Antidive/Antisquat: how much do I want?

[uninformed]

My COG was from LR data for the cab chassis. I added the 50mm or so. The figure is a min and a max. The max being if you are fully loaded to LR guidelines. That may be where you are getting the 1060 from. Yes i agree I have posted that same point. It is very easy to measure all the other data points with fair amount of accuracy, but the COG is tough….and the most important.

Now I read somewhere that COG is COG and making the track wider will not lower it. It will reduce the tendacny to roll over at a given point but not lower the COG? is this correct?

[bill van snorkle]

My COG was from LR data for the cab chassis. I added the 50mm or so. The figure is a min and a max. The max being if you are fully loaded to LR guidelines. That may be where you are getting the 1060 from. Yes i agree I have posted that same point. It is very easy to measure all the other data points with fair amount of accuracy, but the COG is tough….and the most important.

Now I read somewhere that COG is COG and making the track wider will not lower it. It will reduce the tendacny to roll over at a given point but not lower the COG? is this correct?

Yes Serg, apparently that is correct. widening the wheel track just means that it requires a steeper side slope for the vertical line drawn through the COG to fall outside the lower tyres contact patch.

[De Ranged]

Now I read somewhere that COG is COG and making the track wider will not lower it. It will reduce the tendacny to roll over at a given point but not lower the COG? is this correct?

Yes, centre of gravity is all about weight and where it is.... you can change the shape without changing the CoG, the only time you change CoG is when you move weight(adding your bullbar and winch or cutting down your deck, or extreme like my planed changes to lift the gearbox on my toy)

Imagine a triangle, the wheels are the base of the triangle and the Centre of Gravity is the top, to tip it over you have to move the top point out past one of the outside wheels.... there are a few other things that effect things, spring rates, body roll etc that move the CoG towards the downhill side so in reality you will tip sooner than the triangle will suggest lol

body roll is an interesting one and a main part of that is roll centre height, if you can get it to the same height as your CoG the only body roll you get is from the springs but.... this wont tip you, remember I said you could change the shape without changing the CoG yep lol, the distance between Roll Centre Height and the CoG acts like a lever to tip the triangle over before its reached tipping point if the CoG is above..... now whats really cool is if you could get the roll centre height higher then your body roll would act like a lever to stop you tipping.... I have seen some rock crawlers setup with very high panhard bars to try and do this with not that much gain... the reason why they didnt get much gain is it is down to sprung weight vs unsprung weight now for us with a truck our unsprung weight is 20-25% of the vehicle weight so we have 80-75% of the weight acting on that lever thats why we tip before the triangle happens. Now that rock crawler with over built axles and water in his tyres and no body just a light cage frame his weight ratio is 70-80% unsprung weight with only 30-20% acting on a lever and because the lever is short (its very hard to get above your CoG) the effect wasnt noticeable

why I love suspension lol you have to understand the whole before the individual pieces make sense

I've done a bit more research on the 110 CoG.... interesting what you find in forums, Uninformed did you ever tip your truck over with the tractor lol

I did find a PDF that looks like coach builders plans, looks genuine with a stated 655mm CoG

working off a CoG of 655mm and standard unmodified weight, anti dive works out at 198% with your mods adding a 7% increase to the same score of 213%

been trying to find data on a RRC as a comparison as well, as I used to have one and remember what the handling was like

but that one will have to wait till Ive got time lol

[Daan]

You can work out the height of the COG using corner scales, by jacking the front or the back up, as high as you can and put the wheels on the scales in this elevated position. mathematically, from the difference of these valuse, you should be able to work out how high the COG is. To be accurate though, you need to lock the suspension.

Daan

[uninformed]

no, we ran out of time. Had a plan, a 7t 4wd JD tractor and just the guy to do it……but time got away on that visit.

Daan, I have read the method you describe, it requires fairly detailed measuring and being able to lift the front/rear and sides a certain amount and still weigh them. This would involve some $$$ in equipment, because there is no way in hell any place like a wighbridge is going let let us dick around like that for time and liability reasons.

I am happy to go with LR's specs, Id doubt they would be wrong by much.

[De Ranged]

I have a bar at the shop that is for using a set of bathroom scales so the scales take from half to a fifth of the weight.... allows me to do corner weights problem is they compound there margin of error, I tuned them with 4 20l oil drums filled with water.... its not perfict but at a guess I'd say my margin of error would be around the percentage of the fraction i'm measuring on, so at 1/2 I'm +/- 2% at a fifth Im +/-5% and I can measure slightly over half a ton with them

the tractor method isnt very accurate because you get the body roll so on most trucks will give you a false reading higher than it really is

[uninformed]

tractor would be lifting one side VERY slowly. Would that not also be more "real" world as I don't think you are going to roll over on flat ground with the engine off

[Bush65]

I appreciate your interest in making this public, so far you have given posts with broad generalisations that you know something that I have gotten wrong ... stop pussy footing around and tell me what... so I can do something about it lol... you go on about being all concerned about how I have made mistakes and the damage it could cause yet you come back with another generalisation you do understand this doesnt make you look very credible

stick your neck out and tell us all what I have wrong lol

Come on John we want to learn....

To go beyond those generalisations would be like trying to pick up a tu.. from the clean end.

Do you think I am the slightest bit concerned with how credible you think I might look?

I don't have time to come here very often and then it can only be for a short period and I type slower than a slow thing.

[Bush65]

I have found some time to spend on this post.

Percent anti-squat for live axle (aka solid axle) vehicles such as a Land Rover is fairly simple, both in definition and how it is determined. However because the water has been made so muddy, it will take a fair bit to clear up what is involved, before presenting the simple calculation involving four variables.

Because the suspension control arms of live axle vehicles are subjected to forces resisting torque from the wheels, the method is different to that used for most independent suspension systems that have the differential housing mounted directly to the chassis, or inboard braking systems.

Here I will only be concerned with live axle suspension. Further more, where the discussion concerns anti-squat, the approach is applicable for anti-dive.

Some terminology:

• Vector quantities have a magnitude and a direction. It is good practice to state the direction explicitly, but sometimes it is omitted, and if not obvious must be inferred by the context.
  Some examples of vector quantities are; force, displacement, velocity, and acceleration.

• Scalar quantities have a magnitude but no direction. Speed is an example of a scalar quantity.

• Link is a member (in this discussion a suspension control arm) that is only capable of resisting compressive or tensile force. The direction of the force (vector) is along a line from one end of the link to the other. Neither end connection can resist rotation, and it is not necessary for the link to be straight. Neither a radius arm or swing arm are links, as they have an end that resists rotation. Thus referring to swing arm or radius arm suspension as one link, or 2 link is not technically correct.

• Instant Centre (IC) is a notional point at the intersection found by extending the force direction line between each end of a pair of links. For link suspension, the motion over a very small increment (instant), will be equivalent to what it would be if the links were replaced by a swing arm pivoting about the instant centre. Note: unless the ends of the pair of links are coincident, the instant centre will move as the suspension moves, and the geometry of this equivalent swing arm will change.

• Weight is a particular force due to gravity acting upon a mass. One of Newton's laws of motion tells us that force = mass x acceleration, and so weight = mass x g where g is the acceleration due to gravity. If mass is in kg, acceleration is 9.81 m/s^2, weight is in Newtons (1kg mass has a weight of 9.81 N). Note: weight transfer is a commonly used notion used when a vehicle accelerates, but there is no transfer of weight and it is better to stick with the the physics.

• Inertia is a reaction that results from acceleration of a mass. Another of Newton's laws of motion tells us that every action has an equal but opposite reaction. So the inertia is in the opposite direction to the acceleration.

• Centre of Gravity is a notion that we can replace a body, or a number of connected bodies, with a concentrated mass equivalent to the total mass of the bodies, at a location (C of G), so that the concentrated mass will have the same effect. This notion is useful for simplifying calculations. For percentage of anti-squat calculations, we need to know the location of the centre of gravity of the sprung masses, not the overall vehicle mass. More often than not this C of G is little more than an educated guess and the error has the greatest effect on the validity of percentage anti-squat calculations.

• Moment of a force, about a point, is Force x Distance where the distance is taken from the reference point and is measured perpendicular to the force. Torque is similar, where the force rotates about the centre, and distance is replaced by the radius.

• Equilibrium is implied by Newton's law; every action has an equal but opposite reaction. Every system has to satisfy equilibrium. To check for equilibrium, the usual convention is to separately sum the horizontal and vertical components of all forces and also sum the moments of all forces. Each one of those sums must equal zero.

Anti-squat problems can be simplified by using a 2-d instead of 3-d. I won't complicate the discussion by going into the maths showing why this is valid.

For this discussion assume:

1. The 2-d, X,Z plane is vertical through the vehicle's longitudinal centreline. This plane is called 'side view'.

2. The X-coordinate direction is parallel with the ground, the datum is the rear axle, and the positive direction is toward the front axle.

3. The Z-coordinate direction is perpendicular to the ground, which is the datum, and the positive direction is up.

Further assume the following values:

• The total mass of the vehicle is Mt

• The wheelbase is Lx

• The distance from the rear axle to the centre of gravity in the X-direction is Lc

• The height of the sprung centre of gravity in the Z-direction is Hc

• The height of the SVIC (side view instant centre) in the Z-direction is Hic

• The distance from the rear axle to the SVIC in the X-direction is Lic

• The ground is horizontal

Squat occurs if the force on the suspension springs increase during acceleration (rise is the opposite phenomenon). So what causes squat?

If the velocity is constant, i.e. zero acceleration, the forces in the suspension springs and control arms will be in a state of static equilibrium and the traction force at the tyre contact patch will be equal, but opposite to the resistance forces (rolling resistance, aerodynamic resistance, etc.). Now if we apply an increase of torque to the wheels, the tractive force increases, and being no longer in static equilibrium acceleration occurs.

The force creating the acceleration induces a reaction (inertia) so that the vehicle is in dynamic equilibrium.

Now the acceleration force acts at the tyre contact patch, in the positive X direction, however the inertia acts at the centre of gravity in the opposite direction (- X). The difference in height between these two results in a moment in the counter clockwise direction.

For equilibrium there must be a reaction to this moment in the opposite direction. This results in an increase to the reaction between the road and rear tyre and a decrease in the reaction at the front tyre.

Expressing this mathematically:

The acceleration force is: F1 = increased torque at wheel divided by tyre radius

The resulting inertia force equals – F1 ( negative because the vector direction is opposite).

Taking moments about the front tyre contact patch, the moment due to the inertia equals

- F1 x Hc (where Hc is the height of the C of G)

For equilibrium, the the sum of the moments = zero, so the change in reaction force at rear tyre is:

Fr = F1 x Hc / Lx (where Lx is the wheel base)

Important Note from the above mathematical expression, for any acceleration force, the change in vertical reaction force at the wheel where the acceleration force was applied is always proportional to Hc / Lx This ratio will crop up time and time again and you should remember that it is because it relates acceleration to the change in vertical reaction at the wheel.

Some people might like to visualise this ratio as a slope, rise Hc, in distance Lx. This slope is usually drawn on diagrams of the side view, from the tyre contact patch to a point at the height of the C of G (i.e. Hc) at a distance from the rear wheel equal the wheel base (i.e. Lx), which naturally is at the front wheel. When looking at these diagrams, don't forget it is the slope that is important and the slope can be represented at any point along the line, i.e. how long the line is or where it finishes is irrelevant.

Note also that this increase in the vertical reaction is at the tyre contact patch. Remember that squat is caused by an increase in the force on the suspension springs. If all of the increase in the vertical reaction was applied through the suspension springs, we would have what some call 100% squat. Some call the line in the side view from the tyre contact patch with a slope of Hc / Lx the 100% ant-squat line (but we are getting ahead of ourselves).

The acceleration force is reacted through the suspension arms in order to accelerate the vehicle. Depending upon the geometry of the arms, they may transfer a percentage of the increase in the vertical reaction to the tyre contact patch. This percentage is called anti-squat.

What is percent anti-squat?

It is nothing more than the percentage by which the anti-squat effect reduces suspension spring deflection, compared to what would occur with zero anti-squat (i.e. 100% squat).

Do not confuse anti-squat with anti-pitch. Pitch is the angular change resulting from the difference in front and rear suspension deflection (compression and extension). Unlike percent anti-squat, pitch is affected by front and rear spring rates, and by the distribution of torque (driving and braking) between front and rear wheels.

Before we calculate anti-squat we should examine what happens to the forces in the suspension arms during acceleration, in our side view and in particular the significance of the Side View Instant Centre (SVIC).

Recall I said, with regards to a link; “The direction of the force (vector) is along a line from one end of the link to the other. Neither end connection can resist rotation, and it is not necessary for the link to be straight.”

Imagine if our suspension system in the side view consisted of a link that has one end at the tyre contact patch and the other at the SVIC. I want to show that our suspension system does indeed replicate this concept, and once you grasp it, ant-squat/dive should be easier to follow.

Let's start with a swing arm or radius arm suspension connected to the axle.

Now when the torque at the wheel is increased to produce acceleration, there is a reaction (recall every action has an equal but opposite reaction) at the tyre contact patch opposing the torque. This reaction from the road pushing against the tyre is the acceleration force.

Note that if we froze time at that instant, the torque in the tyre, wheel, half shaft, crown wheel, and pinion resist the acceleration force at the tyre contact patch so that it is transferred through them, to the differential housing, to the axle tube, to the radius arm and finally into the chassis at the end of the radius arm. At that instant it behaves the same as if those rotating parts were replaced by an arm extending down from the axle tube to the tyre contact patch, with the acceleration force applied to the lower end.

This simulates the imaginary link from the tyre contact patch to the SVIC.

It can be shown that a pair of links have an instant centre and behave as though a swing arm pivoting at the IC.

If you can follow that, it should be clear that the acceleration force at the tyre contact patch is transmitted to the chassis along the line of action from the tyre contact patch to the SVIC.

The slope of this line is Hic / Lic (i.e. height to SVIC divided by length to SVIC). It is often drawn on the side view and some call it the anti-squat line.

Now we can calculate the percentage anti-squat from the very simple mathematical equation:

Pas = (Hic / Lic) / (Hcg / Lx) x 100%

It should be obvious that if the slope (Hic / Lic) is the same as the slope (Hcg / Lx), then we have 100% anti-squat. Also if slope (Hic / Lic) is above the slope (Hcg / Lx), then we have greater than 100% anti-squat, …..

Note that neither the value of the acceleration or the stiffness of the springs affect the percentage anti-squat.

Under hard acceleration, if the percentage anti-squat is low, the increase in the reaction between the road and rear tyre will have a slight delay until the spring force is increased by the squat. Thus traction can be lost at launch. With a high percentage of anti-squat the increase is nearly instantaneous because it is applied through more rigid suspension arms, however little advantage will be seen by increasing the ant-squat from 80% to 100%.

Under hard deceleration the opposite occurs and high anti-squat can be detrimental to rear wheel braking performance.

[De Ranged]

Thank you for that, if your not a touch typest that is alot of time in that post, I really appreciate that

If you dont mind answering a few questions

In simple terms

Anti Sqaut is your drive axle pushing forward through the links to the body, (and if there is angle up in the links... forcing the body upward ).... and how much this counters the change in mass due to acceleration

So this means the the upward force is at the point the links attatch to the chassis..... Some thing I've wondered is when you have a multi link setup anti-squat is calculated by the instant centre this bit I understand but the way I see it the links don't go all the way to IC, say they are short.... since they are further away from the CoG shouldnt they lift the rear easier.... same as using a power bar on a stubborn bolt..... except the longer distance from the CoG to the chassis mount is the lever

I understand there are other issues with short links but I like to understand all things so I can make an educated decision

Cheers Reece

[Bush65]

Thank you for that, if your not a touch typest that is alot of time in that post, I really appreciate that

Yes, I ran out of time and didn't get to say all I wanted to about link suspension, however what I said was correct i.e. by finding the IC the link forces resolve to the equivalent of a swing arm and then the anti-squat solution is simplified to that used for a swing arm. I still want to explain this, but don't have time now to go into much detail.

If you dont mind answering a few questions

In simple terms

Anti Sqaut is your drive axle pushing forward through the links to the body, (and if there is angle up in the links... forcing the body upward ).... and how much this counters the change in mass due to acceleration

Yes, but not entirely that simple. One link can angle up and the other angle down. One link can be in compression and the other in tension. If you want to resolve the forces in the pair of links, you must satisfy equilibrium, and the solution will end up with exactly the same result as the swing arm. However I have to leave this to another date, except to say now that if you examine the compressive force in the lower link which is angled up, the value of the compression force depends upon the angle of the line of force and has nothing to do with the length of the link. Thus extending the line to the IC doesn't affect the effect of the force - it should become clearer when I have more time to explain it.

So this means the the upward force is at the point the links attatch to the chassis.....

Again not quite so. Yes if the lower link is in compression and angled up, but you can't forget the geometry and tension in the upper link.

Some thing I've wondered is when you have a multi link setup anti-squat is calculated by the instant centre this bit I understand but the way I see it the links don't go all the way to IC, say they are short.... since they are further away from the CoG shouldnt they lift the rear easier.... same as using a power bar on a stubborn bolt..... except the longer distance from the CoG to the chassis mount is the lever

No.

I understand there are other issues with short links but I like to understand all things so I can make an educated decision

Cheers Reece

My comments in red above.

I will come back to these issues when I have more time.

[elbekko]

A very clear an informative post, thanks for that Bush65!

[De Ranged]

Thank you.....

I understand something about the combination of the links and quite often use the top link to tune my AS% when packaging wants to place the lower links in a bad position

I'm guessing the rest of your answer is going to explain why the position of the instant centre is more important to those that want grip.......

[Bush65]

... calculate the percentage anti-squat from the very simple mathematical equation:

Pas = (Hic / Lic) / (Hcg / Lx) x 100%

...

Before I move to the IC issue, I thought I would put some values into Dan Barcrofts excellent spreadsheet for 4 link suspensions.

Firstly to show that the equation above gives the same answer and also to show that when I get around to the post about the instant centre that the link forces will come out the same.

This pic shows the values that I put into the link calculator, and some of the results.

This pic is the diagram that the calculator produced showing the geometry, etc.

This pic shows some more results, including the link forces.

Note the link calculator uses the height of the vehicle CoG and the unsprung weights to finds the heights of the sprung CoG and another value that it uses for the anti-squat CofG. So I will use that height.

Now Pas = (Hic / Lic) / (Hcg / Lx) x 100%

then Pas = (22" / 60") / (33.52" / 110") x 100%

i.e. Pas = 0.3667 / 0.3047 x 100%

i.e. Pas = 1.20033 x 100%

i.e. Pas = 120.33%

Perhaps De Ranged can use the same link geometry with his method and see how his anti-squat results compare.

[De Ranged]

lol you got me there..... I use a version of Triaiged Calculator, my training is computers, yrs ago I got a early copy of it and pulled it apart to work out how he got his figures, I then spent alot of time on google confirming his calculation ... my original copy didnt use an unsprung weight so it worked off the total CoG

My copy V3.0 only adjusts for rear unsprung weight.... I use the two cells as total weight and unsprung weight, also mine is setup for metric down to the mm so the conversion is one decimal place

My result is 118% near enough to not mater

I'll confess I havent been into the newer version to see how the new unsuspended CoG is calculated, I've been curios, given the original was just based off total CoG height and this was from published texts... I can understand that since the Unsprung weight doesnt effect the reactions happening a more accurate figure can be got by removing them.... what has always been in the back of my mind is what if the original calculation was based off results that didnt differentiate for sprung and unsprung then removing the unsprung bias's the result.... I just went with the fact that the times Ive chatted with Triaged he came across as somebody who wouldnt fall for a simple mistake like this

[B reg 90]

Bush65 /John,

I just skimmed though your posts and it's excellent stuff. I will read in more detail when I get a chance and try and ask a sensible question.

In the mean time thanks for your efforts and I look forward to the next instalment!

Adrian

[Bush65]

Continuing from my earlier one where I said; “Imagine if our suspension system in the side view consisted of a link that has one end at the tyre contact patch and the other at the SVIC. I want to show that our suspension system does indeed replicate this concept, ...”

I want to show that the resultant force from this imaginary link at the SVIC is identical to the resultant force from the upper link (tension) plus lower link (compression) at the SVIC.

It is probably easier to base the calculations on the example link suspension in my last post.

Refer to the diagram in that post and the one below.

Assume the tractive force at the tyre patch creating acceleration (and thus squat) is

Ft = 15000 lb acting in the X direction.

Firstly determine the resultant force at the SVIC from the imaginary link from the tyre contact patch to the SVIC.

Now the distance from the tyre contact patch to the SVIC is Lic = 60” in the X direction and Hic = 22” in the Z direction.

Now the resultant force at the SVIC has an X component equal to the tractive force,

i.e. Fx = Ft = 15000 lb

And the Z component (Fz) is calculated from the X component (Fx) and the slope of the imaginary link, i.e. Fz = (Fx / Lic) x Hic, = (15000 lb / 60”) x 22”

i.e. Fz = 5500 lb

Then, using Pythagoras, the resultant force Fr = square root (Fx ^2 + Fz ^2)

i.e. Fr = sqrt (15000 ^2 + 5500 ^2)

i.e. Fr = 15977 lb and the direction is from the tyre contact patch to the SVIC

Now resolve the forces acting in the actual lower and upper links to find the resultant.

In this stick diagram, the wheel and axle assembly is represented by the cyan colour lever. The red lower link and blue upper link are drawn to scale.

The link heights (Z direction) at the axle end are:

Hla = 12” (lower link)

Hua = 22” (upper link)

The link heights at the chassis end are:

Hlc = 17” (lower link)

Huc = 22” (upper link)

The distances in the X direction from the axle are:

Zero at the axle end of both lower and upper links

and at the chassis end
Llx = 30" (lower link)

Lux = 21" (upper link)

Determine the X coordinate of the force in the lower link:

Take moments about the axle end of the upper link (the sum of the moments = zero):

then Flx = Ft x Hua / (Hua - Hla)

i.e. Flx = 15000 lb x 22” / (22” - 12”)

therefore Flx = 33000 lb and the direction is negative X (i.e. the lower link is in compression)

Now the sum of the forces in the X direction = zero

then Fux = Flx – Ft

i.e. Fux = 33000 lb – 15000 lb

therefore Fux = 18000 lb and the direction is positive X (i.e. the upper link is in tension)

Now find the components of the link forces in the Z direction.

For the lower link:

then Flz = Flx x slope of the lower link

i.e. Flz = 33000lb x (17” - 12”) / (30” - 0”)

therefore Flz = 5500 lb

Now the compression in the lower link, using Pythagoras, is:

Fl = sqrt (Llx ^2 + Flz ^2)

i.e. Fl = sqrt (-33000 ^2 + 5500 ^2)

therefore Fl = 33455 lb compression

For the upper link the slope is zero

therefore Fuz = 0 lb

and Fu = Fux = 18000 lb tension

Now we want to find the reaction from the combined lower and upper links.

For equilibrium the sum of the moments = zero and the only point where this can occur is at the SVIC.

For equilibrium the sum of the forces in the X direction = zero:

then Rx + Flx + Fux = 0

i.e. Rx = 33000 lb – 18000 Lb

therefore Rx = 15000 lb

Also the sum of the forces in the z direction = zero

then Rz + Flz + Fuz = 0

i.e. Rz = 5500 lb – 0 lb

therefore Rz = 5500 lb

Now this result is the same as found by taking the imaginary link from the tyre contact patch to the SVIC.

Now if you look at the following diagram that I took from another site, the upper and lower links connected together with the axle assembly at one end and chassis at the other end combine to form a bent link and the force direction is from 'A' to 'B'.

[Bush65]

For those that prefer to visualise these things, in sketches or in their mind, use the graphical method below to find the percentage anti-squat. This method is equivalent to the simple formulae given earlier:

Pas = (Hic / Lic) / (Hcg / Lx) x 100%

where:

Hic is the height of the SVIC (side view instant centre)

Lic is the distance from the wheel centreline to the SVIC

Hcg is the height of the Centre of Gravity

Lx is the wheelbase

Draw to scale:

• The 100% AS line from the tyre contact patch at a slope of Hcg in Lx

• SVIC at height Hic and distance Lic

• A vertical line from the ground to the SVIC, extending above the 100% AS line

Along this vertical through the SVIC, measure the height Has from the ground to the 100% AS line

Then the percentage anti-squat is:

Pas = Hic / Has x 100%

If you want to find a suitable position for the SVIC to obtain a particular value for the percentage anti-squat, for example 75% AS then draw to scale:

• The 75% AS line from the tyre contact patch at a slope of (0.75 x Hcg) in Lx

• The SVIC needs to be located so that it is somewhere along the 75%AS line. The chosen position of the SVIC along the desired %AS line will affect the slope of the roll axis, which is an important value for suspension performance (roll steer).

For link suspensions plot the upper and lower suspension control arms/links in the side view, so the extended lines through the end joints, intersect at the SVIC.

The methods for determining percentage anti-squat presented in this thread to this point are valid for two wheel drive and can be easily adjusted to account for four wheel drive, when the total traction is distributed between the rear and front wheels. Nearly every transfer case will split the torque evenly between front and rear, up until tyre slip occurs. The most common limit uncounted is practically no traction at the front wheels, and then the two wheel drive method is valid.

Most people want to use the percentage anti-squat value for comparison to another vehicle with known performance in the same terrain type, or for planning modifications to change the behaviour of an existing vehicle. Using the two wheel drive method is suitable for those purposes.

For four wheel drive, the total traction affects the change in the resultant force at the rear and front wheels, just as for the two wheel drive case. The change in the resultant force at the rear wheels is what causes suspension “squat”. The change at the front wheels causes suspension “lift”.

The “anti-squat” at the rear suspension depends not on the total tractive force, but on the percentage at the rear wheels. Similarly the “anti-lift” at the front suspension depends on the percentage of tractive force at the front wheels.

For an example of calculating percentage anti-squat (or anti-lift) assume the distribution of tractive force is 50% rear and front.

For the rear suspension, percentage anti-squat is calculated from:

Pas = (Hic / Lic) / (Hcg / {0.5 x Lx}) x 100%

where:

Hic is the height of the rear suspension SVIC (side view instant centre)

Lic is the distance from the rear wheel centreline to the SVIC

Hcg is the height of the Centre of Gravity

Lx is the wheelbase

0.5 is the percentage tractive force at the rear wheels / 100%

For the front suspension, percentage anti-lift is calculated from:

Pal = (Hic / Lic) / (Hcg / {0.5 x Lx}) x 100%

where:

Hic is the height of the front suspension SVIC (side view instant centre)

Lic is the distance from the front wheel centreline to the SVIC

Hcg is the height of the Centre of Gravity

Lx is the wheelbase

0.5 is the percentage tractive force at the front wheels / 100%

It should be obvious how to apply the graphical method for both of these four wheel drive cases.

During braking we have “dive” (same as squat, but normal convention for braking is to call it dive) at the front suspension and “lift” at the rear suspension.

The method for calculating “anti-dive” and “anti-lift” during braking is similar to the four wheel drive case above. The distribution of total braking traction between front and rear wheels depends upon the “brake bias”, which may be 60% or more at the front wheels.

For example assume the brake bias is 60% front and 40% rear, then:

For the front suspension, percentage anti-dive is calculated from:

Pad = (Hic / Lic) / (Hcg / {0.6 x Lx}) x 100%

where:

Hic is the height of the front suspension SVIC (side view instant centre)

Lic is the distance from the front wheel centreline to the SVIC

Hcg is the height of the Centre of Gravity

Lx is the wheelbase

0.6 is the brake bias percentage at the front wheels / 100%

For the rear suspension, percentage anti-lift is calculated from:

Pal = (Hic / Lic) / (Hcg / {0.4 x Lx}) x 100%

where:

Hic is the height of the rear suspension SVIC (side view instant centre)

Lic is the distance from the rear wheel centreline to the SVIC

Hcg is the height of the Centre of Gravity

Lx is the wheelbase

0.4 is the brake bias percentage at the rear wheels / 100%

[De Ranged]

Just from a theoretical point of view, if your calculating 4wd AS/AD wouldnt the bias be based on sprung weights and % of total .... I say this because for most terrain we are in grip is better due to load so there for your front axle in a "normal" 4wd will have more grip than the rear

This brings up the question of weight balance but that is a whole different subject lol but very important!

John could you please expain how too much AS% causes hop

I have issues with High AS%, I have watched very similar trucks/buggies perform on the same obstacles and the ones with higher AS% bounced around on steep climbs... worst case it will lift the nose off the hill.... I have always felt it was because the base line was gravity and the trucks wb etc got shorter, I'm inclined to believe john that I'm wrong..... if this is the case then what is causing this

I know part of it will be due to the AS transfer of weight to the links and not the springs

As your AS% goes up more of the weight transferred onto the axle from acceleration is transferred down the links instead of back on the body and down through the springs to the axle..... that means this weight is unsprung a bad thing if you still want your suspension to work

lets use my Lada as an example (since I know the figures for it)

A couple of things first the rear springs are liner rate @ 26kg per cm of travel, now I have calculated the AS% at rest at near to 60%but for this example I'll cheat and say its 50%

my sprung mass is 560kg, unsprung is unknown but I'll say 400kg (200kg per axle) and I have a 60/40% weight bias to the front

Now if I was to use a zip tie on my shock spear so I can work out the amount of weight transfer to the rear of the axle under acceleration, I accelerate on a level road for a base line.... with 3 cm of sqaut (zip tie has moved 3 cm down shaft)

Now for some math

3 cm x 52 kg (there are two springs) = 156kgs of weight transfer onto the springs.... now because half (remember the AS% is 50) has gone to the axle via the links so your rear axle has increased in weight to 356kgs lets consider the same truck but with 100%AS the rear axle now has a mass of 512kgs

This is all good till you are going forward and trying to keep your axle tracking, remember we arn't on a nice smooth drag strip... we are accelerating across ruts roots ditches bogs etc

How well your wheel stays in contact with all this is based on how well the springs and shocks can keep it down there now if you consider your sprung to unsprung ratio as an indicator of how well this is going

At 50% AS

Back sprung weight is 40% of my 560kgs = 224kg + that from weight transfer 156kg / unsprung axle 200kg + weight transfer 156kg

So we end up with 380kg trying to control 356kg of axle

At 100% AS

You have 224kg of body etc trying to control 512kg worth of axle.....

Now as a small kid at school getting in fights with the first 15 I very quickly learned if they are bigger than me.... I cant make them do what I want! lol

big bully's aside that axle isnt going to track into the terrain as well as the truck with less AS

Now I can see this being more and more of an issue on a hill because as you poke the nose up a hill the rear links are tilting up more carrying more and more weight straight to the rear axle before you add acceleration this helps with grip but not with suspension performance

[Bush65]

Just from a theoretical point of view, if your calculating 4wd AS/AD wouldnt the bias be based on sprung weights and % of total .... I say this because for most terrain we are in grip is better due to load so there for your front axle in a "normal" 4wd will have more grip than the rear

In all cases of friction force, it is only at impending slip that the force reaches the maximum value equal to (Coefficient of Friction x Normal Force).

With a conventional transfer case, if tyres are not about to slip at either front or rear, the tractive force will be distributed 50-50. To do otherwise will require a torque biasing transfer case such as the Ferguson system tried for Formula One many years ago. I don't know about current Freelander, but the first model used different front to rear gearing to achieve a torque bias.

If just the front tyres slip, then it is reasonable to assume the two wheel drive case for anti-squat. If rear tyres slip, then you are left with front wheel drive and anti-lift.

The above assumes a reasonably steady state, but in practice the situation is more dynamic, and slip/stick combined with uneven ground can see the traction vary unpredictably. The traction will always be greater when you are closest to a achieving a steady state.

John could you please expain how too much AS% causes hop

Squat and anti-squat are effects that can only occur during acceleration. Many cases of hopping during a climb are the result of the wild fluctuations in the “jacking” forces induced in the suspension links/arms.

The terrain resistance varies, and so the tractive force at the tyres varies. Traction varies and so the tractive force can rise or fall very rapidly. That can, but will not necessarily, result in acceleration/deceleration and squat, lift dive and the anti's.

Take for example a steep slope with the front tyres against a ledge. There is a large component of the vehicle weight acting down the slope, against the forward travel direction. On a 30 degree slope that component is sin30 x vehicle weight = 0.5 vehicle weight. Then there is the resistance from the ledge acting against the front tyres. That resistance will have a very large component acting down the slope.

As a result there is a large net force down the slope at the front tyres and to simply move forward with no acceleration, the rear tyres have to provide an opposing tractive force. There is no acceleration so no squat or anti-squat, but if the suspension geometry is set up to provide a large percentage of anti-squat, it will create a high “jacking” force that will cause pitching.

The actions/forces in the rear suspension springs and links/arms behave similar to the case for anti-squat and when they are resolved they will also act along a line from the tyre contact patch to the instant centre. In this chosen example the net force at the front tyres is in the opposite direction and the suspension will be behaving as it would for anti-dive.

However the situation is very dynamic and there can be many scenarios in as many seconds:

the front wheels could get over the ledge and suddenly the additional resistance reduces and we get some acceleration.

the front wheels could loose traction and the net force down the slope increases further

the rear wheels can loose traction and there will be a sudden reduction in the jacking force

the rise in the suspension due to the jacking can result in a change of the instant centre

bumps change the suspension height

If speed is needed, the “mass moment of inertia” has plays a great part in how well the suspension can control pitching. This is the distribution of mass away from the C of G and the closer the heavy items to the centre the better, Mass/weight in front of the front axle or behind the rear axle is particularly bad.

Also with speed, the effective stiffness due to suspension geometry that provides high %AS, as you mentioned, has an adverse affect. In a similar way, high roll centres adversely affect “yaw” at speed.

I can only speak for the conditions I am familiar with, mostly steep, rutted or rocky tracks with steps/ledges and a mixture of high and low traction, sometime off camber, and often requiring high torque and wheel speed. In those conditions suspension that provide a low percentage anti-squat and moderate roll centre height and low roll axis work best. You also can't leave out long flexy springs, good dampers and balanced articulation from the package.

I'll have to come back another day to address the rest of your post, where you have made some serious errors.

[De Ranged]

Thanks for that, you realise you answered the next question I was going to ask, lol... why the trucks with the higher AS% rear up, I see it now.... indirectly it is caused by high AS% but not by the AS, the easiest way to get higher AS is to angle the links up to chassis, this angle is making it more suseptable to the jacking

Oh and trust me it doesn't matter how steep they are still accelerating lol, 400+hp in a mid engine, low CoG buggies that weigh less than a ton!

Im still curious about what causes hop, when I was in the drag scene you could tune the car to hop or the other extreme reer up and the idea was to get it in the middle where it launched

My idea was the high AS loaded up the tyres to the point it was greater than the grip of the tyre.... as the tyre slips it looses the torque that is forcing the rear up so the weight transfer from the acceleration causes the back to drop... as the back drops the geometry improved on the links allowing tyre to grab again... up it goes and we have a cycle...... but lol from what I understand now the bit about grabbing traction again doesn't work... so please explain

I agree with the low roll axis angle, but I try to get the roll centre height as high as I can without any silly issues... generally at or just below tyre height is where I aim....

Looking forward to your next bit

Hey O Teunico sorry for taking over your thread lol

[o_teunico]

Hey O Teunico sorry for taking over your thread

Oh no, no need to apologise, I´m really learning a lot!

[Ex Member]

Keep in mind a lot of hop is caused by flexing within the suspension links and not by squat/dive movement in the springs.

[Ex Member]

Also, keep in mind that the whole squat/dive movement is only during acceleration/deceleration do that it is more of something to think about when taking off and not during a normal climb. Most change in tire force during acceleration is simple weight transfer. The extra difference caused by the squat is small and short term with values in a reasonable range.

The biggest concern for our application is how the values change with travel. If the AS is rising quickly as the suspension extends, you can get some bad dynamic behaviour as wheel moves in the rough terrain. This tends to be why short links and long travel are a bad idea.

[De Ranged]

Keep in mind a lot of hop is caused by flexing within the suspension links and not by squat/dive movement in the springs.

Why then with the same links, chassis in fact everything the only difference was the holes on the chassis end that the links are bolted to.... one setting hopped and the other didn't ......