DH Mt Bike - Optimizing Drop Landing Ramp Profile

[erobz]

I do downhill Mt Biking - and this question popped into my head; Is there a kind of "optimal" landing profile mathematically for a drop feature that is not obvious in the sense that it is not what we "intuitively design" for landings on these features?

My idea is that the bike leaves the top of the drop with a certain velocity $v$, and angular velocity $\omega$ that are related through Newtons Laws and the bike geometry. Ignoring drag, these parameters remain constant throughout the remaining projectile motion(s).

I "feel" as though I would like the landing profile $f(x)$ slope at any point of landing (at particular set of initial conditions on $v$ and $\omega$) to coincide with the angle through which the bike will rotate until that point is achieved.

This is not what is done in practice. The idea in the field is that the feature builder would want you to get the speed "right enough" to land somewhere around "here", and put some slope on it (estimated from past experience) in that region and let the shocks do the rest... hopefully well enough.

For reference the typical landing looks similar to this:

we don't anticipate the rider basically "falling off" from riding in real slow which would be a very large angle of rotation for the short trajectory, but also we don't expect the overshoot; a large rotation from excessive hang time on a typical landing profile. Both of these scenarios happen from time to time, even with experienced riders.

This mathematical problem seems challenging to me, I don't know if I'm really rusty, or if I've ever solved one quite like it. I'm expecting some differential equation to describe it.

Any thoughts/opinions appreciated.

Also, for the sake of simplification ignore the bike geometry w.r.t. the landing profile, I certainly don't want to get that in depth.

 

[A.T.]

I "feel" as though I would like the landing profile $f(x)$ slope at any point of landing (at particular set of initial conditions on $v$ and $\omega$) to coincide with the angle through which the bike will rotate until that point is achieved.

Is the sole criterion that the bike lands with both wheels simultaneously? I would assume that in practical terms you also want to limit the impact speed.

 

[Baluncore]

An analytic solution would be difficult, but there is a numerical solution. Start with the highest possible take-off velocity and compute the points along the curve, working back towards the take-off point at lowest velocity.

You know the rate of bike rotation about the centre of mass, due to the front wheel departing the edge before the rear, and you know where the front wheel will hit the flat ground beyond, so you can work out where the rear wheel will be at that same instant. Join those two points with a straight line approximation.

Lower the simulated speed very slightly, and work out where the front wheel will hit the previous velocity's straight line. Mark that point, and where the back wheel will be, then construct another line through the two points. Repeat that process until the line segments reach the point of the take-off. Then you have a line of dots that define the curve of the landing ramp.

You can refine the curve by adjusting the position of the body centre of mass between the wheels, and by raising the rear wheel contact point, so bikes land on the rear wheel just before the front wheel, if that attitude is what you want.

Also, simplify the initial curve generation by computing the wheel axle positions, ignoring the wheel diameter(s).

You might feather the curve into the track ahead, for the highest possible computed velocities. Also consider a vertically-uncurved track, for the ramp just before take-off, as that will stabilise the entry into the flight during the initial computations.

 

[.Scott]

I would thing that the optimal landing slope profile would create the same delta-V on impact - regardless of the exact horizontal velocity. The actual bike pitch moment can be controlled by the biker at the moment the bike crosses over the top of the cliff.

Basically, on impact, the vertical velocity created by the distance dropped like this:
g: gravity acc;
h: vertical distance dropped;
v: vertical velocity;
v=sqrt(2gh)

Upon striking the slope, the horizontal velocity component can remain unchanged, but the vertical component changes to follow the slope of the slope. Curving the slope down at the right rate would keep that velocity difference constant. The curve would be like a parabola - but it would take some work to figure out exactly what it is.

 

[berkeman]

I don't do much in the way of jumps or dropoffs on my MTB now (mostly XC and climbing), but obviously did lots of jumping in my motocross days. A good goal is to generate about 2-3g of upward acceleration on landing, IMO. Less upward force on landing makes it too much of a "floating" landing which is kind of hard to feel and control, and more landing force generates too much of an "oof!" type landing.

 

[Baluncore]

For a biker, braking after a launch would increase the forward rotation, they would need to accelerate during the initial flight to maintain attitude.

Ski jumping has solved the problem. Jumpers are able to control rotation by changing body profile. They do that by spreading legs or arms to raise drag, with a loss of range. Ski jumpers have the added advantages of a very high launch speed, and a single point of contact, their foot as a fulcrum on the ski.

 

[A.T.]

An analytic solution would be difficult, but there is a numerical solution.

@erobz Another numerical solution approach:
- Define a vector field, by assigning each point in space the orientation of the bike during a jump from a given launch point.
- Use a streamline algorithm on that vector field, to compute the landing ramp profile that is always parallel to the local bike orientation vector.

 

[haruspex]

Is the sole criterion that the bike lands with both wheels simultaneously?

My guess is that is not the ideal. Seems better to spread the vertical impulse by landing the wheels one at a time.

 

[Baluncore]

My guess is that is not the ideal. Seems better to spread the vertical impulse by landing the wheels one at a time.

The rear wheel first will align the bike with the flight, before the front wheel lands, and can then follow a pursuit curve to the track.

 

[erobz]

Is the sole criterion that the bike lands with both wheels simultaneously?

Basically that is what I was hoping for. Ignoring the actual bike geometry (treating it as a point), that the tangent of the landing profile matches the angle of rotation of the bike at that point of landing so we aren't landing nose or rear heavy over a range of reasonable takeoff velocities.

I would assume that in practical terms you also want to limit the impact speed.

Are they independent? At any rate it seems like that make the problem more difficult, does it not?

 

[berkeman]

My guess is that is not the ideal. Seems better to spread the vertical impulse by landing the wheels one at a time.

That's not my experience. Landing with both wheels at the same time seems to give the best control and impact absorption. One exception to that is in motocross when you are landing a jump and pinning the throttle to get maximum acceleration. Pinning the throttle just before landing lifts the front wheel (conservation of angular momentum as the rear wheel spins up).

 

[A.T.]

Are they independent? At any rate it seems like that make the problem more difficult, does it not?

Instead of defining the initial ramp drop h, you could also assume some maximally possible initial speed v, and specify how far down you want the bike to drop for that extreme case. It's just another way to constrain the problem, that might be more useful in terms of ensuring safety limits.

 

[berkeman]

 

[A.T.]

My idea is that the bike leaves the top of the drop with a certain velocity v, and angular velocity ω that are related through Newtons Laws and the bike geometry.

What is a reasonable way to express the relationship between v and ω based on the bike geometry? I guess that it's an inverse relationship (the smaller v, the greater ω), but that there is a maximal ω for very small v.

 

[haruspex]

What is a reasonable way to express the relationship between v and ω based on the bike geometry? I guess that it's an inverse relationship (the smaller v, the greater ω), but that there is a maximal ω for very small v.

Taking the wheelbase as 2L, mass centre at midpoint, and assuming it can only rotate through a small angle before the rear wheel becomes airborne, I get the final ω as $\frac{2g}v\frac{mL^2}{I+mL^2}$.
If v is small then the angle change becomes significant.
$\omega^2\leq \frac{2mgL}{I+mL^2}$

 

[berkeman]

 

[A.T.]

Taking the wheelbase as 2L, mass centre at midpoint, and assuming it can only rotate through a small angle before the rear wheel becomes airborne, I get the final ω as $\frac{2g}v\frac{mL^2}{I+mL^2}$.
If v is small then the angle change becomes significant.
$\omega^2\leq \frac{2mgL}{I+mL^2}$

Thanks! However, from the videos posted by @berkeman I'm getting the impression that it's usually not a passive ride over the edge. The riders seem to be actively countering the front wheel drop when going slow. This complicates the task the OP is trying solve.

 

[Baluncore]

The riders seem to be actively countering the front wheel drop when going slow.

Yes, while only the rear wheel is on the ramp, they lift the front wheel to eliminate rotation.

I think an interesting question is, what profile would a jump have, such that a beginner rider without any skills, and at any speed, would land on both wheels.

 

[haruspex]

Yes, while only the rear wheel is on the ramp, they lift the front wheel to eliminate rotation.

Does that mean they push down on the handlebars while the front wheel is still grounded, throwing the body up and back, so that they can then pull up on the handlebars when the front wheel leaves the ground?

I think an interesting question is, what profile would a jump have, such that a beginner rider without any skills, and at any speed, would land on both wheels.

Yes, interesting. I'll have a go at that.

 

[erobz]

Yes, while only the rear wheel is on the ramp, they lift the front wheel to eliminate rotation.

I think an interesting question is, what profile would a jump have, such that a beginner rider without any skills, and at any speed, would land on both wheels.

The thing is you want to do less of this as the drops get large. You want to get the speed correct and let the physics do the rest for you...ideally.


I don't ride this intensity, but that video gives you some scope of the drops that can be encountered and may point you to why I posed the question in the first place.

So yeah,

what profile would a jump have, such that a beginner rider without any skills, and at any speed, would land on both wheels.

Was my main motivation. Perhaps I didn't word it well.

 

[erobz]

Taking the wheelbase as 2L, mass centre at midpoint, and assuming it can only rotate through a small angle before the rear wheel becomes airborne, I get the final ω as $\frac{2g}v\frac{mL^2}{I+mL^2}$.
If v is small then the angle change becomes significant.
$\omega^2\leq \frac{2mgL}{I+mL^2}$

I'm getting $\omega = \frac{L^2 mg}{(mL^2 + I)v }$ otherwise ( the factor of $2$ )we align on the approximation.

Never mind. I see we have more time for the applied torque than I accounted for. You are correct.

 

[Baluncore]

Does that mean they push down on the handlebars while the front wheel is still grounded, throwing the body up and back, so that they can then pull up on the handlebars when the front wheel leaves the ground?

More like a small wheel stand, where they put additional torque on the rear wheel through the pedals while lifting the handlebars, to delay the front wheel falling.

 

[erobz]

More like a small wheel stand, where they put additional torque on the rear wheel through the pedals while lifting the handlebars, to delay the front wheel falling.

Usually we don't do this unless the drop landing in not developed (or underdeveloped). For instance if you are jumping from the top of a rock to a flat landing, it's a slow-moving technique. Not the set up I'm pondering.

 

[berkeman]

Usually we don't do this unless the drop landing in not developed. Like if you are jumping from the top of a rock to a flat landing, it's a slow-moving technique.

I think some videos of your riding group are needed here...

 

[haruspex]

while lifting the handlebars

I don’t get how you can do that without first pushing down on them in order to give the body the necessary posture and or angular momentum.

 

[Baluncore]

I don’t get how you can do that without first pushing down on them in order to give the body the necessary posture and or angular momentum.

I think you assume a front wheel hop from a body bounce, while holding the handlebars. I assume the front wheel lifts due to pedal pressure and acceleration.

When you push down on a pedal, you must hold the handlebars to oppose the force. The force on the pedal is applied along the ground, through the rear wheel. Since the traction force is not aligned with the higher CofM, it generates a backward torque that takes the weight off the front wheel during the initial takeoff.

 

[haruspex]

I assume the front wheel lifts due to pedal pressure and acceleration.

Professional cyclists achieve astonishing acceleration but don’t flip over backwards. It would seem perverse to try to do the lift entirely by pedal power when it would be so much easier to get a start using the body’s inertia. No doubt pedal power comes into play thereafter.

 

[Baluncore]

Professional cyclists achieve astonishing acceleration but don’t flip over backwards.

Flipping over backwards is totally unnecessary.

As an amateur cyclist wearing cleats, I could lift the front wheel just off the ground and accelerate away while leaning towards the handle bars. That was the maximum acceleration possible on a road bike.

 

[haruspex]

Flipping over backwards is totally unnecessary.

As an amateur cyclist wearing cleats, I could lift the front wheel just off the ground and accelerate away while leaning towards the handle bars. That was the maximum acceleration possible on a road bike.

I am not sure how to read that.
Are you saying that when accelerating as fast as you could you could not prevent the front wheel rising, and that limited how fast you could go, for fear of flipping backwards? Or that you deliberately raised the front wheel a little because that reduced the rolling resistance and so permitted greater acceleration?
Or something else?

 

[Baluncore]

Are you saying that when accelerating as fast as you could you could not prevent the front wheel rising, and that limited how fast you could go, for fear of flipping backwards?

In a way, but not to reduce rolling resistance. There is no fear of flipping backwards since you regulate body position and pedal force to keep the front wheel from rising significantly. As the FW rises, you get reduced acceleration, which is not what you want. To maximise traction, all your weight needs to be on the rear drive wheel, with your body as far forward as practical during the burst of acceleration. That is a balance of the two torques, forward body weight and RW torque.