A well-known problem at recumbents with rearwheel-suspension is, that they move upwards and downwards (pogo) through the suspension, because of the varying pedalforce of the rider. This is not only annoying, it also causes loss of energy at the damper of the suspension. Specially at low gears at the mountains pogo and accompanying losses can be considerable.
A lot of solutions have been suggested for this problem. Some people see the solution at the chainline, which should run just above, straight through or just below the pivot of the swing arm.
Others see the solution in using a (Rohloff) gear hub (chain line is always the same for every gear), middle chainwheel mounted at the swing arm (RazzFazz) or desperately removing the rear suspension (combined with a fat rear-tire).
In this article we construct a solution, where the position of the middle chainwheel is crucial. As less as possible formulas are used. As usually all the mass of bike and rider will be situated at the center of gravity which is approximately located at the rider's umbilicus. Rearwheel and swing arm do not have their own mass.
The pedalforce of the cyclist is transported by chain and rearcog to the rearwheel.
Finally it is the friction force from the road to the tire, which pushes the bike forwards.
This force (Fd) causes accelleration of the bike.
Mass of bike and cyclist resist (inertia) with a force which attaches at the center of gravity.
This force is equal to Fd, but has the opposite direction.
The consequence of this force is that the frame will try to rotate.
To avoid rotation, there will be an extra pressure on the rear wheel and an extra reaction force Fw from the road at the wheel.
The forces Fd at the center of gravity and Fw at the rear wheel result into a torque at the contact point of the front wheel at the road.
To avoid rotation of the bike, the sum of these torques should be zero:
Fd * Hcg + Fw * Lwb = 0
where Hcg is heigth of center of gravity and Lwb is length of wheelbase.
If we have a fixed frame there is no problem to transport both forces through the frame to the center of gravity. But if the bike contains rear suspension the forces will compress suspension and cause pogo and loss at the damper, if the bike is not well designed. To avoid this, all forces, needed for acceleration should pass through the chain or the swing arm, and no force should pass through the suspension.
We have a bike with a center of gravity at the (front) frame without front suspension. There is some suspension (not drawn at the picture) from front frame to rear axle to avoid rotation of the frame around the front axle because of the static gravity forces. We introduce a strong bar which runs from the comtactpoint of the rearwheel at the road to a point vertical above the frontaxle at the height of the center of gravity. At the front it is fixed to the frame using a pivot, at the rear it can move without friction on the road. We accelerate the bike by pushing the bar with a horizontal force Fd. Because the bar has an angle with the road this will also cause a vertical force Fw.
The bar can freely rotate at both sides and can only transport forces, which have the same direction as the bar. So the sum of the forces must have the same direction or: Fd / Fw = Hcg / Lwb .
The vertical force Fw at the front cause no movement of the frame as long as this force is smaller than the (static) gravity force.
The force Fd at the front is horizontal and attaches the frame at the same height as center of gravity. There is no torque relative to the center of gravity and the (front) frame will not rotate. Acceleration causes no force at rear suspension and no pogo.
In the next figure we shorten the bar which now is attached to a pivot at the (front) frame somewhere between the wheels. The bar keeps the same direction as the longer bar.
The forces Fd en Fw at the front of the bar now attach the new pivot at the frame. However size and direction of the vector sum of these forces are still the same. Both forces still do not cause pogo.
We replace the bar with a long bar(vertical) with two side-bars which form a triangle and are attached to the pivot of the frame. The bars are also attached to each other by pivots, so there is no internal torque.
The triangle is locked, the construction behaves like the straight bar, there will be no pogo. At both sides still there are the same external forces..
Notice the triangle is stable for any angle of the upper bar. There is a pulling force at the upper "horizontal" bar and a pushing force at the lower "horizontal" bar. The direction of the forces is the same as the corresponding bars, because the bars freely rotate around the pivots.
The last step is obvious but changes a lot. The triangle is replaced by real components. The rearwheel with rearcog will replace the vertical bar. the swing arm will replace the lower "horizontal" bar. The chain atttaches the rearcog and replaces the upper "horizontal" bar and is tempoarily locked at the pivot at the frame. This last action is possible because there is only a pulling force at the upper bar. Therefore the chain in this case is equivalent to a real bar. We push the rearwheel forward at the bottom of the wheel and assume the wheel has no friction to the road.
Comparing both figures learns that internal and external forces will have exactly the same size and direction. This construction will have no pogo. If we put the middle chainwheel at the pivot on the frame and change the size of the rearcog, the triangle will change its form. Above we saw, also this action will give no pogo.