Hallzy
Well-Known Member
Do I need to change torsion bars after big block swap if weight of the big block will be equal or lighter than the stock small block?
Aluminum heads intake rad water pump glass bumper etc....also it's a low deck 400Not if they are correct now.
Unless it's an all aluminum BB and an all cast iron small block I don't know you'll save all that weight.
Not if they are correct now.
Unless it's an all aluminum BB and an all cast iron small block I don't know you'll save all that weight.
The diameter of the bar has to do with the energy it can store. A smaller ID bar with a lot of preload will store more energy, allowing the nose to rise faster as part of weight transfer on a drag car. Larger diameter bars will not need as much preload to hold the same weight, so there is less stored energy and therefore more neutral handling in a street driven car.
Do you have to change - no, no you don't. Six cylinder bars will be fine with a big block, but to get the ride height normal you have to preload them more, and more preload means less neutral handling. Look up the weights of the engines in factory stock form.. You'll note there's not much of a difference between them, certainly no more than the weight of a passenger or two. So "have to" is a little misleading IMO. Having built several 6-into-8 big and small blocks, there's not much of a handling difference until you really start driving the car hard.
Posts 11 and 13 are dead wrong. Dead wrong.
Evidently they have never scaled a torsion bar car, use preload or understand spring rates.
The lighter the spring rate, the more compressed height (think coil spring since it's easier to visualize) there has to be to hold the car at the same height.
Let's consider an example that is easy math.
You have two springs. Spring one has a rate of 100 pounds per inch, and spring two is 500 pounds per inch. Extreme numbers but it makes easy math.
The 100 pound spring will be to be compressed down to a length of 10 inches to get 1000 pounds of spring load and hold the car at ride height. Let's say the spring has a free height 20 inches. That means it has an installed height of 10 inches.
The 500 pound spring will need to get the same 1000 pounds of spring load. To do that the spring is only 15 inches long in free height, and installs at 13 inches as it takes only 2 inches of spring compression to get the same ride height as the 100 pound spring.
No matter what you do, considering shock settings are the same, the 100 pound spring will have more stored energy. So...that means when you mash the throttle and weight transfer begins, for every inch of front end travel in the extension direction will only require removing 100 pounds for every inch of spring travel. The 500 pound spring will require removing 500 pounds to make the spring extend just one inch.
It's simple math.
It is simple math, but you got the physics wrong.
First, your example is clearly for a coil spring set up. Free height and installed height are totally meaningless on a torsion bar suspension, and that matters because unlike a coil spring set up you can't realistically preload the torsion bar. If the UCA isn't pinned to the upper bump stop then the weight of the car is the amount of stored energy. If you change the corner weights on a torsion bar car you're changing the height in the front.
Second, there is no difference in stored energy, not even in your example. The 100 lb spring compressed 10 inches and the 500 lb spring compressed 2 inches have EXACTLY the same amount of stored energy, 1000 lbs. They have different compressed heights, but the same exact stored energy.
So, why do drag racers want light springs? Well it has nothing to do with stored energy. It's about weight transfer. Which is about traction. To go back to the part of your example that you got right, when you mash the throttle you start weight transfer because of acceleration. And the car with 100 lb springs will rise 1" for every 100 lbs worth of transfer you get due to acceleration. With a 500 lb spring you only get .2" of rise for that 100 lbs of transfer. That means that the car with 100 lb springs will be higher in the front, which moves the center of gravity of the car further back, so, more weight on the rear wheels. Which means more traction.
But please note, the same amount of energy was released from the spring. The difference is the change in front end height and the additional weight transfer that occurs because of that change in height moving the CG, not a difference in the stored energy of the spring. The 1" rise in the 100 lb spring and the .2" rise in the 500 lb spring release the same 100 lbs of energy stored in the springs.
Regardless of type, each spring is holding the same weight and storing the same energy.The difference is in how far the different rate springs are compressed or twisted.
On launch the weaker spring will have to rise higher before it is unloaded and is probably only a couple of inches difference.
A spring is a spring is a spring. And, you are wrong.
I used a coil spring (as I said) because it's easier to visualize, but a torsion bar acts exactly the same way.
A /6 cranked up to the same ride height as a bar that is .880 diameter will have more stored energy. It HAS to. My physics is correct. Your understanding is what's wrong.
A
I'll say it again. If you have a spring with a 100 pound spring rate, it takes 100 pounds to compress that spring 1 inch. Very simple no?
Now look at a 500 pound spring. It takes 500 pounds to compress the spring 1 inch. You there yet?
Therefore, if what I wrote above is true and correct (and it is) then the exact opposite must be true. To extend the spring 1 inch on the 100 pound spring you must REMOVE 100 pounds from it. To extend the 500 pound spring 1 inch you must REMOVE 500 pounds from it.
You CANT possibly argue that.
A
Therefore, if you have two springs (IDGAF what type of spring it is, it is math and physics that has been proven forever and you can't change it) of different rates installed to the same height, the spring with the lower rate will have more STORED energy than the spring with the higher rate.
Therefore, it's always EASIER with a spring of a lower rate to move the front suspension with a lower rate spring. Period.
Let's say I need to move the front suspension 4 inches to get the best hook. With the 100 pound spring, I only need to remove (transfer) 400 pounds from the front end. Simple math. Can't argue that.
With the 500 pound spring, to get the same 4 inches of movement (transfer) you'd need to remove 2000 pounds from the front end. That's simple math. And you don't get to change it.
I can post the same math 5 more ways and you'll still argue because you don't get it. That doesn't mean you get to tell people something incorrect just because you don't know it.
The amount of front end separation is controlled by the rate of the spring. The RATE at which it separates is controlled by the shocks extension setting. FACT.
So what I've posted is 100% correct. Chris Alston agrees with me. Dave Morgan agrees with me. Tim McAmis agrees with me. Jerry Bickel agrees with me. Science and math agrees with them.
Sorry, but facts is facts. The simple of it is...the lower spring rate compressed to the same installed height has MORE stored energy. PERIOD. If you don't like my explanation, get the "Doorslammers" book by Dave Morgan. He explains it in more detail.
Not understanding spring rates and shock settings is why so many torsion bar guys leave ET on the table.
A spring is a spring, that's true. But where you go wrong is that different suspensions don't perform the same way. A slant 6 bar cranked up to the same ride height as a larger bar is storing the same amount of energy. That stored energy is equal to the weight of the car.
You think I got the physics wrong? Explain how you put more energy than the weight of the car into a torsion bar. You have no preload. This is why your coil spring example is a bad idea, because you fundamentally don't understand how the torsion bar suspension is working. You're not compressing a 20" tall spring into a 15" space and creating preload. All the torsion bars are the same length, you can't preload them that way. The diameter being different doesn't change the preload, it changes the ride height, and that is made up by different offsets on the hex.
Yes, if you have a 100 lb/in spring rate it takes 100 lbs to compress that spring 1". And if you have a 500 lb/in spring rate it takes a 500 lbs to compress that spring 1". Thats true. And the same is true in reverse.
The problem is, no matter which spring you use, the weight being removed is the same. It's the weight of the car that's being transferred back. You're thinking about this all wrong, you need to stop comparing equal heights and work with a fixed weight transfer, because that's not changing.
When you launch, the same amount of energy is being released regardless of which spring you use. The engine is doing the work, not the springs. So if you transfer 100 lbs, the car with 100 lb/in springs will rise 1", the car with 500 lb/in springs will rise .2", and the same exact 100 lbs of stored energy is released from the springs.
This is simply false for a torsion bar suspension. The stored energy is equal to the weight of the car on that corner. There is no pre-load. That's why you're getting this concept wrong, because this is not a coil suspension with preload.
Your height changes are correct. The problem is, you're comparing different amounts of weight. And that's NOT what's happening. If you're transferring 400 lbs, the 100 lb/in springs will rise 4", assuming you have that much suspension travel. The 500 lb/in springs will move .8". And in both cases, the stored energy that's been released is 400 lbs, despite the difference in height. Again, that's why drag racers use light springs, because the change in height creates more weight transfer. But the damn springs are releasing the same amount of energy. You can't lift more weight than what the car weighs.
You can't post the math "5 different ways". You can use 5 different examples, but in each case the math is exactly the same. The amount of rise is controlled by the rate of the spring, but the amount of weight being removed is the same in both cases, so there's no difference in stored energy release.
What you've posted is not 100% correct. What you've said about the spring rates and how they relate to height changes is correct. But everything you've said about stored energy is incorrect for a torsion bar suspension.
The big names you're referring too are great, but again, not torsion bar suspensions, and they don't agree with you. I get it, you understand what you have to do in order to get a better launch. But you don't understand WHY, and your explanation of what's happening shows this.
There is no difference in stored energy. The difference in height is because of the difference in spring rates, but the amount of energy released by the springs is the same, even if the height is different. You're looking at the suspension rise from completely the wrong perspective.
I've been thinking about this;And I got a poser
Suppose you had 6-banger bars in your 71 Swinger,and no shocks on it, and you cranked the bars to somewhere in the middle of the range and measured the ride height.
And suppose you had several buddies at 200pounds apiece. So one buddy sits on the hood and the car goes down say 2 inches. So you take notes and pull the bars swapping in teener bars, and crank it up to the same height again. And now it takes two buds to drop the front the same amount. And you take notes.
Next you install the 340 bars and reset the ride-height and now it takes 3 buddies to get the same 2" drop., and finally the BB bars and now 4 buds are required.
Ok so that sets the stage for part II. Now,Lets lift the car :
How many buds will it take to lift the car 4 inches with the BB bars in it?, and How many buds to lift it the same 4 inches, with the 6-banger bars? Aw com'on the guys are all equally powerful.
And what do the results mean?