Free Body Diagrams
We're all familiar with the use of arrows to represent forces. We're constantly being told to "push here" or "pull here" with an arrow showing the direction of the push or pull. Similarly, the knob on your air compressor pressure control probably has a curved arrow to indicate rotation for increased pressure. Engineers call the force arrows "force vectors," but you don't need to talk like an engineer. The meaning is obvious. And, it's equally obvious that, when two arrows are in the same direction, they add in effect and, when in opposite directions, they subtract. Engineers would call this "vector addition and subtraction," but, again, the meaning is obvious.
When constructing what they call a "free body diagram," engineers use the straight arrows to represent forces and the curved arrows to represent moments. ( A "moment" is the action of a force about a point and, when two parallel forces are equal and opposite in sense and not acting on the same line, it is called a "couple." And, of course, you've used a "torque" wrench. So, whether it's a moment, couple, or torque, the result is the same: Something is being twisted.)
The first problem an engineering student is given is usually a horizontal flag pole sticking out of a building. To make a free body diagram of the flag pole, he "removes" the building and replaces it with the arrows representing what the building was doing to hold the flag pole in place. There had to be a vertical force to support the weight of the flag pole, but there also had to be a moment. So, the engineer would draw a vertical arrow to represent the force and one of those curved arrows to represent the moment.
But, we're interested in cars and not flagpoles, so let's go right to a free body of a car. Consider the car as seen from the United States driver's side. The car is parked on level ground, so we will be considering what are called "static" forces. The only contacts the car has with the rest of the world is through the four tire patches. These tire patches can only support vertical and horizontal forces. No twisting is available. And, since the car is parked, we know that there are no horizontal forces. As with the flag pole, the ground is "removed" and replaced with force vectors. These 4 arrows will be vertical and pointing upwards. Yes, you're way ahead of me: There's a 5th vector, acting vertically downward through the center of gravity, which represents the total weight of the car. The 4 tire patch vector magnitudes can be determined with the use of wheel scales. Suppose we've done this with a car and the two front values are each 825 pounds and the two rear are each 675 pounds. As we view the car from the side, it's both convenient and practical to simply consider the car as if it were a "bicycle" and add the two front forces and the two rear forces. So, we end up with a 1650 pound upward force at the front and a 1350 pound force at the rear.
You've probably done this before and you now know that the total car weight is 1650 plus 1350 or 3000 pounds. But, before we leave the obvious, let's attach a fancy name to this simple addition. This is called, by an engineer, a "vertical force balance." Before an engineer considers his free body diagram complete, there must be a force and moment balance for each of the three planes (side view, top view, and front view).
So far, this is a very simple free body, but we haven't located that total weight arrow. We know it's pointing down, but where is it? Suppose the wheelbase is 100 inches. We must now perform a "moment balance." To do this, we must first arbitrarily pick a point. It is important to realize that this point's location is entirely arbitrary, but, at the same time, it must be realized that some points are far more convenient for the purpose than others. We'll use the front tire patch as the point. The distance, on a line perpendicular to the line of action of the force at the rear tire patches, is the wheelbase or 100 inches. The moment provided by that force is the product of the wheelbase and 1350 pounds or 135,000 inch pounds. In this case, that moment is acting in a counter-clockwise direction and we'll consider that CCW direction to be positive. An equal and opposite (CW) moment is provided by the weight of the car. If we consider the distance, on a line perpendicular to the line of action of the total weight vector, to be "a," we then know that 3000(a) is equal to 135,000 inch pounds. Solving, we see that the CG (center of gravity) is located 45 inches behind the front tires. You can satisfy any doubts about the initial point selection by making similar calculations using, say, the rear tire patch and a point halfway between the front and rear.
But, of far greater interest are the "dynamic" free body diagrams. These are diagrams which portray ONLY those forces and moments which occur as the object is being accelerated. Continuing with the car just described, no dynamic elements exist. But, suppose the car is rear wheel drive and is being accelerated forward. We now must have a horizontal arrow, pointing forward, at the rear tire patches to indicate the thrust accelerating the car. If we are to have the necessary horizontal force balance, there must be an equal and opposite force. This is the inertial force, acting through the CG. (This is the "F=ma" which is the centerpoint of an engineer's existence.) The inertial force is always opposite in sense to the acceleration. In this case, the acceleration is forward, so the inertial force is backward. Let's say the car is accelerating forward at 1 G, so the total rear wheel thrust equals the inertial force equals the total weight of the car equals 3000 pounds.
Okay, we have a force balance, but what about a moment balance? The two horizontal forces constitute a couple, the value of which is 3000 times the CG height. We need another number, so we'll say the CG height is 20 inches. This yields a couple value of 60,000 inch pounds. Time for another moment balance: Again, we'll use the front patch for the point, resulting in a net negative moment of 60,000 inch pounds. For balance, there must be a positive (CCW) moment of the same magnitude. The only way this can occur is if there is a vertical upward force at the rear tire patch. Further, we know that its value, when multiplied by the wheelbase, must equal 60,000 inch pounds. So, the value must be 600 pounds. This, of course, is what we call the "weight transfer." Finally, in order to have a vertical force balance, there must be a force downward, at the front wheels, of 600 pounds.
It might be difficult to visualize a downward force at the front tire patches. But, if you'll recall, the static free body indicated the presence of an upward force at the front tire patches, so a vector subtraction is occuring. So long as the dynamic downward force is less than the static upward force, the car is still being supported by the front wheels.
But, what if that is not the case? Suppose the dynamic force exceeds the static. This would mean that a moment balance no longer exists and the car would start to rotate about the rear tire patches. Or, put simply, you've got a wheelstand.
The basis for a free body diagram is the principle that a body's dynamic status can only be affected by its contact with the outside world. The arrows replace and represent the effects of the outside world. In this case, we've considered the entire car as "the body." It's only contact with the outside world is through its tire patches. Within the car, there are many forces as it accelerates, but these forces are all cancelled (balanced) within the car. A simple example is that which occurs when you pull up on the handlebars of a bicycle in order to apply more force to the pedals. These forces are balanced internally and do not affect the tire loadings. Similarly, changes in suspension geometry can change force magnitudes within the car during acceleration, but they do not affect the analysis of a free body diagram.
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.
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.
Please, continue with the anti-squat description! Understanding how these forces interact definitely helps understand the how's and why's of the parts and their designs.
(BTW, when I saw the title of the thread, I almost passed it by. Then I saw the author and knew it wasn't about 'Get your FREE! Body Diagrams here -- just $19.95 shipping!' :) )
I will take this opportunity to correct my description of the Pythagorean Theorem. I said, in my last post:
"...you could determine the total force by taking the square root of the sum of the two numbers."
That should have been:
"...you could determine the total force by taking the square root of the sum of the SQUARE OF EACH OF THE two numbers."
Sorry about that. I make every effort to avoid errors, but...even with a careful read before posting...they still show up!
Also, I would point out that most books (including those used in courses for automotive engineering students) state that the 100% anti-squat line has a slope equal to the CG height divided by the wheelbase. I only recently realized that this must be multiplied by the ratio of the total weight to the value of the total weight less the weight of the rear axle assembly. In other words, the books are making the same simplification that I made in the anti-dive post (that the weight of the rear axle assembly is zero). This is a very small error, but, again, I try to present things accurately. (Your wife would probably sympathize with my wife for having to live with such a nit-picker.)
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