hand-wavey here, both because I'm trying to keep the concepts clear and because I'm short on time.)

The horizontal motion of the bike is unimportant here, so lets imagine the bike and rider being dropped vertically from the same vertical height that the bike reaches during its jump. The instant before the bike makes contact with the ground, it is moving at some speed in the vertical direction $v_i$. Thus, it has momentum $p_i=mv_i$ and some kinetic energy $K=\frac{1}{2}mv_i^2$. During the collision, the vertical velocity of the bike is reduced to zero, so there must be some impulse exerted which equals the momentum of the bike. Impulse is force times time, and the force being exerted here is going to be varying strongly with time, so we're going to need an integral:

$$p_i = \int_0^{t_c} F(\tau) d\tau$$

where $t_c$ is the time it takes for the collision to happen, and $\tau$ is just a dummy variable of integration.

The kinetic energy of the bike will also be reduced to zero, so the work done on the ground by the bike must equal the kinetic energy. Again we have an integral:

$$K=\int_0^{x_f} F(\xi) d\xi$$

where $x_f$ is the equilibrium deformation of the ground/bike/shock absorbers etc. and $\xi$ is again a dummy integration variable. Notice that this requires the collision force to come in as a function displacement, while the momentum relation requires it as a function of time. We can get one from the other by assuming that the whole system can be treated as a spring with effective spring constant $k_{eff}$, so

$$F=k_{eff}x(t)$$

which means we ultimately need to find the deformation of the system as a function of time to give us both functions.

Now, it would be quite reasonable here to make some approximations, and assume that the force is a constant equal to the average value of the time dependent force, but we are still left with this unknown $k_{eff}$ which will really be needed to solve our system of equations. In practice this parameter is very hard to know, or even guess at. It will also depend on the bike frame and, worst of all for those of us who like nice rigid mathematical certainty, will depend on how the rider uses his limbs to help cushion the impact.

A very rough estimate (technique suggested by @dmckee) would be to just pick some reasonable value for $t_c$ and use that to calculate the average force, then double it to estimate the peak force. Using this technique, and estimating a quarter second for the collision ($t_c=0.25 \textrm{sec}$) we get:

$$\begin{eqnarray} mv_i &= &\overline{F}t_c \\ \frac{mv_i}{t_c} &= &\overline{F} \\ \frac{mv_i}{0.25 \textrm{sec}} &= &\overline{F} \end{eqnarray}$$

where $\overline{F}$ is the average force. This is a very rough estimate, and will really only get you the right order of magnitude, but without a lot of measurements to determine $k_{eff}$ it isn't really possible to do much better.

EDIT: I realized that I never explained how to get $v_i$. @dbrane included this in his answer, but I will plug the result into my equations for completeness. Assuming the bike drops from a height of $h$, whether this is the apex of a jump, or simply the height from which the pike is dropped, the velocity at the ground will be

$$v_i=\sqrt{2gh}$$

where $g=9.8 \frac{m}{s^2}$ is the acceleration due to gravity. Plugging this in to the result above, we get:

$$\begin{eqnarray} \frac{mv_i}{0.25 \textrm{sec}} &= &\overline{F}\\ \frac{m\sqrt{2gh}}{0.25 \textrm{sec}} &= &\overline{F}\\ \end{eqnarray}$$

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I've noticed that something is incorrect where I handle the energy conservation aspect, and I should also explicitly write the system of equations that would need to be solved under the constant force approximation, but I'm out of time. I've got no objection to downvotes for factual errors, but if the energy conservation problem is your reason for a downvote, please point out my error so I can make a quick edit later. :) – Colin K May 11 '11 at 16:08
Your explanation looks amazing and since I'm no physicist, I wouldn't even dare to question your statement, but with all the respect I need to ask the following: I think I remember that if you push an object so that it slides on a thin ice, it will keep going forward, not breaking the ice, but when it stops, it will begin to press with its full weight on the ice and will break it. Therefore I figured that objects when moving have less "vertical mass" than when standing. Why doesn't this rule apply in this scenario, if ever? Or am I completely mistaken here? – Richard Rodriguez May 11 '11 at 19:09
@Rimmer no, the object will not push less on the ice while moving. It may be that the ice doesn't break when something slides quickly along it, but not because it fails to push on the ice. The ice breaks when it deforms sufficiently, which is a different issue than the force on it. If the object is sliding fast enough, the force may be sufficient to break the ice, but the ice doesn't have time to break. – Mark Eichenlaub May 11 '11 at 19:10

The most crucial factor here is of course the nature of the solid you land on. If you're landing on a mattress, your bike can handle a much larger height of fall than if you're landing on concrete (much much larger height if it's water). This is because the average force your bike endures during contact is inversely proportional to the time of contact, i.e, concrete will reduce your speed from $v$ to zero in milliseconds, whereas the mattress will take slightly longer $$F\approx m \frac{v}{t} = m\frac{\sqrt{2gh}}{t}$$ Dividing this by $g$ will give you the effective mass the bike feels during contact. But the formula is only useful if you know what the contact time $t$ is and I'm not sure if that's easy to determine for different materials.

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I was expecting concrete, so yeah, I guess no mattress or anything like that. – Richard Rodriguez May 11 '11 at 15:03
Yeah, I know, I just wanted to point out the dependence on the contact time. Maybe others will have better ideas. – dbrane May 11 '11 at 15:05
I'd add that the effective weight isn't F/g, but F/g + mg because the effective weight depends on the normal force from the ground, not the net force. – Mark Eichenlaub May 11 '11 at 17:40
that should have been that the weight is F + mg. Dividing by g is never necessary to find the weight, which is synonymous with force – Mark Eichenlaub May 11 '11 at 19:12
@Mark I'm not sure if one needs to add $mg$ right away. Shouldn't the $mg$ contribution be added only once the bike is at rest on the ground? The normal force is precisely what is being calculated in my answer. $mg$ is what the normal force finalizes into once the bike is at rest, no? – dbrane May 12 '11 at 11:50

Without any ramps that would convert forward momentum into vertical momentum or vice versa the forward movement can be ignored. The gravitational potential energy from the fall must therefore be absorbed during the collision as the final vertical kinetic energy is zero.

$$G.P.E. = m\,g\,h$$

Where m is that mass of the bike and rider, g is acceleration due to gravity, and h is the height of the drop.

To minimize the peak force, the optimum is a constant de-acceleration

In the case of a constant deceleration force $F$ over a stopping distance $s$ the energy absorbed would be

$$E=F \, s$$

Thus:

$$m\,g\,h = G.P.E. = E = F \, s$$

$$F=m\,g\frac{h}{s}$$

Since $m\,g$ is just weight $w$:

$$F=w\frac{h}{s}$$

Now this presumes a constant deceleration which would mean from the instant the tire touches the ground its supporting a constant force. This is rather unrealistic, so lets instead model it as a spring saying the force increases linearly with distance.

$$F(y)=F_{peak}\max(\frac{s-y}{s},0)$$

$$E=\int_0^sF(y)\,dy=\int_0^sF_{peak}\frac{s-y}{s}\,dy = \frac12 F_{peak} s$$

Note that the energy absorbed is just half that of a constant force. This just results in a doubling of the peak force required to stop the fall.

$$F_{peak}=2w\frac{h}{s}$$

So for a rider and bike weight of 220 pounds (force) a height drop of 4 feet, and a deceleration distance of 1 foot, the bike would have to be able to handle a tire load of

$$F_{peak}=2\, 220 lbf\frac{4ft}{1ft}= 1760 lbf$$

Now it's very unlikely that you'll find a bike rated for 1760 pounds even if there are plenty of bikes that could safely make this jump. The reason for this is typically products that give weight limits are already factoring in a typical use. For example a jerky elevator car may decelerate at 2 g right when coming to a stop. This would cause the car and cable/piston to need to be able to handle 3 times the weight in the car. So if the elevator was "rated" for 2000 lbs then it would need to be able to handle 6000 lbs as when someone reads the rating they assume that is the load they may safely load into the elevator. Thus a bike rated for 300 pounds is probably designed to handle a 300 pound rider during typical use, which would cover dropping down off curbs and hitting the occasional pothole for hybrid bike, but may include much larger drops for mountain bikes.

In contrast to this, climbing gear is rated at failure load, so the ratings are typically in the tens of kilo-newtons (around 5000 pounds) but if iron man in a 5000 pound suit tried to use the gear it would immediately break.

My point is that if your bike is rated in the thousands of pounds then it is probably the failure strength, but if it is rated in hundreds of pounds then it is probably rated as rider weight during typical use. Unfortunately, without knowing how the engineers determined typical use, this information is mostly useless for exploring atypical use safely.

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The simplest way to look at this: you are accelerating with an acceleration $g$ for a distance $h$, then decelerating with deceleration $a$ over a distance $s$ (however far the rider can move his center of gravity to "absorb the shock").

During the initial acceleration, the rider picks up a speed

$$v = \sqrt{2 \;g\cdot h}$$

(simple conservation of energy).

During deceleration, he has to get rid of this speed over a distance $s$. Again, conservation of energy tells us that

$$v = \sqrt{2\;(a-g)\cdot s}$$

We have the $-g$ term in there because, as the rider is decelerating, he is still subject to the force of gravity.

Setting these two equal to each other, we can solve for $a$:

$$(a-g)\cdot s = g\cdot h\\ a = g\left(\frac{h}{s} + 1\right)$$

So this tells you how much greater the weight of the rider is during the landing. If the distance he can move his body after the drop is the same as the distance he dropped, his weight doubles. As the drop becomes greater, the weight during the landing increases.

Incidentally, you can compare this expression with the result in @Vintage's answer - for a drop of 2 m and a deceleration over 0.3 m, the calculated result was 7.7 g. Which is the same result that this expression gives.

AFTERTHOUGHT

It is no coincidence that you see people who jump their bikes land on the rear wheel. By extending one wheel down, they decrease $h$ (the distance of free fall) and increase $s$ (the distance of deceleration). In fact, by throwing their weight back (or the bike forward) before landing, they can convert some of the linear momentum into angular momentum, and in general give themselves more time to absorb the shock. Which is how you can get these larger jumps. The slope of the surface also helps - landing on a slope (rather than a horizontal surface) means that your velocity normal to the surface you land on is less: this in turn gives you more time to absorb the shock. This is enhanced by having a high forward velocity, and incidentally is why ski jumpers can survive those enormous jumps.

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Why would distance go with acceleration squared? – Rick May 15 '15 at 12:43
@Rick - well that was embarrassing. Thanks for spotting my mistake. I believe the answer is now corrected. – Floris May 15 '15 at 13:23
regarding the $-g$ term, I believe this would be required if s takes place entirely after h, but that would mean h would be the height of your drop minus your stopping distance. If we were looking at a trampoline where the stopping takes place after the fall h, then I think the $-g$ term would indeed be required, but I think in this case it's being double counted. – Rick May 15 '15 at 13:44
@Rick - I take $h$ to be the distance to first contact... – Floris May 15 '15 at 13:44
hmm... interesting point. I think to resolve which model is more realistic we'd have to stop modeling the bike rider system as a point, but it's probably better to just go with the higher estimate. – Rick May 15 '15 at 13:57