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Old 04-26-2007, 06:16 AM
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Originally Posted by grouch
Very interesting. Thank you.

Care to walk through what happens during deceleration, specifically, how anti-dive works? It appears to me that keeping a car from nose-diving in a panic stop either involves a lot of moments and forces or some black magic, somewhere.
Okay. I was going to explain 100% anti-squat at the rear, but we'll look at anti-dive first.

Suppose, instead of the car accelerating forward at 1G, it is decelerating (braking) at 1G. The arrows are simply reversed from those occuring during forward acceleration, but with the difference that braking force is being applied at both front and rear. The two horizontal arrows are then added to reach the 3000 pound value.

It's important to realize that, to this point, anti-dive has not been considered. In other words, the above is true, whether the front bumper is scraping the ground or not.

Most of the braking effort is normally at the front. So, let's say that, in our case, 75% of it is there, meaning the total horizontal force, at the front wheels, is 2250 pounds.

In the first post, I considered only forces that were either horizontal or vertical. These are very easy to add and subtract. Unfortunately, forces can be neither horizontal nor vertical. We can, however, resolve such angled forces into horizontal and vertical components. In other words, an angled force has the same effect as its two resolved horizontal and vertical components.

Suppose a force of 10 pounds is angled up from the horizontal at an angle of 30 degrees. This angled force would be equivalent to a horizontal force of 8.66 pounds and a vertical force of 5 pounds, acting at the same point. These numbers are obtained by multiplying 10 by the cosine and sine of 30 degrees. If, on the other hand, you knew the horizontal component was 8.66 pounds and the vertical was 5 pounds, you could determine the total force by taking the square root of the sum of the two numbers. (This is that right triangle from high school trigonometry.)

Getting back to the car under braking, let's greatly simplify matters by assuming that the front suspension is like a reversed ladder bar. In other words, the entire front suspension connects to the rest of the car at a single point (when viewed from the side) above and to the rear of the tire patch. Also, we'll further simplify by assuming that the weight of the front suspension components is negligible (zero). For the free body diagram, we'll not only erase the ground, but, at the front, we'll erase everything right up to that suspension attachment point. It should be obvious that, with that point behind and above the front tire patch, the braking force exerted on the rest of the car will have an upward vertical component.

It's time to consider another important aspect of working with force vectors: A force vector can be considered to act at any point on its line of action. To illustrate this, consider that first free body diagram of the static loads. If you put larger tires on the car, the CG will be raised, but, of course, the wheel scale readings would not be affected. You could put the car at any height and the wheel loads would remain constant.

So, with the car under braking, the line of action...of that front braking force...would pass through the front tire patch and the suspension pivot point. Now, we could do the force and moment balance with the force at that pivot point, but we might as well take advantage of the ability to move the force along its line of action and make the work a bit simpler. We'll move that force until it's directly above the rear tire patch and then we'll calculate moments about that rear tire patch. The vertical component of the braking force must equal the vertical force on the rear tires. As we've seen from the previous post, this would be 600 pounds. The horizontal component we know to be 2250 pounds. As we consider moments, we see that we have a CCW (positive) moment of 3000 times 20 (the CG height) and a CW moment of 2250 times the vertical distance from the rear tire patch to the intersection with the line of action of the force. This means that, to achieve moment balance (100% anti-dive) the intersection must be 26.7 inches above the rear tire patch. In determining the angle from the horizontal, we would reach for our calculators and find the arc tangent of 26.7 divided by the 100 inch wheelbase. This would be slightly less than 15 degrees.

Now, it's time to go back and tidy up some of the simplifications. First, that reverse ladder bar pivot point would be the instant center of the independent front suspension. This is the intersection of lines through those angled inner pivots on a modern suspension. And, since the front suspension components definitely have more than zero weight, the line of action must pass very slightly below the front tire patch and, in addition, the full 2250 pounds is not available, as some of it is used to slow the front suspension components.

As has been mentioned, I believe, in another thread, 100% anti-dive is not used, as it causes severe ride harshness.

Well, I hope that has taken the mystery out of anti-dive suspensions. As I said, I had planned on a post covering anti-squat, which is, perhaps, a bit easier to understand.
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