Simple theoretical suspension question I can't wrap my feeble mind around

So on another forum a guy posed a suspension problem on a go kart. It has a swing arm suspension similar to a motorcycle. The first and most obvious answer was to lengthen the swing arm. Another poster commented that if he lengthens the swing arm he needs to change to a stiffer spring to counteract the additional leverage of the longer arm. At first this seemed correct to me, but the more I thought about it I started to question that logic. Here's why, the suspended weight (the body and driver) stays the same. If the suspension mounting points (where the swing arm attaches to the body, where the coilover mounts to the body and the distance from the swingarm pivot to where the coilover attaches to the swingarm) stay the same shouldn't the suspension work just as it did before? I'm only talking about the springs ability to support the body, I realize that there will be more wheel travel, and the springs would have a harder time keeping the tire in contact with the ground after a bump. I know there are some really smart people here that can explain this to me. Thanks

No, it needs a stiffer spring to maintain the same wheel rate.  The wheel is where the force is acting on the lever arm.  The father away from the fulcrum (attachment point) the end of the lever (wheel) is, the greater the mechanical advantage it has.  This is regardless of where the spring is placed along the lever.

Lengthening the lever arm and leaving the spring in the same location is exactly the same effect as keeping the lever arm the same length and moving the spring inboard toward the pivot point - you need to stiffen the spring to maintain the same wheel rate.

There seem to be several key bits of data missing here.

1. What exactly is the kart owner trying to accomplish?

2. Where does the spring attach to the suspension - is it connected to the axle or to a point in the middle of the swing arm?

3. If the swing arm is lengthened, and the spring attaches to the swing arm, will it be cut between the spring and the  body mounting point, between the spring and the axle, or both?

Think if it this way: if you have a 200lb. bolder, and use a 4 foot lever, it's not easy to lift. If you use an 8 foot lever, it's twice as easy if the distence from the rock to the pivot is the same. The weight of the rock hasn't changed.

I just ran into this with my 3-link setup on a solid rear axle. In my case, the axle is between the spring and fulcrum. I sized the springs assuming the springs act at the center of the axle, but they actually mount a few inches behind it, effectively increasing my wheel rate way too stiff. I just sourced springs with a smaller rate proportional to the increased motion ratio. Or in your case, whatever % you increased lever length, increase your spring rate by that amount to keep the handling the same, unless you want a softer ride.

MadScientistMatt said:

There seem to be several key bits of data missing here.

1. What exactly is the kart owner trying to accomplish?

2. Where does the spring attach to the suspension - is it connected to the axle or to a point in the middle of the swing arm?

3. If the swing arm is lengthened, and the spring attaches to the swing arm, will it be cut between the spring and the  body mounting point, between the spring and the axle, or both?

1) His case is a little strange because the engine is mounted to the swing arm, and it was hitting the body on compression, but that is not pertinent to my question, this post is more of a thought exercise.

2) Lets say the spring mounts in the 6" from the pivot (where the swingarm attaches to the frame), then the axle mounts 18" from the pivot originally, and he wants to move the axle to 24" from the pivot

3) The cut and extension would happen between suspension and the axle. The 6" from the pivot to the suspension would stay the same.

Appleseed said:

Think if it this way: if you have a 200lb. bolder, and use a 4 foot lever, it's not easy to lift. If you use an 8 foot lever, it's twice as easy if the distence from the rock to the pivot is the same. The weight of the rock hasn't changed.

Right, I totally get that, and that was my first reaction, but on further thought I realized that "my hand on the pry bar" is now the tire sitting on the earth, which is going to hold regardless. A good analogy of what I'm thinking is you set the tip of the prybar on a work bench. You set the 200 lb boulder 6" from the tip. Regardless of how long the prybar is the weight of that boulder at where it sits on the prybar is the same regardless of how long the prybar is past that point. So if you put a spring under the boulder it will compress the same amount whether the handle is one foot or ten feet away. Or am I wrong in my thinking? As I stated, I'm sure that if you hit a bump the wheel will now have a harder time staying on the ground, but I don't think the spring will have any harder of a time holding the body up.

Duke said:

No, it needs a stiffer spring to maintain the same wheel rate.  The wheel is where the force is acting on the lever arm.  The father away from the fulcrum (attachment point) the end of the lever (wheel) is, the greater the mechanical advantage it has.  This is regardless of where the spring is placed along the lever.

Usually when I try to figure these things out I picture an extreme change because it is easier for me to visualize the change. So lets say you took a motorcycle swing arm and cut it between the spring and the axle. Then you welded in one mile of tubing. Would the spring be fully compressed afterwards? I don't believe so, because the spring is still holding the same amount of weight. This is my argument against the premise of needing a stronger spring.

depending on the angle of the swingarm, he may not need to stiffen the spring. The increase in ride height may be enough to keep the body from hitting under the same circumstances.

No way to know with the info provided. I assume this is just a play kart and not a racing kart, as you would be more interested in the effect of the lengthen wheel base, and I have never seen a race kart with a motor mounted as sprung weight.

No, because the wheel will have force enacted on it in either of 2 directions: the weight of the vehicle pushing down, and / or the force of a bump rising to push it up.  In both cases, if you have changed the swing arm length (outboard of the spring) from 18" to 24", you have increased the leverage the wheel has to move force in either direction by 33%.  For a fixed spring rate and location, that makes the spring 33% less resistant to transmitting force in either direction.  In a simple system like this, there is no magic about which way the force is flowing.

gearheadmb said:

1) His case is a little strange because the engine is mounted to the swing arm, and it was hitting the body on compression, but that is not pertinent to my question, this post is more of a thought exercise.

2) Lets say the spring mounts in the 6" from the pivot (where the swingarm attaches to the frame), then the axle mounts 18" from the pivot originally, and he wants to move the axle to 24" from the pivot

3) The cut and extension would happen between suspension and the axle. The 6" from the pivot to the suspension would stay the same.

In that case, your suspension will have a longer lever arm acting on the spring. With the original design, 3" of suspension travel will compress the spring 1". With 6" of extension, the same 3" of suspension travel compresses the spring 0.75". So it will behave as if the spring rate is 3/4 of what it was before.

He could get away with not changing the spring rate if he added equal amounts of length to both ends of the swing arm, then moved the chassis side spring mount back to keep the spring at the same angle. 

I think I'm starting to get it. Sitting still the spring compression would be the same as is was before, this is the fact that made me think the spring wouldn't need to be changed. But when you are moving anything that causes the spring to compress will now be multiplied by whatever percentage you lengthen the swingarm, causing more body travel and causing it to bottom out quicker. I'm glad I started this thread, I learned something today.

Wheel Rate = Spring Rate * Motion Ratio^2

The motion ratio describes how far the spring moves per unit of the wheel.  It also describes the ratio between the force at the wheel and the force at the spring.

For example, in the pictured setup, if C is 20", and B is 15", then the motion ratio is 15/20 = 0.75.  If the spring is 100lb/in, then the wheel rate will be 100*0.75^2= 56 lb/in

EDIT: A picture's worth a thousand words! I was a little slow with my tome... Thanks, Matthew Kennedy!

As Duke notes, the spring doesn't care whether the car is pressing down or the wheel is pushing up. It only cares about the distance between its two mounting points.

The important term which Duke also first mentioned above is "wheel rate". The rate of the spring and the leverage ratio (and a very little math) give you wheel rate. One important thing that hasn't been mentioned above is that wheel rate is the spring rate multiplied by the *square* of the motion ratio.

To explain, a MacPherson strut is very close to 1:1, because the amount the spring is compressed is very close to the amount the wheel is moved. 1 squared is 1, so a spring rate of 300 lb/in gives you a wheel rate of 300 lb/in.

By comparison, an arm with the wheel at the end and the spring in the middle has a ratio of 1:2. There are *two* things that happen as a result of that change in leverage. One is that the wheel has leverage, so if the spring is compressed 1" at a seat force of 300 lb, the wheel would be seeing only 150 lb. The other thing that gets missed sometimes in these discussions is that because of that ratio, if we see that spring compressed 1", the wheel has moved 2". For 1" of wheel motion, the spring would only be compressed 0.5", have a seat load of 150 lb, and a wheel load of 75 lb. EDIT: Summary: The "squaring" of the motion ratio is because it has a compound effect, and so is effectively applied twice (or that's how I tend to think of it): Once because it gives the wheel leverage over the spring, and AGAIN because it reduces how far the spring moves compared to the wheel.

To put it in the terms of your example with the spring at 6" and the axle at 18", and picking an arbitrary spring rate of 100 lb/in just to have all the terms, out motion ratio is 1:3, so our wheel rate is 1:9 with the spring rate. So our 100 lb/in spring gives us a wheel rate of 100/9 or about 11.1 lb/in. If we stick another six inches between spring and wheel so that the spring is still 6" from the pivot but the axle is out at 24", our motion ratio is 1:4, our wheel rate becomes 100/16 or about , 6.25 lb/in; a LOT softer.

We haven't talked about dampers at all, but increasing leverage ratio also has a similar effect on dampers, excepting that the reason is that doubling the motion ratio halves the damper velocity relative to the wheel, while having the same reduced leverage. So same principle, only applied to a device (the damper) that cares about velocity instead of position.

"I have a simple suspension question."

"No, you don't."

"What?"

"You don't. You don't have a simple suspension question. Those don't exist."

"Oh. Okay. I have a complicated suspension question..."

snailmont5oh said:

"I have a simple suspension question."

"No, you don't."

"What?"

"You don't. You don't have a simple suspension question. Those don't exist."

"Oh. Okay. I have a complicated suspension question..."

There is a Simple way to Explane this

I don't know what it is though.

Perhaps one of the best things about images of the lever thing is they show visually not only the reduction (or increase) in force provided by a lever, but ALSO that if the force is halved, that the distance is doubled.

So you can set up a 10:1 lever so that you only need to supply 10 lbs of force to lift a 100 lb object, but you'll have to move your end 10" to lift the object 1".