Forces on swingarm of leading-link front suspension of motorcycle

Question

I'm trying to understand the forces on the forks and the swingarm in a leading-link motorcycle suspension during braking. It seems simple at first glance but on closer examination, perhaps a bit less simple... Here's how I break it down—no doubt there is a clearer logic that can be used but let me know if my results are sensible. In short, what are the forces at the axle, caliper, spring, and pivot of the swingarm during braking?

Btw, here's an example of a leading link:

AT REST

Below is a schematic of an ugly motorcycle at rest, with the center of mass halfway between the points where the front and rear wheels contact the ground, so the load is evenly distributed between the wheels. For ease of discussion, the total weight is 1000 N, so the normal force of the ground pushing up on the front wheel is 500 N, and thus the wheel pushes upward on the axle with a force of 500 N.

The wheel radius is 25 cm, for another round number.

The axle is at the end of a swingarm, which is 30 cm long.

The swingarm is attached to the frame of the motorcycle at a pivot.

The spring is connected to the swingarm at a point 20 cm from the pivot.

When at rest (say when riding on a flat and level surface) the swingarm is angled 20 degrees away from horizontal, just to pick another round number (I know that looks different from many real-world examples but that's ok for my purposes).

And when at rest, the spring is along a line 10 degrees away from vertical.

The drawing above breaks down the forces on the swingarm.

The ground's vertical push on the axle has a component that is perpendicular to the swingarm with a magnitude of approx. 470 N, which tends to rotate the swingarm around the pivot.

To balance the torque, the spring needs to be pushing against the swingarm with a perpendicular component of about 705 N; that requires a spring force of about 716 N.

The frame must be pushing against the spring with the same force, and that force from the frame has a vertical component of about 705 N.

Next, I assume that since overall the frame is pushing down on the swingarm with a force of 500 N, it must be pulling upward on the pivot with a vertical force of about 205 N.

And since the frame is pushing on the spring with a horizontal component of about 124 N, it must be pulling rearward on the pivot with a force of 124 N.

Another way of putting it is that the resultant of those forces on the pivot actually tends to bend the bottom end of the forks forward and downward, unless I've gotten off track here.

Do those results make sense so far?

BRAKING

In this drawing, the bike is traveling forward but decelerating due to a horizontal force of 1000 N.

All of the vertical load of the centre of mass is transferred to the front wheel, so the normal force of the ground on the wheel is 1000N. (Greater deceleration would mean the bike would tip forward—the center of mass height here is equal to the distance from the front contact point with the ground.)

The deceleration is caused by a braking force exerted by the brake caliper at the outer edge of the brake disc, which has a radius of 12.5 cm, or half that of the wheel.

As the caliper is clamping the brake pads on the disc, and the disc is fixed to the wheel, the wheel now transfers force to the swingarm via both the axle and the caliper.

The torque produced by the horizontal force of the ground on the wheel results in a force of 2000 N at the caliper, perpendicular to the swingarm.

The force in the spring is the same as it was in the resting case, about 716 N (as it's determined by the length of the spring—assume the swingarm angle is the same as it was at rest. That said, for this particular geometry, braking will produce an "anti-dive" torque that tends to extend the swingarm, i.e., the swingarm won't be in static equilibrium.)

Now I want to find the forces at the axle and pivot. Does the following make sense? Let me know if I'm overcomplicating (or undercomplicating) things.

The frame acts on the swingarm via the spring and the pivot.

Overall, the frame is pushing horizontally on the swingarm with a force of 1000 N; the horizontal component of the force of the frame on the spring is approx. 124 N, so the frame must be pushing horizontally forward on the pivot with a force of approx. 876 N.

And the frame overall is pushing vertically downward on the swingarm with a force of 1000 N; the vertical component of the force on the spring is approx. 705 N, so the frame must be pushing vertically downward on the pivot with a force of approx. 295 N.

The wheel acts on the swingarm via the axle and the caliper.

Overall, the wheel is pushing horizontally on the swingarm with a force of 1000 N; the horizontal component of the force at the caliper is approx. 684 N, so the horizontal force at the axle must be approx. 316 N.

Overall, the wheel is pushing vertically on the swingarm with a force of 1000 N; and the vertical component of the force at the caliper is approx 1880. N upward, so the vertical force at axle must be approx. 880 N downward.

Does that make sense?

One last step: take a reference frame where the axle is fixed in space and find the net torque on the swingarm (since it's not in static equilibrium)—there's about 0.64 Nm turning the swingarm clockwise around the axle, which produces about 21 N of force at the pivot, perpendicular to the swingarm.

That force has a vertical component of about 20 N that lifts the pivot and thus the front end of the motorcycle (and a horizontal component of about 7 N tends to push the motorcycle backward... of course the real effect would be that the overall center of mass of the motorcycle wouldn't translate backward, and instead the front wheel would be pushed forward)

Of course a real suspension will have additional complexities, like damping, but I'm just trying to get a basic understanding here. Thanks! (I also know that good free-body diagrams would be helpful... any guidance there would be appreciated.)

Answer 1

Nope - you can't do that. You have to get the link into static equilibrium first. That means you have to fix the geometry (consider the shock as a rigid link) and compute the force balance on the link. In component form, you have eight unknowns and need eight equations. The angle of the shock to the link is one. Summing x and y forces to zero gets to 3. Moment of link about fork pivot is 4. Component forces at caliper and hub must match external forces on wheel (6). The angle of the caliper gets you 7, moment of wheel about hub = 0 is the 8th.

So some of what you did is okay - getting the caliper forces from the external forces by setting wheel moment = 0. But the remaining six need to be redone.

This is called a quasi-static analysis, and they can be fraught with trouble when you have to introduce these pseudo-forces derived from accelerations. It would be nice if it were called pseudo-static analysis, but it isn't. There are quasi-static force balances as well. By ignoring the angular momentum of the decelerating front wheel, we balance the wheel torque to zero instead of $\dot\omega I_{hub}$. That simplification is a quasi-static approximation.

So go back and recompute the link forces at the hub, shock, and fork pivot such that the entire link is in equilibrium.

A more intuitive approach is to separate this into two linearly independent problems and analyze separately. Notice that the weight transfer during deceleration is purely a matter of wheel base and all the geometry is fixed for this analysis. So just drop it. Anti-dive means it dives less than you expect it to for a given weight transfer. So if there is no weight transfer, then anti dive means the shock must extend and that the force of the shock must decrease. Clearly this is due entirely to the deceleration forces. So just do the analysis for the 1000 N deceleration force, and see if it results in a larger or smaller compression on the shock. Then you can go back and notice that for this geometry, weight transfer is 1/2 decel, and decide how effective the anti-dive is based on that relation. You still have to dig out shock compression as a function of weight transfer, but then you just have a simple ratio to look at.