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Page 12
REAR SUSPENSION
INSTANT CENTER LOCATION
(dragracing)

The following is intended to clear up some of the confusion regarding the location of the rear suspension instant center (IC) of a dragstrip car with a beam rear axle.

This is a rough sketch of the car as it accelerates:

The arrows represent the dynamic forces involved. There is a force forward generated at the rear tire patches and a rearward force, equal in magnitude, which acts through the center of gravity. The latter is called an "inertial force." The weight transfer is represented by the letter "N." It is seen to act upward on the rear and downward at the front.

If we sum the torques acting about the front tire patches, we see that the inertial force is attempting to rotate the car counter-clockwise (CCW) and the weight transfer is attempting to rotate the car clockwise (CW). If there is not to be a wheelstand or a blowover, these two torques must be of the same magnitude. If we call the CG height "h" and the wheelbase "L," this can be expressed as:

Fh = NL

The IC is understood to be the point at which the net forces in the links act on the chassis. With a ladder bar setup, the front pivot for the ladder bar is the IC. With a 4link, the intersection of lines through the upper and lower links is the IC.

In order for this explanation to cover all cases, we'll simply say that a car has P% anti-squat. This would mean that P% of the weight transfer is taken through the suspension links and (100 - P)% is taken through the rear suspension springs. This would look like this:

The IC is located by the horizontal dimension "a" and the vertical dimension "b." Again, we can equate the torques acting about the front tire patches, but, before we do, we'll take that Fh = NL equation, solve for "N," and make that substitution. In other words, N = Fh/L.

So, equating CCW torques to CW torques:

bF + (L - a) hFP/(100L) + hF(100 - P)/100 = Fh

The "F's" cancel out and, with a little manipulation, we end up with:

b/a = Ph/(100L)

In other words, so long as the ratio of "b" to "a" is equal to Ph/(100L), we'll have P% anti-squat. But, that b/a is the slope of a line that passes through the rear tire patches. The dashed arrow represents the total dynamic force acting at the IC. It is the sum of the horizontal and vertical forces. Note that this total force's line of action is along a line that passes through the rear tire patch. Any line of constant P% anti-squat, then, passes through the rear tire patch.

(Actually, when the weight of the rear axle assembly is taken into account, the line passes about an inch below the rear tire patch, but this small difference is normally ignored.)

Note, also, that (L - a) becomes negative when "a" is greater than the wheelbase. This would be expected as the vertical force, acting at the IC, is attempting to rotate the car CCW when the IC is forward of the front tires.

So, suppose the wheelbase is 105 inches, the CG height is 21 inches, and the car has 50% anti-squat. This would mean that b/a = 21/(210) = 0.1 and "b" might equal 5 and "a" might equal 50 or, on the other hand, "b" might equal 500 and "a" might equal 5000. The torques acting on the chassis remain the same.

Note that the relationship between the horizontal location of the IC and the center of gravity of the car is not considered. All that is important is the location of the IC relative to the rear tire patch, or, specifically, the tangent of the line passing through the IC and the rear tire patch.

So, does this mean that the distance forward, from the rear axle to the IC, makes absolutely no difference? Well, not exactly. With a 4link, the IC moves vertically as the fuel load changes and as the tires travel over the track surface irregularities. While this might not amount to a large change, it would still be best to minimize it. For this reason, it is desirable to have the IC as far forward as conveniently possible. There are practical limitation with ladder bars, of course, but, with a 4link, it is usually possible to place the IC a large distance forward. With 4bar linkage, where the bars/links are always parallel (to each other, NOT to the ground), it must be recognized that parallel lines meet at infinity. In other words, if a 4bar is adjusted so that the links are parallel to the 100% anti-squat line, the IC will be located at an infinite distance ahead of the car. And, of course, the car will neither squat nor rise on acceleration.

(A few words on that which is occurring at the front of the car, during launch, are in order: Since, with a rear wheel drive car, no tractive forces are acting through the front suspension links, all of the weight transfer must occur through the front suspension springs. In other words, the front of the car is going to rise, according to those forces shown in the first picture above, and there's nothing that can be done about it. The nominal amount of front end lift is a function, again, of those forces indicated in the first picture AND the front suspension spring rates. Forces internal to the picture...such as those acting at the rear suspension instant center...can, however, slightly increase or decrease the value of front end rise, with low values of anti-squat increasing the rise and high values decreasing. The notion that a rear suspension instant center forward of the center of gravity will somehow aid in lifting the front of the car is simply not true. If it were, it would logically follow that an IC located an infinite distance forward...as with a parallel 4bar arrangement...would result in a violent blowover! Again, all that is important is the location of the IC relative to the rear tire patch, or, specifically, the tangent of the line passing through the IC and the rear tire patch.)

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