I have a question about putting a stoker motor into a 1992 Z28. If I do a stroker motor I have heard they have put out alot of torque.Now I was wondering if this kind of a motor would be good for the street but take it onto the strip. Because I was thinking about doing this and then I wonderd if I just put a 355 insted of a storker motor would the 355 beat it because the stroker would just be sitting there spinning the tires?? Any Info you can give me I would apreciate very much.

 

anyone at all????

 

Well it is said that torque is what accelerates a car.A longer stroke engine such as a 350 stroked to 383 has the ability to develop excellent torque but is somewhat RPM Limited due to its longer stroke .A shorter stroke engine may achieve a higher RPM
So you could build a High RPM screamer or a torque monster and gear it to your needs.For a street car lots of cubes and a broad torque range work well ( 383 or 400 ).Or you could do a shorter stroke engine 327,350,377 high rpm with some deep gears and run just as fast or faster but not that streetable.

 

Torque is a twisting force that comes from the equation torque=force x distance. The shorter your driveshaft or axle, the more torque you are going to develop at the wheels. With the case of engine cranks, at lower engine speeds vaccum creates torque and the more stroke you have, the more vaccum is present which means more torque.

 

And exactly how bored were you when you made all this up?

AJ

 

Build a stroker for your car. I have a 383 in my 92 RS with a 4L60 or 700 R4 whichever you prefer and 2.73 limited slip. I wouldn't believe there were 2.73's in the rearend if I hadn't seen them my self, it is that radical. I can leave dual burnouts for half a block and the front of the car raises way up in the air. With 3.73's and slicks at the track I think it will pull the front tires. Now as for streetable, if you drive normal you can't even tell there is a beast under the hood, but when you punch it hold on. Traction is only a problem if you get on it hard in 1st gear from a dead stop other than that it is wonderful. It is hard to believe that much torque is coming out of a smallblock.

 

Ya know what?

I shoudn't have posted that reply the way I did. Sometimes I get in a sarcastic mood, and I'm just kidding. But I can see why someone would take it as a flame.

I guess what I should have really said is;

"I've never heard that.

"I was taught that torque is measured in lbs/ft. by some man (can't remember his name ) who determined that lifting one pound exactly 1 ft off the ground, at sea level, required a certain amount of work.

"The twisting force you are referring to is the same amount of work that is needed to lift 'X' amount of lbs. 1 ft. off the ground @ sea level.

"The longer the stroke of an engine is, allows the piston/connecting rod to obtain more leverage. By using more leverage, more work can be done. Kinda like it's easier to loosen a bolt with an 8" wrench than it is with a 4" wrench, assuming that you hold both of them at the end."

I'm sorry. No flame intended. I was being an inexcusable smartass.

AJ

BTW, work is equal to the product of the force and the distance through which it produces movement. I.E., work = force x distance, not torque.

 

I have a question for Jamesburn. Do you race your car. And if you do, do you have alot of problems with off the line traction I would suppose you do if you can leave burnouts like that. But do you think with some LCA'S, and things like that I could get the thing to hook?

 

Not bored at all and didnt make it up; I'm in high school physics 12. Haha I read your post again and you make me laugh. Work is equal to force x distance yes. The net force is determined from Force=mass x acceleration. The Work done by a force acting on a body is equal to the product of the force and the distance through which the force acts, provided that F and s are in the same direction; and remember that Work is a scalar quantity; meaning that no direction is associated with it. The equation for Torque I'm pretty sure was invented by James Watts (the units for work are Watts). And you're description of the wrench is using Torque NOT work because you have a pivot point (i.e. nut) on the end of the wrench. I was wrong though with the shorter your driveshaft is the more torque you develop though; the longer it is the more you develop (i.e. T= force x distance T=1lbx1ft T=1ft/lb (use whatever units you want, most countries use SI; Newton.Meter) T=1lbx2ft T=2ft/lbs) So really the more distance you are away from the wrench the more actual torque you will have at the pivot point (nut). You actually use work to determine horsepower; using the Power equation. (although horsepower is actually intially derived from torque; its just a 3 dimensional measurement) Hence work = force x distance / time and 746 watts equals one horsepower ( a kilowatt equals 1.34 horsepower).
In the case of a driveshaft, you would use Torque in the equation Torque=(moment of inertia) x (angular acceleration) where: T=(mass of driveshaft/9.8m/s2) x (radius2) that's the easier way of going about Torque or you can go: T = (mass of driveshaft x 2(pi)radius, then substitue that into: distance / time(time of one rotation (or period)) intial - final change in velocity) to get acceleration then times that by the diameter which is 2(pi)radius. Yes Work and Torque share the same equation but Torque isnt scalar; it measures the effectiveness in turning the body about a certain pivot point (again i.e. nut incorporating the wrench). Torque was never determined by someone lifting 1 lb off the ground 1 foot because that would make Torque linear when actually it's an angular measurment. I believe if you wanted to
figure out how much work each piston makes you could go: work = (mass of piston x (distance / time) velocityfinal - velocityintial / change in time (again final - intial) all multiplied by the actual stroke of the crank.
It might be confusing but that's how it works. I just wanted to clear up any inaccuracies in the post; I also hope that I covered everything cause i didnt want to go upstairs and grab my physics textbook.

 

Yea, I race when I can find one. I haven't done a thing for traction devices yet, but with some of those I am quite certain you could hook it up. Even with it spinning the tires like that the car is hauling good enough to beat most people and it catches fine if you don't floor it from a dead stop. What I have done is remove a lot of weight from the frontend and the car transfers really well. Good rear tires are a must, my Firehawks are very sticky and work quite well so far, but those street slicks would probably be needed with a steeper gear. Going to the race track in a couple of weeks and I will post times, I guess low 13's right now. Good luck!

 

I'm sorry. You are right. It's not 1 pound/1 ft. off the ground, it's moving 1 lb. a distance of 1 ft. That is what the measurement ft/lbs. is equivilent to.

And as far as torque being greater by using a longer driveshaft is not correct. One thing your formulas (and just about any other formulas for that matter) does not take into acount is the fact that a portion of the twisting force will be lost in the flexing motion of the metal (aka torsional loss) that the driveshaft is made of.

And I'm sorry that you don't truly understand why the longer stroke of an engine is the primary cause of the increased torque. It's caused from (as I said once) increased leverage. The recipricating motion of the piston/con. rod is transfered to centrifical motion. The effeciency of that transfer is determined by the leverage involved. Sorry I don't know the 'formula' off hand. To me it's called common sense.

I love book worms who read about what should happen, and the mathmatical formulas that determine the output based on other 'fixed' variables. The truth be known, I'd love to see you crank a 3/4" nut onto a rusty bolt to 190 ft.lbs using a 6" wrench. That will give you a who new light on the subject, and teach you about leverage.

What they say will happen, and what really happens are two different things.

I'm not saying that you aren't learning well, and I wish I had "Physics 12" in high school. I think they were still working on "Physics 1" when I was in school. But you are trying to apply these formulas to something that is easier to explain by simple logic.

BTW, you are using a different term of foot-pounds than should be used here. Time is not part of the formula when speaking of torque on an axis.

foot-pound, a unit of work or energy in the customary English gravitational system; it is the work done or energy expended by a force of 1 pound acting through a distance of 1 foot. It is equal to 1.356 joules. The term foot-pound is also used to designate a unit of torque that is sometimes called the pound-foot to distinguish it from the energy unit. A force of 1 pound applied 1 foot from and perpendicular to the direction to an axis of rotation produces a 1 foot-pound (or pound-foot) torque at the axis.

One more question? How is vacuum a determining factor in creating torque? Can you please show me the formula for that?

 

Whoops you're right I gotta correct myself again using my own equation: You're get more torque to your rear wheels using a fatter driveshaft, one with a larger diameter; NOT a longer one.

I dont recall ever saying that the longer stroke of the crank didnt make more torque. I said it did mostly at lower engine speeds; it all comes with rod ratios too. The reciprocating motion of the piston/rod is transferred to centripetal force. mass x 2 x pi x r divided by period squared.

I'm not really a book worm; I just found the torque unit interesting so I paid attention. As for using formulas with fixed variables to figure things out, I just like to think that it actually would work in an ideal situation.

I know time isnt part of the torque formula. You use time to get the net force. I dont know how I can be using the wrong foot pounds; there's only one type of use and when you combine 2 feet and 4 lbs it makes 8 ft/lbs. Kinda makes sense.

I dont know about the vaccum thing; I read it in a Chevy racing book and it just kinda stuck in my mind.

 

Well I can tell you that 8ft/lbs. is a measurement. Even though torque is a force of twisting, it's still measured by using linear movement. I guess because the guy (Mr. Watt as you said ) figured it up that way. Kinda like horsepower is a measurement of horses working. If he (that guy) would have used cats instead of horses, then it'd be catpower instead of horsepower. :sillylol:

But when you use the term 8ft/lbs. it is commonly misunderstood because that is the incorrect way of describing it. It should be 8lb/ft. The 'ft' part is always to expressed as one linear foot. Moving an 8 pound object 1 ft in any direction ( I used 'off the ground' earlier because draging or pushing it across a plane would impart friction, and that is not part of the foot-pound, AKA pound-foot measurement) is 8 pound-foot of torque. It's pounds-per-foot.

By saying 2 ft and 4 pounds (what you said was 8ft/lbs.) would require the same work as moving 8 pounds along a one foot distance. So it would be 8lb/ft of torque. Kinda like reducing a fraction to it's simplest term.

So I guess you and I are both right. Just you say po-tate-o, and I say po-tat-o I was just not taught the same way you were. I had to teach myself.

AJ