Understand Rear Axle Forces in Car Suspensions w/ Instant Center Plot

[Tom Rauji]

When I read pages or look at calculators for car suspensions it seems they have things wrong. They seem to consider both the upper and lower arms as a straight line intersection where the tire contact patch effectively "pushes" through the intersect point. I'm trying to understand the forces at work here. If the upper arms (UCA) had a net pushing force, I would agree. The UCA's however pull. Only the LCA's push forward.

Wouldn't both upper and lower arms have to push with equal forces to plot an effective single net force line crossing where those arms intersect? It seems to me since the upper arm pulls backwards, it does not have the same effect as the lower arm and cannot simply be plotted to a single point where they cross. They are separated a fairly large distance on the chassis and both ends swivel on joints.They do not joint at one point.

This is a rear axle in a car. The axle is driven by a ring and pinion gear. The axle is anchored only by four control arms, I'm looking at just one side. If I assign some arbitrary torque on the axle, like 1000 pound-ft, the upper arm would be pulling backwards and the lower arm alone would push forward. Wouldn't the forces have to be treated that way, instead of treated like a sum at a point where an imaginary extension of the lines would meet?

I can't get my head around this. Can anyone help me with the forces in this system assuming some normalized torque at the axle that is working against the housing along with the normal reaction to the tire thrust on the road?

This is a link to an Internet example of an instant center plot.

http://performancetrends.com/graphics/4 Link Plus Main.gif

 

[.Scott]

Relative to the road, the tire is rotating about that patch. And the entire forward push from the tire is through the axle.
So, by the time you get to the UCA and LCA, you can ignore the rotation of the tire.
The axle is pushing its housing forward. So you will have push on both the UCA and LCA.

The flaw in your reasoning is that you are treating the torque to the axle as coming from the housing. Of course, it does not.

 

[Tom Rauji]

Relative to the road, the tire is rotating about that patch. And the entire forward push from the tire is through the axle.
So, by the time you get to the UCA and LCA, you can ignore the rotation of the tire.
The axle is pushing its housing forward. So you will have push on both the UCA and LCA.

The flaw in your reasoning is that you are treating the torque to the axle as coming from the housing. Of course, it does not.

I'm pretty sure I am treating it correctly. The upper control arms actually rip backwards out of the floor. They tear the welds and the floor as they pull rearward. This fots perfectly with what I calculate, although I'm not sure of the exact sum of forces.

The lower arms push against the mounts, because they tend to crush the torque boxes forward.

This is what it looks like when I add up numbers and it agrees with how parts bend or fail. The upper arms absolutely pull out of the floor. No question at all about that at all.

This drawing has slightly different dimensions, but it is an example of the forces.

 

[.Scott]

The upper control arms actually rip backwards out of the floor. They tear the welds and the floor as they pull rearward. This fits perfectly with what I calculate, although I'm not sure of the exact sum of forces.

That should only happen during braking - assuming that braking brakes the wheel against the housing.
I am assuming that the axle conducts only torque, so that the entire translational forces from wheel to car are conducted through the arms. If this is the case, then the total force through the arms must add up to the force against the road (assuming the wheel, housing, and a portion of the axle are relatively low mass compared to that of the entire car).

 

[cjl]

The torque should actually be transmitted to the car through the differential mounting. If that is also mounted in a way that is connected to the control arms, then yes, you are treating it correctly (I haven't worked with solid axles, just independent, where it's connected directly to the rear subframe and thus transfers the torque there).

 

[Ranger Mike]

short answer is that the rear links merge at an instant center in front of the axle. The location both forward and side to side impact on t he amount of traction delivered to the rear wheel/tire combination. Same goes for handling braking forces. the long answer is to read Race Car suspension class (see above ) for the physics of the tire / wheel / suspension interaction. This applies to the independent rear suspension as well as solid differential type.
will take you a long time to read it but should explain all you ask.
rm

 

[Tom Rauji]

The torque should actually be transmitted to the car through the differential mounting. If that is also mounted in a way that is connected to the control arms, then yes, you are treating it correctly (I haven't worked with solid axles, just independent, where it's connected directly to the rear subframe and thus transfers the torque there).

It interacts on the arms exactly the way I treat it. The axle housing has a rotating force opposite the axle and the same torque as the axle, and that must add through some sort of vector addition to the tire thrust. Since the control arms are only about 4 inches out from center line and the tires are 13-14 inch radius (they squash on the bottom) the force multiplication has to be leveraged to something like 13.5/4 = 3.375 times the wheel. This should add to the lower arm and subtract from the upper arm, making the upper arm pull backwards.

This makes sense since in the real world all Mustangs with any power and good tires will tear the upper arms backwards out of the floor. In my car the force is so great it actually cracks the floor pan, but every one I've looked at using big tires and good power is ripping the upper arms backwards. Usually it cracks the welds and starts to pull the mounting boxes out.

I'm just not sure how the forces are transmitted through the vehicle and what the forces actually are. I'm an electrical engineer, not mechanical, so I have no experience in how to treat this. I need help with that. Plotting this like all the arms "push" to find instant center makes no sense to me. It cannot push when it is pulling backwards.

 

[cjl]

The axle housing definitely does not have that torque applied there. The location where that rotating force is applied is at the rear differential, since that is the other end of the drive axles. This will be transferred to the car through the differential mounts.

EDIT: After looking at some diagrams of solid rear axle suspension layouts though, it appears that this force will be transferred to the axle housing, since the diff is rigidly mounted to it. As I said, I previously only have worked with independent rear suspension vehicles, where it would be transmitted directly to the rear subframe through the diff mounts.

 

[Tom Rauji]

short answer is that the rear links merge at an instant center in front of the axle. The location both forward and side to side impact on t he amount of traction delivered to the rear wheel/tire combination. Same goes for handling braking forces. the long answer is to read Race Car suspension class (see above ) for the physics of the tire / wheel / suspension interaction. This applies to the independent rear suspension as well as solid differential type.
will take you a long time to read it but should explain all you ask.
rm

I understand that is what everyone says. I just disagree with it. Plotting that way makes no sense to me when the upper arms are pulling backwards and only the lower arms are pushing.

If it was a ladder bar that was really long axle rotation would transfer into a lifting force that vectors with the wheel's forward push from the contact patch, but this opened up link system is not pushing at all on the top. It is pulling. We know that for a fact because of how it diagrams and because of how it rips the upper torque boxes backwards. I just don't see how a forward force plot that treats upward and lower arms as having equal forward force is valid.

 

[Tom Rauji]

The axle housing definitely does not have that torque applied there. The location where that rotating force is applied is at the rear differential, since that is the other end of the drive axles. This will be transferred to the car through the differential mounts.

EDIT: After looking at some diagrams of solid rear axle suspension layouts though, it appears that this force will be transferred to the axle housing, since the diff is rigidly mounted to it. As I said, I previously only have worked with independent rear suspension vehicles, where it would be transmitted directly to the rear subframe through the diff mounts.

...and with an integrated differential and housing the axle housing has to transfer torque from the housing as a differential force on control arms. One way to look at this is the ring gear is working against the housing, pushing on the bottom and pulling backwards on the top arms.

I think I can do a better job tuning my suspension once I get a handle on the force vectors, but I am not a mechanical engineer. I've also had limited mechanics like this in physics classes and that was 40-50 years ago. :)

 

[Tom Rauji]

Thanks everyone! I greatly appreciate everyone responding. Disagreements are valuable because they help me look at my idea critically when considering alternatives.

I just don't find anything that tells me everyone is doing instant center plots correctly with this suspension and geometry.

 

[Ranger Mike]

This is my last post on your subject. You do not want to read the recommended posts and i can not make you so my time is devoted to assisting those who will.

you are forgetting a few major things. You are thinking 2D not 4D as required.

You have zero traction without down force to plant the tires. it is all about tires, Tires, TIRES. Learn about down force and the traction circle. Linkage length and instant center is very important to safety and proper acceleration. . The angle of each link makes a huge difference in traction as well. This is why dirt track race cars use 4 link rear suspension. Asphalt cars use 3 link to kill off the huge traction available. The answers are there for you to see.

 

[Tom Rauji]

This is my last post on your subject. You do not want to read the recommended posts and i can not make you so my time is devoted to assisting those who will.

you are forgetting a few major things. You are thinking 2D not 4D as required.

You have zero traction without down force to plant the tires. it is all about tires, Tires, TIRES. Learn about down force and the traction circle. Linkage length and instant center is very important to safety and proper acceleration. . The angle of each link makes a huge difference in traction as well. This is why dirt track race cars use 4 link rear suspension. Asphalt cars use 3 link to kill off the huge traction available. The answers are there for you to see.

Thanks! I obviously am not communicating well.

I'm trying to understand **why** instant center calculators treat the upper and lower arms like they both equally push on the chassis, when in the real world the upper arms pull back and only the lower arms push forward.

I understand how a ladder bar, where the upper and lowers merge at a point and axle housing rotation is transferred to the chassis as lift, lifts at the attachment point as the tire patch pushes forward, and how that forward push adds a second force at an angle to that chassis connection upward lift. I'm simply looking at why the upper and lower control arms are treated as both pushing when they don't act that way.

 

[Tom.G]

Sounds like the conflicts in understanding are due to Solid Rear Axles vs. Independent Rear Suspension (i.e. with U-joints in the axles, as in Front Wheel Drive vehicles). The OP has the Solid Rear Axle configuration. @cjl stated this difference in his post above but was appearently overlooked.

 

[Tom Rauji]

Sounds like the conflicts in understanding are due to Solid Rear Axles vs. Independent Rear Suspension (i.e. with U-joints in the axles, as in Front Wheel Drive vehicles). The OP has the Solid Rear Axle configuration. @cjl stated this difference in his post above but was appearently overlooked.

This may be my fault because I initially said:

"This is a rear axle in a car. The axle is driven by a ring and pinion gear. The axle is anchored only by four control arms, I'm looking at just one side."

I probably should have said: "This is a SOLID rear housing in a car. The axles are driven by a ring and pinion gear IN THE HOUSING. The axle HOUSING is anchored only by four control arms. I'm looking at just one side."

(Very few higher power drag cars have the differential carrier anchored to the body with U-joints in the axles. Axle torque is enormous. I start with 900 lb/ft on the engine, the torque converter is over 2:1 multiplication at stall, the trans is 1.68, and rear is 4.10 so axle torque is 12,400 lb-ft or more divided between two axles or about 6,200 lb-ft per tire. The tires sometimes actually don't spin with that hit so there is a great deal of stress as the car starts to move. That's why independent suspensions aren't common in drag cars, and why upper arms pull backwards so hard. I don't think that force should be ignored in calculations.)

 

[jack action]

So forget about torque for now. Forget about drag also and just focus on moving the vehicle's body (the inertia $ma$):

Both red forces must be equal and opposite. Each set of blue forces acting on the suspension links must be equal and opposite to their respective red force. The upper and lower blue forces may not be equal, but let's say they are just for the sake of discussion.

Now let's add a torque on the axle:

Now the red forces stay the same. They have to. On the suspension links, the sum of the blue forces stays also the same, since it's equal to the red force. What changed is that the upper link force has decreased and the lower link force has increased. If the upper link force reduces enough to the point of changing direction, the only difference will be that the lower link force will increase even more to compensate for it.

So as you can see, whether it is pushing or pulling of either link, the resultant force from the axle - going through the links - must be pushing against the body (the red forces must be one against the other one). That is what is important.

 

[Tom.G]

So forget about torque for now.

But that is the problem definition. There is no other way for the reaction torque of the differential to couple to the car frame. Perhaps @Tom Rauji can supply some photos of the upper control arm coming thru the floorboard.

The upper control arms actually rip backwards out of the floor.

The lower arms push against the mounts, because they tend to crush the torque boxes forward.

Jack, could you apply some labels to the nice drawings you did? I can't quite sort out what is what.

p.s. Here are some videos showing rear leaf spring windup (aka axle wrap) under acceleration.
This first one is the best. Especially at the 1 minute mark.

 

[jack action]

But that is the problem definition. There is no other way for the reaction torque of the differential to couple to the car frame.

The problem, as I understand it, is here:

They seem to consider both the upper and lower arms as a straight line intersection where the tire contact patch effectively "pushes" through the intersect point. I'm trying to understand the forces at work here. If the upper arms (UCA) had a net pushing force, I would agree. The UCA's however pull. Only the LCA's push forward.

Wouldn't both upper and lower arms have to push with equal forces to plot an effective single net force line crossing where those arms intersect?

My answer is to show why it doesn't matter whether one is pushing or pulling. All that matters is the sum of both forces (which must «push»).

Jack, could you apply some labels to the nice drawings you did? I can't quite sort out what is what.

• The square box is the car frame with the reaction force ($ma$, in red) at the CG;
• The round is the axle with the thrust force (in red) at the tire contact patch;
• The green lines are the upper and lower links of the suspension, split in two to show the internal forces within the links (in blue).

So there are 2 simplified free body diagrams on each drawings: one for the body, one for the axle. Though, I did not draw the torque on the second drawing (on either the frame or the axle).

 

[jack action]

I just thought of something else that may add to the conversation.

Although the Instantaneous Center of rotation (IC) may be found the same way for a vehicle with the differential bolted to the frame or one within the axle, the effective swing arm is not the same for both:

One begins at the tire-road contact patch, the other at the wheel center. The difference between the two being the thrust force times the tire radius, or the equivalent of the wheel torque.

 

[Tom.G]

Ahh, OK. We are all looking at different aspects of the situation then. I agree that both control arms contribute to accelerating the vehicle. The reaction torque is then added to these vehicle accelerating forces on the arms, thus there is a resultant vector tending to rotate the control arms 'upward' and/or 'rearward.'

I suspect the online calculators and descriptions the OP has seen are 'simplifying' the description by ignoring the reaction torque, and concern themselves only with the acceleration forces. For a street car this may well be reasonable. However, the OPs situation is extreme enough that the reaction torque can not be ignored, as evidenced by failed welds and torn sheetmetal!

The crushed torque boxes for the LCA may be occurring after the UCA mounts fail.

A solution could be to design the UCA mounts to handle the full reaction torque in addition to at least 1/2 the vehicle accelerating force. This may also require beefier UCAs.

Comments?

p.s. Just as I was going to 'post reply', Jack beat me to it. I think we are saying the same thing now.

 

[Tom Rauji]

So forget about torque for now. Forget about drag also and just focus on moving the vehicle's body (the inertia $ma$):

View attachment 216531​

Both red forces must be equal and opposite. Each set of blue forces acting on the suspension links must be equal and opposite to their respective red force. The upper and lower blue forces may not be equal, but let's say they are just for the sake of discussion.

Now let's add a torque on the axle:

View attachment 216532​

Now the red forces stay the same. They have to. On the suspension links, the sum of the blue forces stays also the same, since it's equal to the red force. What changed is that the upper link force has decreased and the lower link force has increased. If the upper link force reduces enough to the point of changing direction, the only difference will be that the lower link force will increase even more to compensate for it.

So as you can see, whether it is pushing or pulling of either link, the resultant force from the axle - going through the links - must be pushing against the body (the red forces must be one against the other one). That is what is important.

And what you describe above is exactly what happens in real life. You got it.

The housing bolts have about 7.5 inches of spread. The tire surface allowing for flattening from weight is about 13 inches or so (depending on air pressure and weight and distortion) so let's call it 13.5.

This means the lower to upper bolt spacing has a leverage of 13.5/7.5 = 1.8 : 1. So if the tire was pushing with 1000 lbs the housing would be twisting with 1800 pounds on that spread. Do I have this correct??

Thanks for your time.

 

[Tom Rauji]

Ahh, OK. We are all looking at different aspects of the situation then. I agree that both control arms contribute to accelerating the vehicle. The reaction torque is then added to these vehicle accelerating forces on the arms, thus there is a resultant vector tending to rotate the control arms 'upward' and/or 'rearward.'

I suspect the online calculators and descriptions the OP has seen are 'simplifying' the description by ignoring the reaction torque, and concern themselves only with the acceleration forces. For a street car this may well be reasonable. However, the OPs situation is extreme enough that the reaction torque can not be ignored, as evidenced by failed welds and torn sheetmetal!

The crushed torque boxes for the LCA may be occurring after the UCA mounts fail.

A solution could be to design the UCA mounts to handle the full reaction torque in addition to at least 1/2 the vehicle accelerating force. This may also require beefier UCAs.

Comments?

p.s. Just as I was going to 'post reply', Jack beat me to it. I think we are saying the same thing now.

This is exactly what I suspect they are doing, simplifying. Because the housing torque forces are so large the simple calculation must have errors. Even if everyone is doing it that way and has done it for years, it can still be wrong.
I need to understand how the bar angles interact with the car.

 

[jack action]

This means the lower to upper bolt spacing has a leverage of 13.5/7.5 = 1.8 : 1. So if the tire was pushing with 1000 lbs the housing would be twisting with 1800 pounds on that spread. Do I have this correct??

It might be worst than that.

Because the tire is pushing 1000 lb, the links must also push 1000 lb net. Say both links each pushes 500 lb, then you add the 1800 lb couple, which leads to 2300 lb (= 500 + 1800) on the lower link and -1300 lb (= 500 - 1800) on the upper link. The couple is added to the net force (2300 + (-1300) = 1000).

This is of course an ideal case in 1 dimension, but you would have to consider 2-D, where you will also take into account jacking forces in the vertical direction (anti-squat) which may absorb part of the torque reaction.

It is really just doing a free body diagram to find the reaction forces at the suspension links, springs and tires.

 

[jack action]

This means the lower to upper bolt spacing has a leverage of 13.5/7.5 = 1.8 : 1. So if the tire was pushing with 1000 lbs the housing would be twisting with 1800 pounds on that spread. Do I have this correct??

It might be worst than that.

Because the tire is pushing 1000 lb, the links must also push 1000 lb net. Say both links each pushes 500 lb, then you add the 1800 lb couple, which leads to 2300 lb (= 500 + 1800) on the lower link and -1300 lb (= 500 - 1800) on the upper link. The couple is added to the net force (2300 + (-1300) = 1000).

This is of course an ideal case in 1 dimension, but you would have to consider 2-D, where you will also take into account jacking forces in the vertical direction (anti-squat) which may absorb part of the torque reaction.

It is really just doing a free body diagram to find the reaction forces at the suspension links, springs and tires.

 

[Ranger Mike]

Instant center on the rear end simply shows the front pivot point of the linkage. Do not get hung up on its calculation. Simply be aware that longer trailing arms are preferred as they produce less angular change as the car experiences compression on one side and rebound on the other side.

There are four different movements that happen under acceleration. Side to side, fore to aft, Pitch about the rear end axle axis and Yaw rotation about the differential centerline vertical axis that goes up and down as viewed from above . You have these 4 forces happening over a very brief period of time.

Think 4-D at this point. As you spool up the engine and the green light pops, the engine rotary motion converts 90 degrees to rotational motion thru the wheels and tires to the pavement. If you are really hooked up, the tire contact patch resists the rotary force and we have # 6 in attached photo lifting the front of the vehicle. If great enough we get a wheelie. Typically we want maximum front end lift as we compress the rear suspension. Squat is described as weight transfer force to the rear end under acceleration. The term “weight transfer” is used and I hate it as it is not really what is happening but is a good visual. The front end lifts and the rear end gets “ heavy” and we have additional down force on the tires. Downforce is traction ( to a certain point). Additionally, the rear bottom links are trying to crawl under the car adding to more traction. This is why the attack angles on the links are critical to fine tune the suspension on different tracks or the same track at different times ( think temperatures).

 

[Tom Rauji]

Instant center on the rear end simply shows the front pivot point of the linkage. Do not get hung up on its calculation. Simply be aware that longer trailing arms are preferred as they produce less angular change as the car experiences compression on one side and rebound on the other side.

There are four different movements that happen under acceleration. Side to side, fore to aft, Pitch about the rear end axle axis and Yaw rotation about the differential centerline vertical axis that goes up and down as viewed from above . You have these 4 forces happening over a very brief period of time.

Think 4-D at this point. As you spool up the engine and the green light pops, the engine rotary motion converts 90 degrees to rotational motion thru the wheels and tires to the pavement. If you are really hooked up, the tire contact patch resists the rotary force and we have # 6 in attached photo lifting the front of the vehicle. If great enough we get a wheelie. Typically we want maximum front end lift as we compress the rear suspension. Squat is described as weight transfer force to the rear end under acceleration. The term “weight transfer” is used and I hate it as it is not really what is happening but is a good visual. The front end lifts and the rear end gets “ heavy” and we have additional down force on the tires. Downforce is traction ( to a certain point). Additionally, the rear bottom links are trying to crawl under the car adding to more traction. This is why the attack angles on the links are critical to fine tune the suspension on different tracks or the same track at different times ( think temperatures).

Right! I've had a real 4 link and a few ladder bar cars over years. I have a good grasp of real ladder bar and four link cars.

The issue I have is how people are plotting Mustang Fox "four links" with the short upper arm tied to the floor by the package tray and the lower to the frame rails. I suspect they are doing anti-squat in a meaningless way. My car reacts a mile off from calculators that have no allowance for axle wrap forces. I'm trying to learn the correct way to do this because going on the back bumper at high speed can be lethal.

 

[Tom Rauji]

It might be worst than that.

Because the tire is pushing 1000 lb, the links must also push 1000 lb net. Say both links each pushes 500 lb, then you add the 1800 lb couple, which leads to 2300 lb (= 500 + 1800) on the lower link and -1300 lb (= 500 - 1800) on the upper link. The couple is added to the net force (2300 + (-1300) = 1000).

This is of course an ideal case in 1 dimension, but you would have to consider 2-D, where you will also take into account jacking forces in the vertical direction (anti-squat) which may absorb part of the torque reaction.

It is really just doing a free body diagram to find the reaction forces at the suspension links, springs and tires.

This is great. This is very helpful.

I'm an electrical engineer, not mechanical, so I'm a little stupid on this.

So I treat the axle wrap as a couple and it just adds into other forces as a vector? Can you recommend a good book that deals with this? Not necessarily for cars, but the physics or mechanics involved? The issue I have to solve is I plant the tires really well but once I apply over a certain threshold of power the car gets unstable. It does a nice small lift in the front over a wide range of power up to a threshold, but once beyond that it will go right on the back bumper. Something rapidly changes in the geometry at a certain front end lift or rear extension down out of the body, and up it goes. My real four links and ladder bars especially were not
unstable like this. My gut feeling it is because of upper arm length and location

.

 

[jack action]

So I treat the axle wrap as a couple and it just adds into other forces as a vector? Can you recommend a good book that deals with this? Not necessarily for cars, but the physics or mechanics involved?

The basics for it is statics analysis. I found this website which seems to give good tutorials about it. The chapters about moments and trusses should relate closely to a suspension analysis.

The issue I have to solve is I plant the tires really well but once I apply over a certain threshold of power the car gets unstable. It does a nice small lift in the front over a wide range of power up to a threshold, but once beyond that it will go right on the back bumper. Something rapidly changes in the geometry at a certain front end lift or rear extension down out of the body, and up it goes.

Now we're discussing a specific issue. From the OP, it seemed like you were asking for some general knowledge, but you really have a problem to solve. Give us some specific about that problem. What do «the car gets unstable» means?

You are right about the suspension changing and that is really important to understand. I'm not sure what you already understand and what are your references for your info about suspensions so let me try explaining what I think you need to know.

Here is a simple trailing axle:

The motion of this axle - with respect to the frame - is easy to analyze: The whole axle can only rotate about the joint IC.

The trailing axle has many default as a suspension, thus most suspension designs have multiples links:

The motion of this axle is neither pure translation nor rotation as for the trailing axle. Instead, it will travel along some curvy path.

But a curvy path is only a succession of arcs with a given location for their centers. So, for every position of the suspension links, you can locate that center (Instantaneous Center - or IC - since it is always changing). There are solutions already found for most typical suspension designs. For example, for our four-link suspension design:

So when the suspension is in that precise position, we know about what point the axle is in pure rotation (no translation).

Why is that important?

If the axle is in pure rotation, so is any point on that axle, like the tire-road contact point where the reaction forces are applied. So we could «re-position» our arm this way and still get the same result:

On the previous figure, there is also the thrust $T$, the normal force $N$ and the reaction forces at the IC, which are equal and opposite.

If you just look at $T$ and its horizontal counterpart at the IC, you can intuitively see that this couple will try to rotate the axle counterclockwise.

If you just look at $N$ and its vertical counterpart at the IC, you can intuitively see that this couple will try to rotate the axle clockwise.

The moments for each couple can be calculated easily: $-Th$ and $NL$ (I assumed the clockwise direction is positive).

If both moments have equal magnitude, then they will compensate each other and the virtual link will be in pure compression and it will not try to rotate in either direction. So the ratio $\frac{h}{L}$ becomes significant, which can be further simplified to only one value, i.e. the angle of our virtual link since $\tan \theta = \frac{h}{L}$.

Now imagine $\theta = 90°$ for an instant. Then the normal force of the car would be entirely taken in compression by the virtual link, meaning the suspension spring would be useless. On the other hand, if $\theta = 0°$ (IC is at the ground level), then the full thrust of the tire is taken by the virtual link. Remember that even if this link is virtual, it still means that the actual suspension links are taking the loads in compression or tension somehow.

Anti-squat

When you accelerate (because of the thrust force $T$), there is a weight transfer from the front to the rear. This means $N$ will increase. That normal force increase will be visible by a compression of the rear springs (and similarly by a decompression of the front springs). We can easily calculate that increase $\Delta N$ this way:
$$\Delta N = T\frac{h_{cg}}{L_w}$$
Where $h_{cg}$ is the center of gravity height and $L_w$ is the wheelbase length.

For now, note that the higher the $cg$, the larger the weight transfer toward the rear axle, something that is desirable when accelerating.

Back to the suspension design, you can position the IC such that there will be a counterclockwise (negative) moment that will compensate for the additional weight transfer. If you perfectly achieve that, the rear springs will not compress because the axle will try to lift the frame while using the ground as a pivoting point. So:
$$Th = \Delta N L$$
But we already know the relationship between $\Delta N$ and $T$, so:
$$\frac{h}{L} = \frac{h_{cg}}{L_w}$$
With this simple geometric relationship, no matter the thrust, no matter the weight, we know the rear suspension will not squat under acceleration. We call that achieving 100% anti-squat. By moving the IC location, we can achieve partial anti-squat or we could even create a moment so high that the rear end would lift under acceleration.

Real Life

I'm getting closer to your situation. Say you position your IC such that you want to achieve 100% anti-squat and you expect the rear end stable, right? Not quite.

Once you will accelerate, your front end will lift (although you could technically put anti-lift in the front suspension too if it was an AWD, but you most likely won't do that). If the front end lifts, 2 things happen:

1. The $cg$ height will increase, thus increasing the weight transfer;
2. With the front end lifting, it means the car frame rotates about the rear tires' contact patch, thus the IC location moves, thus $\theta$ increases.

So your whole geometric relationship $\left(\frac{h}{L} = \frac{h_{cg}}{L_w}\right)$ can change as well. If you don't have 100 % anti-squat, then $\theta$ will change even more. If the change is noticeable quickly (when you have a short swing arm length, for example), then unexpected things can happen rapidly. So it's not only where your IC is that matters, but also how fixed it remains as the suspension and frame move all around.

If that is not enough trouble, if the suspension is too rigid (high $\theta$, i.e. a lot of anti-squat), the suspension doesn't do its job efficiently as it cannot absorb vertical wheel motion very wheel (rigid links instead of springs). So it is possible you will see wheel hop appearing.

In your case, when «it goes right on the back bumper», it is because you have enough thrust to transfer all the weight from the front end to the rear end - according to your (effective) $cg$ height and wheelbase. Once you go past that thrust threshold, then the car body is in pure rotational acceleration and it will lift up until it flips over. (And remember that the $cg$ height is still increasing as the front end lifts up!) You can control that by adjusting the thrust (like what bikers do when they ride on a wheelie).

Long story short, your $cg$ height is too high at some point during the initial acceleration. This can be adjusted by making sure your front and/or rear suspension will not lift as much as they do right now.

 

[Tom Rauji]

Thanks much. This is the most helpful exchange I have had on this topic. Most appreciated.

I'm going to see if I can interleave my responses...I'm not sure how this forum works for that.

The basics for it is statics analysis. I found this website which seems to give good tutorials about it. The chapters about moments and trusses should relate closely to a suspension analysis.

Now we're discussing a specific issue. From the OP, it seemed like you were asking for some general knowledge, but you really have a problem to solve. Give us some specific about that problem. What do «the car gets unstable» means?

Up to a certain level I had back tire separation and could plant the tire hard and get the weight on the back tires as the rear was still extending a little. The front tires would barely be off the ground. I could ride out maybe 200 feet as the front slowly settled with just enough contact for steering. The entire car would be slightly lifted.

Now I got in a pickle. I changed rear shocks from single adjustable to double and first pass it wheel hopped. I tightened the shock rebound (extension remained at half) a little tighter and next pass it went up on the back bumper. It hasn't done that in years. The only change was the rear shocks.

Back to the suspension design, you can position the IC such that there will be a counterclockwise (negative) moment that will compensate for the additional weight transfer. If you perfectly achieve that, the rear springs will not compress because the axle will try to lift the frame while using the ground as a pivoting point. So:
$$Th = \Delta N L$$
But we already know the relationship between $\Delta N$ and $T$, so:
$$\frac{h}{L} = \frac{h_{cg}}{L_w}$$
With this simple geometric relationship, no matter the thrust, no matter the weight, we know the rear suspension will not squat under acceleration. We call that achieving 100% anti-squat. By moving the IC location, we can achieve partial anti-squat or we could even create a moment so high that the rear end would lift under acceleration.

The rear normally lifts about an inch or two. I'm trying to do two things: 1.) Plant the tire hard at the start 2.) Lift the back more to have less front lift

I'm getting closer to your situation. Say you position your IC such that you want to achieve 100% anti-squat and you expect the rear end stable, right? Not quite.

Once you will accelerate, your front end will lift (although you could technically put anti-lift in the front suspension too if it was an AWD, but you most likely won't do that). If the front end lifts, 2 things happen:

1. The $cg$ height will increase, thus increasing the weight transfer;
2. With the front end lifting, it means the car frame rotates about the rear tires' contact patch, thus the IC location moves, thus $\theta$ increases.

So your whole geometric relationship $\left(\frac{h}{L} = \frac{h_{cg}}{L_w}\right)$ can change as well. If you don't have 100 % anti-squat, then $\theta$ will change even more. If the change is noticeable quickly (when you have a short swing arm length, for example), then unexpected things can happen rapidly. So it's not only where your IC is that matters, but also how fixed it remains as the suspension and frame move all around.

It may have gone past that point. I don't really have a video or data logging.

If that is not enough trouble, if the suspension is too rigid (high $\theta$, i.e. a lot of anti-squat), the suspension doesn't do its job efficiently as it cannot absorb vertical wheel motion very wheel (rigid links instead of springs). So it is possible you will see wheel hop appearing.

In your case, when «it goes right on the back bumper», it is because you have enough thrust to transfer all the weight from the front end to the rear end - according to your (effective) $cg$ height and wheelbase. Once you go past that thrust threshold, then the car body is in pure rotational acceleration and it will lift up until it flips over. (And remember that the $cg$ height is still increasing as the front end lifts up!) You can control that by adjusting the thrust (like what bikers do when they ride on a wheelie).

Long story short, your $cg$ height is too high at some point during the initial acceleration. This can be adjusted by making sure your front and/or rear suspension will not lift as much as they do right now.

I can lower the car 1 inch almost certainly, and maybe as much as 2 inches. I also could build an electronic height measurement system to pull back power at a certain bumper to ground height. That actually might be a good idea because I could tune suspension for minimum ET or maximum power.

I'd like to learn enough to understand what every suspension component does. I don't like to cut and try, and all of the anti-squat stuff I've found does not include the effects of the housing wrapping where the lower arms push more and the upper arms actually pull backwards. I'm going to read the link you posted and think about things a bit. This is my hobby where I can learn and improve things to the very best they can reasonably be. Guessing isn't pleasurable.

 

[jack action]

I have no experience in tuning a suspension for drag race cars, but I'll give my opinion anyway.

I changed rear shocks from single adjustable to double and first pass it wheel hopped. I tightened the shock rebound (extension remained at half) a little tighter and next pass it went up on the back bumper. It hasn't done that in years. The only change was the rear shocks.

How I explain this is as follow:

You rear end lifts up because it has more than 100% anti-squat. By tightening the rebound on your shocks, you slowed down the process. But the front end stayed the same and lift the same way, at the same pace. So, this must lead to a backward rake of the body (compared to the original setting), thus increasing the angle $\theta$. This leads to more anti-squat -> higher rear lift (the shocks slow down the process but they don't change the end effect) -> higher $cg$ -> higher weight transfer -> front wheel goes off the ground.

I would bet that if you tighten the front shock rebound as well, it would slow down the front lift too and once in synch with the rear, the body rake would stay as it was originally while the entire car lifts (the process would only be slower).

The rear normally lifts about an inch or two. I'm trying to do two things: 1.) Plant the tire hard at the start 2.) Lift the back more to have less front lift

I heard this reasoning of raising the rear end to «plant the tire». What people think it does is that the moment caused by the thrust force ($Th$) is «adding» a force to the ground. It does not. It cannot do that. The only thing it can do is create motion for the frame, i.e. raising it. The normal force ($N$) is only influenced by the weight and weight distribution of the car and the weight transfer. In any case, it will never exceed the total weight of the car (when the front wheels are off the ground, and only in that case).

The «added» force can only come from the fact that the $cg$ height is increased with a higher rear end, thus increasing the weight transfer.

It's like using softer front springs to «plant the rear tires». Most people think it releases some extra energy that magically transfer to the rear end. It does not. The only thing it does is lifting your front end higher as the weight transfer begins and thus it raises the $cg$ height more than with stiff springs, leading to more weight transfer.

 

[Tom Rauji]

I have no experience in tuning a suspension for drag race cars, but I'll give my opinion anyway.

How I explain this is as follow:

You rear end lifts up because it has more than 100% anti-squat. By tightening the rebound on your shocks, you slowed down the process. But the front end stayed the same and lift the same way, at the same pace. So, this must lead to a backward rake of the body (compared to the original setting), thus increasing the angle $\theta$. This leads to more anti-squat -> higher rear lift (the shocks slow down the process but they don't change the end effect) -> higher $cg$ -> higher weight transfer -> front wheel goes off the ground.

It's easy to get this confused.

The rebound is the collapse. The extension is the hit when in separation. I left the extension at my normal levels that are on the edge of distorting the sidewall. I only adjusted the rebound to control the upward tire bounce. This would keep the back higher longer, not lower. This is "normal" group opinion when running a real stiff sidewall tire like a radial tire. The shock manufacturer who made the radial valved shocks suggested half on extension and near full on rebound.

I would bet that if you tighten the front shock rebound as well, it would slow down the front lift too and once in synch with the rear, the body rake would stay as it was originally while the entire car lifts (the process would only be slower).

I didn't have dual adjustable on the front, but the fronts were set slow. This adds the front tire and suspension weight and whatever little bit the inertia helps to slow down rise. It really won't do much after one second or so because by then it has extended to the front travel limiters and all the weight is just hanging there solid.It hangs the weight there for about 200 feet (under 3 seconds) until the front settles and the tires start kissing the track..

I heard this reasoning of raising the rear end to «plant the tire». What people think it does is that the moment caused by the thrust force ($Th$) is «adding» a force to the ground. It does not. It cannot do that. The only thing it can do is create motion for the frame, i.e. raising it. The normal force ($N$) is only influenced by the weight and weight distribution of the car and the weight transfer. In any case, it will never exceed the total weight of the car (when the front wheels are off the ground, and only in that case).

I have not quantified this, I hate facts without numbers, but getting a number is on my bucket list. Certainly there is some amount of inertia involved in this that will help hit the tire. I'll post a slow mo video of an earlier launch while I was setting up bar positions.

The «added» force can only come from the fact that the $cg$ height is increased with a higher rear end, thus increasing the weight transfer.

There is a lot of initial slap there before the body can lift.

It's like using softer front springs to «plant the rear tires». Most people think it releases some extra energy that magically transfer to the rear end. It does not. The only thing it does is lifting your front end higher as the weight transfer begins and thus it raises the $cg$ height more than with stiff springs, leading to more weight transfer.

I hear that, too.

Of course it does not store more energy. The car weighs what it weighs. What it did in my old cars from the 1960s-1980's was increase the extension distance where the spring was contributing and allowing lift. Now I have the opposite problem. I want to keep the front down after a few hundred feet so I can still steer. Noty steering at 100 MPH can be bad, as can air getting under the car.

 

[jack action]

The rebound is the collapse. The extension is the hit when in separation.

I've always seen 'rebound' as what you call 'extension' (which I never heard). The opposite - what you also call 'collapse' - has many names though, such as 'bump', 'compression' or 'jounce':

But I still have a theory to explain the phenomena even with stiffening up the compression instead of the extension! Before, your rear end collapsed rapidly under the weight transfer. By stiffening up the shock, the process doesn't go as fast and the $cg$ remains high longer, therefore permitting more weight transfer.

There is a lot of initial slap there before the body can lift.

As soon as there is thrust, there is weight transfer. It is instantaneous, it is proportional. The fact that the body lifts, depends only on the suspension design. A car with no suspension still has weight transfer, even though the body - obviously - doesn't move in any way. In your video, it's probably the difference in air pressure between front and rear that throws you off. The softer rear tires squish a lot more under load than the stiffer front tire expand under the same load difference.

With the live axle, there is also the load transfer from side-to-side to consider. The following video demonstrate the phenomena:

​

This car has no torsion resistance in the rear which leads to extreme torsion of the body. But even with a stiff torsional resistance, the effect is still there. Now, for the car in the video, when both front wheels are off the ground, there is no difference with a more stable vehicle (well, there may be an undesired positioning of the rear suspension links); it only looks crooked, that's all. But when at least one wheel touches the ground, it gives a reaction point to transfer the load from passenger-to-driver side on the rear axle and that can be bad.

This is what happen to the Holden in the next video, which always keep one front wheel on the ground. It always wants to initially steer right because the left rear tire has more traction than the right one:

​

 

[Tom Rauji]

I've always seen 'rebound' as what you call 'extension' (which I never heard). The opposite - what you also call 'collapse' - has many names though, such as 'bump', 'compression' or 'jounce':

View attachment 216642
View attachment 216643
View attachment 216644​

But I still have a theory to explain the phenomena even with stiffening up the compression instead of the extension! Before, your rear end collapsed rapidly under the weight transfer. By stiffening up the shock, the process doesn't go as fast and the $cg$ remains high longer, therefore permitting more weight transfer.

OK. I'll get on the correct terms.
I stiffened the bump or collapse. The rebound is half as it always was.

I have no issues with body roll. The body roll is caused by reaction to driveshaft torque. The engine torque is rotating the rear end up on the right side and down on the left. I have an anti-roll bar. It is a cross bar from frame rail to frame rail with roller bearings at the ends. It has links with heim joints from the rear end housing axle tubes to the bar. This allows the rear end to move freely up and down but forces the car and rear end to be level. It prevents drive shaft torque applied to the pinion gear from tilting the housing with respect to the body.

As soon as there is thrust, there is weight transfer. It is instantaneous, it is proportional. The fact that the body lifts, depends only on the suspension design. A car with no suspension still has weight transfer, even though the body - obviously - doesn't move in any way. In your video, it's probably the difference in air pressure between front and rear that throws you off. The softer rear tires squish a lot more under load than the stiffer front tire expand under the same load difference.

The rear tire has about 16 psi and is 12.5" cross section, 10.2" tread width, and 28 inches tall (with no weight). The rear tires initially flatten about twice (real rough guess) the amount they do with the whole car weight alone when the car is up on the wheels and rolling.

Something beyond weight goes on in the first 10 feet or so. They normally put a very sticky coating on the track as a traction aid. I can go on a non-prep track, using a loose rebound, and hit the tire hard enough (with enough anti-squat) to fully get on the back tires. About 20 feet out it starts to spin from lack of traction and the nose drops and tires go into uncontrolled spin. Something happens in the first few feet. As a matter of fact I can go on bare concrete without compound and if I loosen rebound it flattens the tire and plants, but as soon as the rear extension stops it spins. There is a little something going on with the initial "push" that digs the tires in, and that something goes away if I angle the arms or tighten rebound to not have that start extension. The better the track, the more I can slow the rebound without spinning and the longer the initial hit lasts. On a tight track with tight rebound settings I can get into the 1.1 second range for 60 feet. My issue is keeping the nose down. I'm not sure why it went up in the nose when I tightened the bump. That makes no sense to me. It is opposite what I expected.

 

[Ranger Mike]

Traditionally we used a 90 – 10 front shock set up. This shock takes 10% effort to extend so it is not hindering the front end lift. The 90% is to hold it up and keep it from compressing. This is one fine tune area to explore for several reasons. When you launch you are dealing with body roll in and X,Y and Z axis. You also deal with TIME. Hence , my 4-D comment. You can fine tune the “Weight Transfer” (really the x,y and z acceleration forces) by experimenting with an 80-10 left front shock and 90-10 rt ft shock. Same in the rear end. Usually a 50-50 shock was used on both sides but as the car lifts and twists , the left rear may need more higher setting on compression to counter the twist action. Again Time comes into play as to how long do you want to keep the front end at max lift. How quick do you want the front to settle. Hopefully way before the traps.

 

[Tom Rauji]

Traditionally we used a 90 – 10 front shock set up. This shock takes 10% effort to extend so it is not hindering the front end lift. The 90% is to hold it up and keep it from compressing. This is one fine tune area to explore for several reasons. When you launch you are dealing with body roll in and X,Y and Z axis. You also deal with TIME. Hence , my 4-D comment. You can fine tune the “Weight Transfer” (really the x,y and z acceleration forces) by experimenting with an 80-10 left front shock and 90-10 rt ft shock. Same in the rear end. Usually a 50-50 shock was used on both sides but as the car lifts and twists , the left rear may need more higher setting on compression to counter the twist action. Again Time comes into play as to how long do you want to keep the front end at max lift. How quick do you want the front to settle. Hopefully way before the traps.

Because of power levels and tires today, 50/50 rears and 90/10 fronts aren't useful on faster cars. Racers generally use two distinct rear shock settings, extension faster on radials and slow on bias sidewalls. They use two different rear shock "bump" or collapse settings, slow on radials and medium on bias. My car would go on its back at much less power if I used a loose rise front shock. The body front would have so much momentum it would be going up to the sky more than forward.

The fronts stay about the same regardless of rear tires in faster cars. Extension is slowed to prevent wheelies and collapse or "bump" is slowed to hold the front stable. Only extension (I guess I should call that rebound but the mode is actually extension from resting) is adjusted to reduce front end rise or increase rise on poor tracks. I could just copy what others do, but I want to understand exactly what happens physically. I don't think everything commonly accepted is factual, like energy stored in soft tall springs or instant center calculations that ignore effects of axle wrap forces. For example I know for a fact my upper control arms pull backwards and do not push forward at all. They do not react like a real ladder bar or four link (which I've had) when I change them, so I want to learn why the upper arm act differently.

My general interest in racing is the actual way things work, not how people say things work. I want to understand it in detail.