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Something To Think About

3K views 29 replies 11 participants last post by  theHIGHLANDER 
#1 ·
Just like some reactions on a little procedure I've been promoting for the last year or two.

Suppose a horizontal chain was hooked to the back of a car, at about the CG height, and the other end attached to a power winch. The chain is centered between left and right tires. Some wheel scales are placed under the front wheels ONLY, where they record 880 pounds for the left front and the same for the right front BEFORE the chain is tightened. The automatic transmission is placed in PARK. As the chain is tightened, left front and right front loads are noted and recorded. These loads are (838.5, 850.1), (797.0, 820.2), (714.0, 760.4), and (548.0, 640.8). As the chain tension is increased, the sum of the two loads decreases. The difference between the original sum of 880 and 880 and the sum with the chain under tension represents the "weight transfer" which occurs during acceleration. The difference between the left front and the right front increases as the weight transfer (or, in other words, the acceleration) increases. The left front is always "lighter" than the right front. Is all of this exactly what you would expect? Although there are no wheel scales under the rear tires, could you determine their loads at each chain tension?
 
#27 ·
I'm back to mess with your minds a bit more.

Suppose you used this traction dyno on something without a suspension. Suppose, for instance, that you started pulling back on a AA/FD with, again, its engine prevented from rotating. As you pulled, you'd see that same left/right difference developing and, if you pulled hard enough, the left front would leave the shop floor first, just as it does during a launch at the dragstrip. That torque necessary to lift that left front is also acting at the rear of the dragster to cause problems with the rear tire loading. And, of course, that isn't good.

So, suppose you built some preload into the chassis. Suppose that, if you supported the front at the middle so that both fronts were off the ground, the left front would "sag" below the right. If sufficient sag was present, rear wheel loading could actually be equal just before both fronts lifted on launch.

So, use of the traction dyno would also be beneficial to the constructor of dragster chassis.
 
#28 ·
My mind isn't messed yet...however, why do you pull back on the chain instead of forward? We're moving our cars forward, right?

My guess is that as the car is "lifted" at a point determined by the weight exercises discussed earlier we'd see some interesting things happen as well.


Super stock cars need to do gargantuan wheelstands to transfer load to the tires. Enough trial and error has proven lower elapsed times by loading the tire to the max.

Top fuel cars as well as funnycars use the same loading principles, or "sag" as you refered to it already. Many front motored combinations (funnycars and nostalgia slingshots) have loading bars in strategic locations near the rear to help load the car straight on launch as well as down track. The accell forces on these cars never quits until the end of the run. Most of us can only imagine what dealing with that kind of power encompasses.
 
#29 ·
theHIGHLANDER said:
[W]hy do you pull back on the chain instead of forward?
The inertial force appears ONLY when an object is being accelerated. Since the car isn't being accelerated as it sits on the shop floor, I'm simulating the inertial force with the force on the chain. The inertial force is ALWAYS acting in a direction opposite to the direction of acceleration. As NASA attempts to accelerate a rocket up into space, the inertial force is acting in a direction to keep it back on the launching pad.

As the chain is tensioned, an equal, but opposite, force develops at the rear tire patches. I think this is the forward force for which you're looking. It is because of this force at the rear tire patches that the engine must be prevented from rotating. It is this same rear tire patches force that generates the driveshaft torque which, in turn, causes the force imbalance measured at the front wheels.
 
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