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8 second OEM 4 link drag race chassis setup

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The differing anti-squat numbers through the range of rearend travel may cause a slight oscilation, but it should be minimal. It should also afford a slight amount of dynamic roll control.
 
There seems to be conflicting opinions among the experts as to whether the length of the IC is important or not, assuming the IC is located on the same AS% line.
One opinion is that as long as the IC is located anywhere on the same AS% line, whether the IC length is 36" or 100', the launch reaction of the car is the same.
Another opinion has it that there is an advantage to having the IC length at a certain value according to the performance potential of the car. Even if the same AS% is being used.

A long IC length will cause lower AS% fluctuation with suspension travel, since the AS% lines diverge as the length becomes greater.
A long IC length can contribute to body roll? Don't really have my head wrapped around that one, yet. Maybe that might be the affect of the previous statement. A shorter IC length would cause higher AS% fluctuation with suspension travel, and would provide more dynamic roll control.
 
Note: The length of the rear axle housing swing arm has an affect on braking forces. A short swing arm length would tend to cause rear wheel hop during braking. With a short swing arm, the axle housing torque needed to counter braking forces can raise the rear wheels off the ground.
A recommended minimum swing arm length is 42 inches.
 
One thing I've come to realize as I was playing with the 4link program. The closer the upper and lower links are to parallel with the anti-squat 100% line, the less AS% variance there will be through the normal range of rear suspension movement.

A link angle change that results in more AS% variance through the normal range of rear suspension movement is lowering the UCA frame mounting.

A link angle change that results in less AS% variance through the normal range of rear suspension movement is lowering the LCA axle mounting.
 
There seems to be conflicting opinions among the experts as to whether the length of the IC is important or not, assuming the IC is located on the same AS% line.
One opinion is that as long as the IC is located anywhere on the same AS% line, whether the IC length is 36" or 100', the launch reaction of the car is the same.
Another opinion has it that there is an advantage to having the IC length at a certain value according to the performance potential of the car. Even if the same AS% is being used.

I'll add my .0002.
I do not agree with the below statement.
One opinion is that as long as the IC is located anywhere on the same AS% line, whether the IC length is 36" or 100', the launch reaction of the car is the same.

It easy to get caught up in AS and IC "values" when looking at a graph but one thing that never gets looked at is the actual measurements of the control arm mounting points relative to the axle centerline or how the forces act relative to the CG of the car.
As we all know the axle housing rotates backwards "around" the axle centerline. If the UCA mounting points were 14" above the centerline this would have MORE "pull" on the chassis than if they were 3" above the centerline.
The same type of push/pull affect is being done with the LCA's too. So the "distance" of the mounting points from the axle CL has a huge affect on how the suspension reacts on the chassis.

Now, all these pushing/pulling forces from the control arms are directed in straight lines "through" the UCA/LCA's. For simplicity I'll just talk about the forces through the LCA's below.

Lets say the Center of Gravity (CG) is located in the middle of the radio. The LCA's produce a "pushing" force in a straight line forward, past the radio. If the LCA's are parallel with the ground then the force is about 10" off the ground and parallel with the ground.
If the radio is 24" off the ground then the distance the "LCA Force" is from the radio is 14" (24"-10"=14").
Because the LCA force is points UNDER the CG, it tries to ROTATE the car UPWARDS (weight transfer). Kind of like pushing on a refridgerator. Push really hard below the CG it will fall backwards on top of you.

If you move the LCA's to 4" off the ground it will have MORE leverage to rotate the car (weight transfer).

If the LCA's point UP so their force is directed ABOVE the CG then the force is trying to lift the REAR of the car and push the nose of the car down.

The same type of forces are in the UCA's but instead of PUSHING on the chassis they are PULLING on the chassis. The UCA "forces" need to be considered along with the LCA forces to get an "overall" affect on how the chassis will react.

The reason why I disagree with the above quote is because the control arm mounting holes can be located at some crazy positions and it's these positions and angles of the control arms that determine the forces (and thier directions) that's exerted on the chassis.

For example:
Using your program create some geometry that has an IC length of around 25" that lies on the Neutral Line (100% AS).
Now create new geometry that as an IC length of 300" and 100% AS.

Now compare the ANGELS of the control arms. The first example will load the tires a LOT more than the second example but it won't have squat for leverage to lift the nose of the car. The second example won't load the rear tires at all but will have a lot more leverage to lift the frontend, compared to the first example.

Some people ask, "why not have a long IC if that's what lifts the frontend?"
Answer: The car can't lift the frontend UNTIL it gets traction and the rear tires need to be loaded first before the frontend can be lifted. That's where a compromise between the UCA/LCA angles need to be found which is done by testing and experimenting.

ks
 
I'll add my .0002.
I do not agree with the below statement.
One opinion is that as long as the IC is located anywhere on the same AS% line, whether the IC length is 36" or 100', the launch reaction of the car is the same.

It easy to get caught up in AS and IC "values" when looking at a graph but one thing that never gets looked at is the actual measurements of the control arm mounting points relative to the axle centerline or how the forces act relative to the CG of the car.
As we all know the axle housing rotates backwards "around" the axle centerline. If the UCA mounting points were 14" above the centerline this would have MORE "pull" on the chassis than if they were 3" above the centerline.
The same type of push/pull affect is being done with the LCA's too. So the "distance" of the mounting points from the axle CL has a huge affect on how the suspension reacts on the chassis.

Now, all these pushing/pulling forces from the control arms are directed in straight lines "through" the UCA/LCA's. For simplicity I'll just talk about the forces through the LCA's below.

Lets say the Center of Gravity (CG) is located in the middle of the radio. The LCA's produce a "pushing" force in a straight line forward, past the radio. If the LCA's are parallel with the ground then the force is about 10" off the ground and parallel with the ground.
If the radio is 24" off the ground then the distance the "LCA Force" is from the radio is 14" (24"-10"=14").
Because the LCA force is points UNDER the CG, it tries to ROTATE the car UPWARDS (weight transfer). Kind of like pushing on a refridgerator. Push really hard below the CG it will fall backwards on top of you.

If you move the LCA's to 4" off the ground it will have MORE leverage to rotate the car (weight transfer).

If the LCA's point UP so their force is directed ABOVE the CG then the force is trying to lift the REAR of the car and push the nose of the car down.

The same type of forces are in the UCA's but instead of PUSHING on the chassis they are PULLING on the chassis. The UCA "forces" need to be considered along with the LCA forces to get an "overall" affect on how the chassis will react.

The reason why I disagree with the above quote is because the control arm mounting holes can be located at some crazy positions and it's these positions and angles of the control arms that determine the forces (and thier directions) that's exerted on the chassis.

For example:
Using your program create some geometry that has an IC length of around 25" that lies on the Neutral Line (100% AS).
Now create new geometry that as an IC length of 300" and 100% AS.

Now compare the ANGELS of the control arms. The first example will load the tires a LOT more than the second example but it won't have squat for leverage to lift the nose of the car. The second example won't load the rear tires at all but will have a lot more leverage to lift the frontend, compared to the first example.

Some people ask, "why not have a long IC if that's what lifts the frontend?"
Answer: The car can't lift the frontend UNTIL it gets traction and the rear tires need to be loaded first before the frontend can be lifted. That's where a compromise between the UCA/LCA angles need to be found which is done by testing and experimenting.

ks
I think you are touching on another area of study where it comes to understanding the forces placed on a drag chassis. Kinematics? I have not looked into this much. All I've come across on that subject is what you touched on with the placement of the link end in relation to the axle centerline. I understand that it's not best to have the link end any closer than 3" from the axle centerline due to the high loads placed on the link and link ends as you move the link end location closer to the axle centerline. I can see how that would be with the LCA, but you state that with the UCA, as you move the link end further up from the axle centerline, the pull on the link increases? Which end of the link are you referring too? I would think that if it were the link end at the axle housing, the pull would decrease as the link end was moved away from the axle housing. A simple leverage problem, I thought. Please explain.
BTW, thanks for the input.
 
I can see how that would be with the LCA, but you state that with the UCA, as you move the link end further up from the axle centerline, the pull on the link increases? Which end of the link are you referring too? I would think that if it were the link end at the axle housing, the pull would decrease as the link end was moved away from the axle housing. A simple leverage problem, I thought. Please explain.
BTW, thanks for the input.

This is where things get confusing b/c it needs to be thought of as "work" being done ON THE CHASSIS/FRAME.

Think of the distance from the UCA hole to the axle CL as a fishing pole where the axle CL is the handle and the tip of the pole the UCA hole.

When you get a bite and try to set the hook, a 12 foot fishing pole will have a lot more pull on the hook than a pole 2 inches long even with the same force that you yank on the fishing pole.
It's the same affect with the control arm spacing from the axle CL.

ks
 
Come to think of it, is there really a pull going on at the UCA during acceleration, or is it a compressive load? If it were compressive, then I would understand the leverage on the rearend housing with the upper rear link end mounted further from the axle centerline.
 
Come to think of it, is there really a pull going on at the UCA during acceleration, or is it a compressive load? If it were compressive, then I would understand the leverage on the rearend housing with the upper rear link end mounted further from the axle centerline.

The "pull" is due to the pinion rotating and trying to climb up the ring gear. As long as the pinion is spinning faster and faster (accelerating) then the UCA's are being pulled.

ks
 
This is where things get confusing b/c it needs to be thought of as "work" being done ON THE CHASSIS.

Think of the distance from the UCA hole to the axle CL as a fishing pole where the axle CL is the handle and the tip of the pole the UCA hole.

When you get a bite and try to set the hook, a 12 foot fishing pole will have a lot more pull on the hook than a pole 2 inches long even with the same force that you yank on the fishing pole.
It's the same affect with the control arm spacing from the axle CL.

ks
This is how I think of it. Imagine you're standing with a bucket full of water on the floor next to you and you have to pick it up and hold it to shoulder height. Is it easier to pick up the bucket of water with your arm bent to give a shorter distance from your body to the bucket, or would it be easier to pick the bucket up with your arm stretched out at full length and the distance between the bucket and yourself being longer?

I remember doing an exercise where you lifted dumbbells out to your sides with your arms locked at full extension. It was not the easier way of lifting the dumbbells.

You also have to remember that a fishing pole bends with added pull to give you a shorter distance from hand to tip of pole. If it didn't, and you were holding the pole at 90 degrees to the force pulling on the pole, the fish would have the leverage advantage.
 
This is how I think of it. Imagine you're standing with a bucket full of water on the floor next to you and you have to pick it up and hold it to shoulder height. Is it easier to pick up the bucket of water with your arm bent to give a shorter distance from your body to the bucket, or would it be easier to pick the bucket up with your arm stretched out at full length and the distance between the bucket and yourself being longer?

I remember doing an exercise where you lifted dumbbells out to your sides with your arms locked at full extension. It was not the easier way of lifting the dumbbells.

You are correct and this is the confusing part b/c you're looking at it from the opposite end of the force diagram. You're thinking about it more from "conserving power" "what's easiest to do" instead of exerting the most power TO the object.
If you used the whole length of your arm to lift the buck you will lift the bucket farther and faster than if you only used half of your arm.

The power or torque the rearend is rotating is already a given at 12000ft/lbs of torque (or whatever it is), this won't change no matter where the control arms are mounted b/c this is derived from the power of the engine turning the driveshaft. Now using that power, what will produce the most affect on the chassis? A longer "lever" or shorter one?

ks
 
The "pull" is due to the pinion rotating and trying to climb up the ring gear. As long as the pinion is spinning faster and faster (accelerating) then the UCA's are being pulled.

ks
I understand how the pinion wants to climb the ring gear, and rotate the differential carrier and axles. The tire patch, when in contact with the ground is resisting this rotation. It's this resistance setup by the tire on the ground that causes torque on the axle housing. If the car was up off the ground on a lift and the driveshaft was turned, the tires would freely turn, and not the axle housing. No push/pull force on any link. Maybe a minute amount contributable to the mass of the ring gear, differential carrier, axles, wheels and tires initially resisting acceleration due to the pinion input.
 
You are correct and this is the confusing part b/c you're looking at it from the opposite end of the force diagram. You're thinking about it more from "conserving power" "what's easiest to do" instead of exerting the most power TO the object.
If you used the whole length of your arm to lift the buck you will lift the bucket farther and faster than if you only used half of your arm.

The power or torque the rearend is rotating is already a given at 12000ft/lbs of torque (or whatever it is), this won't change no matter where the control arms are mounted b/c this is derived from the power of the engine turning the driveshaft. Now using that power, what will produce the most affect on the chassis? A longer "lever" or shorter one?

ks
My first thought is a shorter one. Just as a smaller gear has an easier time turning a larger gear. Reverse the situation and the larger gear has a harder time turning a smaller gear.
 
Kevin. Please go back and read post 90. I edited it.

My first thought is a shorter one. Just as a smaller gear has an easier time turning a larger gear. Reverse the situation and the larger gear has a harder time turning a smaller gear.

You're still correct in that it requires less power to lift it but we're not conserving power. We are trying to take what power we have and do the most work with it.
In the fishing pole example, pretend none of the poles bend at all. The longer pole will affect the chassis more than the shorter one.

How about this example:
A baseball bat hitting a ball. If you wanted to hit a homerun (apply as much force as you can) you want to hit the ball towards the end of the bat not near the handle where your hands are, even though your swing is the same. Your body will feel the impact more when the ball hits at the end of the bat but the ball will go farther.
Same example with a golf swing: short club or long club.

ks
 
I understand how the pinion wants to climb the ring gear, and rotate the differential carrier and axles. The tire patch, when in contact with the ground is resisting this rotation. It's this resistance setup by the tire on the ground that causes torque on the axle housing. If the car was up off the ground on a lift and the driveshaft was turned, the tires would freely turn, and not the axle housing. No push/pull force on any link. Maybe a minute amount contributable to the mass of the ring gear, differential carrier, axles, wheels and tires initially resisting acceleration due to the pinion input.
Now with the tire contact patch in the picture, it changes how the forces are transmitted through the upper and lower links. We all understand that the lower link wants to push forward, but the lower link is between the axle and the tire patch. Without the upper link in place, the axle housing would push forward and onto the lower link, as long as the lower link had a mass that resisted the push forward by the axle housing.
 
You're still correct in that it requires less power to lift it but we're not conserving power. We are trying to take what power we have and do the most work with it.
In the fishing pole example, pretend none of the poles bend at all. The longer pole will affect the chassis more than the shorter one.

How about this example:
A baseball bat hitting a ball. If you wanted to hit a homerun you want to hit the ball towards the end of the bat not near the handle where your hands are, even though your swing is the same. Your body will fill the impact more when the ball hits at the end of the bat but the ball will go farther.

ks
In the baseball case, you are swinging the bat. You have built up a momentum. There is power in that momentum. Much more than simple leverage on another object.
If you are trying to make the case that there is enough clearances in the suspension components that allows a momentum to build in the rotating axle housing, then I'm starting to understand. But if everything is solidly mounted, I don't see how this momentum has a chance to start building.
 
Going to bed. I hope we can pick this up again, later.
Thanks for the input, Kevin.
 
Bell went off in my head. OK. One last try. Are you saying that you are trying to do the most work at the expense of some of your power?
 
In the baseball case, you are swinging the bat. You have built up a momentum. There is power in that momentum. Much more than simple leverage on another object.
If you are trying to make the case that there is enough clearances in the suspension components that allows a momentum to build in the rotating axle housing, then I'm starting to understand. But if everything is solidly mounted, I don't see how this momentum has a chance to start building.

It has nothing to do with momentum b/c the momentum of the bat is the exact same in both examples and no matter where the ball hits the bat. It has to do with what example will produce the most affect on the ball (or chassis).

It's getting late here 1:30am and time for bed.:o

ks
 
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