Bush65

BTW, the SAE definition of roll centre is the point where you can apply a load on the spring mass and not cause it to roll, or something to that effect.

Roll centre depends upon the suspension geometry and has nothing to do with speed or time. The statements about roll centre location are not strictly correct. It is not correct to say that the roll centre for A frame suspension is the ball joint. If the ball joint happens to be on the transverse vertical plane through the centre of the wheels, then, by coincident, the roll centre just happens to be where the ball joint is. Admittedly this is a likely scenario, but it it doesn't make the statement correct. The roll centre for suspension with a pan hard bar is not at the centre of the pan hard bar, unless the centre of the pan hard bar happens to be on the transverse vertical plane through the centre of the wheels. However if you are looking at a front view, and the roll axis happens to be horizontal, then the roll centre will be directly inline with the centre of the pan hard bar, but on the transverse vertical plane through the centre of the wheels. The transverse vertical plane through the centre of the wheels is used for the roll centre definition because the front and rear roll centre heights affects the transverse load shift distribution to the tyres during roll. This is important because it affects the traction and determines under/over steer. The suspension roll axis passes through the roll centre, and its slope determines bump/roll steer. As for the point where the pan hard bar is attached to the axle, that is not the roll axis, it is the instant centre for the pan hard bar. Consider if it was (which it isn't), and consider if the pan hard bar was horizontal with the ends inline with the springs and the left end attached to the axle. Now consider what would happen when the chassis/body rolls to the right. The right end of the pan hard bar will force the chassis/body to move down in an arc. There will be a large vertical downward and small horizontal to left displacement at that point in the chassis. Now if we look at a point on the chassis at the perch for the left spring. This point will also swing in an arc about the roll centre. There will be a large displacement to the right and a small displacement vertically downward. How can you reconcile these displacements? Now consider the loads on the springs. The right spring will compress a large amount and the left a small amount. This is a significant nett increase in the total load on the pair of springs. Where will that extra load come from? Unless you have a good explanation to those questions, it should be obvious that the roll centre can not be at the end of the pan hard bar.

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. 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.

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%

I don't see a problem using the 2WD method to compare behavior, when comparing apples with apples, so to speak. If you want to be pedantic the %AS, %AD method/value are easily adjusted for 4WD, but when your being pedantic, what do you do when you reach the limit of traction at the front?

But if you are going to consider anti-lift with a 4wd, then you should not be using the usual 2wd theory to determine %AS.

Because it should be obvious, I'm not sure if I correctly understand what you are getting at when you said: The front to rear weight distribution directly affects the fore & aft position of the C of G. By definition C of G is a notional point where a concentrated mass equal to the total mass will produce the same effect. For equilibrium the sum of all of the moments must equal zero. If: the weight at the rear wheels is Wr the weight at the front wheels is Wf the total weight is Wt = (Wr + Wf) the wheel base is Lw the distance of the C of G ahead of the rear wheels is L1 the distance of the front wheels ahead of the C of G is L2 Taking moments about the rear wheels gives: L1 = (Wf x Lw) / Wt then L2 = Lw - L1 Alternatively for the same results take moments about the front to give: L2 = (Wr x Lw) / Wt then L1 = Lw - L2 The above is an example finding C of G give 2 known weight/masses. Note when taking moments about the rear wheels the moment of the weight on the rear wheels is zero (Wr x 0.0 = 0.0) so it wasn't included in the 1st example ..... The method for finding the C of G for a large number of masses is the same i.e. add the moments for all of the masses, then divide by the total mass.

Bill, the important thing is the path of the line from the tyre contact patch through the side view instant centre (SVIC). So you must consider both the height and for & aft position of the SVIC, i.e. don't consider one in isolation. Now plot that line and see where it passes the sprung C of G, more importantly how close/far it is from the sprung C of G and on which side. I can't see the clips, Photobucket gives a message that this is temporary. %AS numbers are only as good as the data values used to determine them, and the height of the sprung C of G can often be a wild guess that introduces significant errors. Your statement about using the %AS to compare what works well for you with one build and use it again with another build for similar/same use is spot on. This is also valid for using the same %AS as someone else if it works well in the same circumstances. What is not valid is taking values that work well for completely different terrain and traction conditions and expecting them to be ideal in all conditions. So it is a meaningful measure when used appropriately. Hills don't change the C of G or wheel base. You are making the mistake of not changing your reference frame with the slope if you think that, and you are probably getting %AS wrong if you haven't adjusted the reference frame. The change is suspension height, front and rear does change on hills (and with articulation) and that may, or may not, change the position of the SVIC, relative to the tyre contact patch. In a 2wd vehicle the %AS doesn't change on hills. In a 4wd vehicle it does because the traction, front and rear, changes. We have squat at the rear wheels, and its opposite, lift at the front (%anti-squat and % anti-lift (for a 4wd only) are calculated the same way, but the proportion between rear and front traction needs to be taken into account for a 4wd, similar as for braking). Link suspension has an advantage over radius arms in that there is more freedom in where to locate the SVIC, it can even be where there is an empty space. Another big advantage, assuming there is nothing in the way, is they can allow the %AS to reduce when the suspension rides, thus creating a more stable system vs unstable system if the %AS rises. For %AS, radius arms will behave the same as link suspension with the same SVIC, and vice versa at a particular suspension height. Compared to radius arms, link suspension allow more freedom to adjust the roll axis. However there are more ways to get link suspension wrong compared to radius arms. Don't judge either link or radius arm suspension by comparing a good example of one to a poor example of the other. Also remember good link/radius arm suspension can be let down by poor springs or dampers.

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'.

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.

My comments in red above. I will come back to these issues when I have more time.

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: The 2-d, X,Z plane is vertical through the vehicle's longitudinal centreline. This plane is called 'side view'. The X-coordinate direction is parallel with the ground, the datum is the rear axle, and the positive direction is toward the front axle. 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.

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.

Bill is extremely good at observing what happens, what works, and what doesn't, and using that experience to build stuff that does what he wants. You seem to take a similar approach, but what has disturbed me is that some of your explanations are clearly wrong. I choose to ignore them, but many others will read it and take as gospel. It is clear to me that you don't understand some fundamental concepts of physics and maths but use them in explanations. For example vectors and weight transfer some recent posts. Concepts of squat, dive and their anti's is simple stuff (most should have learned enough arithmetic, trigonometry, and physics, i.e. three of Newton's laws of motion) yet people seem bent on making it complicated and stuff up the simple underlying physics.

As Michele said. IIRC his wife was into horse sports while Sam was involved with 4x4. He made that switch a few years back and I haven't heard if there has been any change. Serg is closer to him and would know more. Yes the video link was after Sam disposed of mogrover. The last pic doesn't tell the full story of the events. He certainly knew what he was doing and the old strangerover, and later version modified with mog axles, were both extremely capable in the rocks. In the last pic what can't be seen is the waterfall he was attempting to climb. The front wheels barely reach over the top while the rears are in the pool at the base of the fall. That attempt was a fail as the pic shows, but it was driven later and repeated successfully.