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01-14-2006 11:37 AM THE TRACTION DYNO
In my comments on WEIGHT TRANSFER, I described an "inertial force." This force, if you'll recall, has 3 important properties: Its line of action is through the center of gravity, it's proportional to the product of the car's weight and its acceleration, and its sense (direction) is opposite to that of the car's acceleration.

So, a 3000 pound car, launching at 1G of acceleration, would have a horizontal inertial force of 3000 pounds, acting through its CG (center of gravity), and pointing to the rear.

This inertial force causes the car to undergo certain changes. The front of the car rises. The loading on the rear tires increases (weight transfer). And, with some cars, a very noticeable event occurs at the front of the car: The LF (left front) tire begins to leave the track surface before the RF.

But, is it necessary to have an inertial force to bring about these changes? Of course not! The car can't distinguish between an inertial force and any other kind of force. As Mr. Ed might say, "A force is a force, of course, of course." (I'm sorry. I couldn't resist that.)

And, with this realization, we have the basis for a "traction dyno." Just as an engine dyno loads the engine to simulate the conditions when the engine is installed in the car, the traction dyno loads the chassis to simulate the conditions during launch.

So, let's proceed to a "for instance." We'll start with a top fuel dragster, which, of course, lacks those components which we normally associate with a suspension. Yet, as we'll see, it responds to the inertial force in much the same manner as a suspended car.

To provide the inertial force, a horizontal chain is attached to the rear of the dragster. It is centrally located (between the rear tires) and is at a height equal to that of the CG. The engine must be prevented from rotating by some means. The rear tires and wheels, however, are free and able, when the chain is tensioned, to provide a torque into the driveshaft. If, with such an arrangement, the chain is tensioned sufficiently, the LF tire will be seen to leave the shop floor...before the exactly the same manner as it does at the strip during launch.

But, we need to quantify our results, so we'll ease the chain tension, jack up the front wheels, and place wheel scales under the two front tires. (If you value your wheel scales, DO NOT attempt to also place scales under the rear tires as the chain is tensioned. The rear tires must sit directly on the shop floor. By observing the front wheel scale loadings, it's easy enough to calculate what's happening at the rear tires.)

Before beginning a traction dyno "run," wheel scales are used to determine the corner weights. Then, the scales at the rear are removed and the tensioning chain attached. With the LF and RF scale readings recorded, jack the front up, tension the chain slightly, and east the car back onto the front wheel scales. The total of the LF and RF readings will now be less. This difference in totals is the weight transfer. But, even with this first reading, you might notice that more weight has come off the LF than the RF. This trend will continue as you repeat the procedure, each time jacking up the front, shortening the chain, and then easing the car back onto the scales.

It's not necessary to continue the run until the car is hanging on the chain. Actually, only a few hundred pounds of chain tension is necessary. Since the results tend to be very linear, 3 or 4 "points" of data will give you all the information you need.

But, what do we have? Well, if, for each data point, you subtract the difference in weight removed from the RF from that removed from the LF, you'll see that this difference, when plotted against the total weight transfer, falls on a straight line. And, if you're an engineer or just "mathematically inclined," you might want to complete that exercise. But, for the average racer, all that's a waste of time! The important thing to realize is that the front tires are not being unloaded equally and that's BAD!! Why? Well, if the front tires are not being unloaded equally, that means the rear tires are not being loaded equally and that means both performance and safety are being compromised. As any oval track racer will tell you, maximum performance from a tire pair is achieved when they're equally loaded. As for safety, it's pretty obvious that a car has a better chance of staying in its lane when the tires are equally loaded.

And, here's where the traction dyno becomes a very useful tool. Instead of studying video and 60 foot time slips, you can make your suspension adjustments in the shop without even starting the engine.

Returning to the top fuel dragster, what could be done to equalize rear tire loadings during launch? Since there's no suspension, a totally satisfactory "fix" is not available, but we certainly can do something to improve the situation. We know that most of the run (I'm talking now of the "run" at the dragstrip, not the traction dyno "run"-) will be made with very little load on the front tires. So, we'd like to see the front wheel scales with the same readings under those conditions. To achieve this, it would be necessary to build in a certain amount of chassis "droop" at the LF. In other words, when the front is lifted with a centrally located jack, the LF should droop. When the car is lowered, this means that a torque preload is put into the frame. If the proper amount of droop is used, the rear wheel loadings can, indeed, be nearly equal during most of the strip run.

With suspended cars, there are means of achieving equal rear tire loading with ANY value of driveshaft torque. If anybody's interested, I might comment on these at another time.

  [Entry #2]

12-28-2005 06:10 AM WEIGHT TRANSFER
It's possible to define "weight transfer" in very few words, understandable by the average dragracer:

Weight transfer is the increase in vertical loading of the rear wheels that occurs during launch.

But, this definition doesn't begin to answer many of the questions that arise when dragracers start benchracing.

For a very basic starter: What causes it?

To adequately answer that apparently simple question, we have to go back to a time when there were no dragstrips. There weren't even any cars. There was, however, a very bright young man by the name of Isaac Newton. The popular story is that he was lying under an apple tree, in about the middle of the 17th century, when an apple fell from the tree and hit him on the head. Now, the only idea I would have gotten would have been to gather more apples for a pie, but young Isaac was prompted, instead, to develop his Laws of Motion. While the apple story is most likely a fable, the incident would have demonstrated one of his maxims: The force on an object is equal to a change in momentum of the object. But, we need to put that in other terms before applying it to dragracing.

Long before Isaac and his apple came along, engineers had a pretty good understanding of the forces encountered in a structure and in the use of a lever. But, it wasn't until about a hundred years after Isaac Newton that another man demonstrated that these ancient principles could, with a simple "tweak," be used with Newton's Laws of Motion. His name was d'Alembert and that which is known as "d'Alembert's Principle" is the idea that problems involving the acceleration of bodies can be reduced to relatively simple static problems with the application of a little imagination; specifically, the application of an imaginary force.

Suppose, for instance, that a force is accelerating a block of material. Instead of considering the block in motion, we can consider it to be stationary and that the accelerating force is acting against an equal and opposite force, acting through the block's CG (center of gravity). Another way of expressing this force is to say that, according to Newton's Laws, its magnitude is proportional to the product of the block's mass and its acceleration. This force doesn't actually exist, so we must consider it "imaginary," yet, to a dragracer, the concept should be so evident that it's difficult to consider it as anything but real. This force is called the "inertial" force.

It's important to realize that this added force is ALWAYS in a direction opposite to the direction of the acceleration. In the normal dragstrip situation, the rear tires are exerting a force in the direction of the acceleration. Using d'Alembert's Principle, the reaction is a force, acting through the CG, in exactly the opposite direction.

The forces are equal, but in opposite directions. This is one of those principles developed by those ancient engineers: Forces must be "balanced" on a body at rest. That is, if you consider an arbitrary direction for a body at rest, the external forces acting in one sense (of that direction) must be equal to those acting in the other sense. For instance, the sum of the four wheel loads must equal the weight (gravitational attraction) of the car.

Another ancient principle is that the sum of moments about any arbitrary axis of rotation must be zero. To illustrate this, we might as well go directly to the car during launch.

We've already established that the forward thrust of the rear tires is equal and opposite to the inertial force. But, the tire thrust force is at the strip surface level while the inertial force is acting through a point some distance up from the strip surface. Two equal, opposite, and parallel forces, acting at different locations (in this case, at a spacing equal to the height of the center of gravity), constitute a "couple" or "moment," the magnitude of which is the product of one of the forces and the spacing. If we select, as an axis of rotation, a line through the front tire patches, that line would appear as a point in the side view. Consider the car in the illustration. The situation described would provide a CCW (counter clockwise) moment. But, it's been stated that the ancients established that the sum of moments, about any point, must be zero. So, we must somehow find a CW moment of the same magnitude. And, this is where the "weight transfer" comes in.

If equal and opposite vertical forces are acting at the front and rear tire patches, we have a second couple with a spacing equal to the wheelbase. So, if the magnitude of this couple equals that of the couple just described and if we associate a negative value with CW (clockwise) couples, a sum of the two couples results in the desired value of zero.

Again, looking at the illustration, we see a force acting upward at the rear tire patch. (This is, of course, actually two forces acting at the two rear tire patches.) You should immediately recognize this as that which we call the "weight transfer." The force at the front tire patch is in the opposite direction, which might seem puzzling, as it would appear the the front of the car is being pulled down. This is because, when we think of vertical forces, we tend to think of the force that the car exerts and not the force being exerted on the car. At the front, the couple force is down, but the force exerted by the strip surface is up as it supports the weight of the front end of the car. So, that downward couple force is going to simply subtract from the static upward force. At the rear, the forces will be additive.

But, what if the front end of the car weighs less than the couple force? Not to worry! We simply add another imaginary inertial force...a vertical one this the CG to maintain the couple balance. This represents what will eventually be a "blowever" unless the forces change in a manner that halts the upward acceleration of the car.

Note that, in determining the value of the weight transfer, no consideration is given to the "internal" workings of the suspension. Suspension design parameters affect weight transfer only when they affect CG height and tractive force.

An example of the former would be the use of low rate front springs. An example of the latter would be any effort to equalize rear tire loading for maximum tire effectiveness.

Artwork courtesy of "Willys36." (With that profile, who else would it be?)

(click photo to enlarge)
  [Entry #1]

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