X64v Posted November 28, 2007 Share Posted November 28, 2007 This is carried over from 1 fast z's dyno thread, to help keep that one on topic. The posts so far, in order: I think 600hp in s30 is alot easier on the tranny since you are accelerating alot lighter vehicle. So more power is wasted on accelerating the parts then moving the vehicle vs a heavier car. Agreed. F=MA, less mass equals less force. Make the car lighter, makes parts stronger. Sorta like lighter pistons make stronger rods, etc. I politely disagree with the above statements. Force is dictated by the output of the engine, not the mass of the vehicle. Less mass equals less force for a given acceleration, but not for a given engine output. If you put 600ft.lbs. of torque through a tranny, it does not care what it's trying to move. You're not moving as much mass, but you're going to get better acceleration, making the stress on the tranny the same. The lighter pistons analogy is not the same, as acceleration is held constant, dictated mostly by engine rpm. The point is moot, as you're using that tranny no matter what, but none of us like incorrect information floating around. No, I dissagree. If you make a car lighter, the DRIVETRAIN COMPONENTS ARE STRONGER, it IS the same as if a lighter piston is on a rod, it makes the rod stronger. You take a car that weighs 10,000 lbs, with 500 torque, and get sticky tires and launch, then take the SAME motor in a car that weighs 1000 lbs, the car that is ligther, will be less force on the drivetrain. Take a slege hammer, leave it on the ground, then with one arm, take it and as fast as you can swing it 360 degrees over your head. Then take a tack hammer and to the same thing. Post your results back when your done with the hammer testing. Unless the 1000lb car accelerates 10 times as fast as the other car, which I think may have been X64v's point. There is a reason transmissions are rated in torque. Assuming that 100% of the torque is used (and not lost in say, tyre slip > 1), both cars will pop the transmission at the same time. Dave Your piston and rod analogy is true, but does not relate in the same way. That car that weighs 1000lbs will accelerate 10x faster than the 10,000lb car, because the force is the same. The hammer example is the same as the piston and rod thing. You're holding acceleration constant while varying the mass, which varies the force. This is not the same as the two different mass cars with the same force applied. The force on the car (generated by the engine, transmitted through the drivetrain) is the same in both cases. Cut the weight in half and you double the acceleration, not cut the force in half. Edit: Dave beat me to it. Take and tie your car to a tree, and if you could have 4 wheel drive and 100% traction, then rev it to 10k and drop the clutch, this simulates if your car weighed TONS, see what happens. Then take that same car and un tie it. Less FORCE on the trans/diff/axles. Take a go kart with 500 hp, and all use a small dinky trans, will work fine. Take that same motor and trans, in a 10000 lb car, and see if that trans last. It all goes back to F=MA. Less mass, less force, I want a scientific/mathmatical equation showing otherwise, untill them, I will create my own equation to show the difference. Ok let me break it down M = weight of car A = how fast car is accelerating F = forward force acting on car, and a direct translation of torque produced by the engine and tyre traction. Let us first off assume tyre traction is perfect (slip ratio = 0), and let us also assume the same engine is being used, thus F is the same. If you change M, all you are changing is A. It sounds like you're talking about M `magically' affecting F without a change in A. Show me how that works? Dave I may be wrong, but here is how I'm looking at it...if we're talking about breaking transmissions and rear differentials it comes down to shock loading, not a constant force or acceleration. Say that you have a 500 HP engine and you drop the clutch on a 1000 pound car with no tire (or clutch) spin. I think it would be harder on the drivetrain (a bigger shock) to drop 500 HP on a 10,000 pound car under the same conditions. I think this is basically what Brian is saying...in general lighter is better...less mass to accelerate is less force on the components. Here is another analogy: Who would win a 400m dash...a fat guy or a skinny guy? lol...if it were a tie, I would think it would be much harder on the fat guy's bones (components) than the skinny guy's bones. Maybe it's not a perfect analogy, but this thread could use some more lightening up! Somewhat of a good analogy, but if you slowly crept upto full-taught with the rope or whatever you're tying the car to the tree with, and then started a launch with the rope already very tight, the only thing that would happen if you got 100% traction is your clutch would just spin freely after your bumper or tow-hooks took the brunt of the launch along with the tranmission. but your analogy does make sense... but there are way too many variables to tell for real. not that you're gonna tie this car to a tree anyway, but whatever. Yes, a shock load, if a roll is done, theres not much difference, thats why you break things on the line at the drag strip and not so much half track, although it can happen. Take a 2000lb z to the track with slicks, then take the same drivetrain with a 3000lb z to the track with slicks, you will be harsher on the 3000lb car off the line, and more likely to break the drivetrain. I am speaking from first hand experience fellaws. I dont just pull this stuff out of thin air. Noone is arguing that a heavier car isn't harder on the drivetrain (atleast not me). I am arguing your reasoning. The reason is that the heavier car has more load on the tyres, which makes them less likely to spin, in addition to requiring more torque to move it. Experience is worthless to us here if you can't explain it properly. Noone is going to take your word for it. Dave Where, am I talking about tires helix? I am talking MASS, thats always the varible I am talking about. What do I know right. Well Ive always been the one to think and do outside of the box, if not, we wouldnt even have a twin cam L series head to talk about in this dyno thread. But next time you go to the drags, take and put 1000lb's of lead in your spare tire well, and launch your car with 10 inch wide slicks at 8k RPM, lets see if your trans holds up. But untill then, I will be installing the Z32 transmission, as the Z32's were the heaviest of the z cars untill 96, so nissan but a bigger trans for a reason. Clifton and I were talking last night, were gonna go back to the dyno before the year is over, I will use the turbos one last time, and see what we get. As people have said before, no one is doubting that a heavier car will be harder on the drive train. And I certainly am not doubting your expertise, skill, or experience, all of which are clearly far above mine. But F=MA isn't a robust explanation for the higher shock loads. By saying that F increases with M, you assume that A remains constant. This would undermine the advantage of bringing a lighter car to the track. The force that breaks drivelines isn't the required force to accelerate the car, but the maximum applied torque before the tires lose grip. This is defined not by the weight of the car necessarily but by the weight on the rear tires. The traction of the tires has a positive, but not necessarily 1:1 linear relationship with the weight loading on the them. Picture a Z with 1000lbs of lead directly on top of or in front of the front axle. It would require no more force for this Z to spin its tires than a standard Z, and thus would be no harder on the drive train. A Z with 1000lbs directly on the rear axle would have significantly more traction as well as a higher traction to overall weight ratio. As a result the required force to spin the tires would be higher and thus it would be harder on the driveline. just food for thought.... Where, am I talking about tires helix? I am talking MASS, thats always the varible I am talking about. In chronological order: You take a car that weighs 10,000 lbs, with 500 torque, and get sticky tires... Take and tie your car to a tree, and if you could have 4 wheel drive and 100% traction... Take a 2000lb z to the track with slicks... Dave MASS IS THE ONLY VARIBLE, your freaking nuts man, Look up what a variable means PLEASE, DAMN. Haha.. ok. Meh.. I cbf retyping it when zero summed it up perfectly. Dave Quote Link to comment Share on other sites More sharing options...
X64v Posted November 28, 2007 Author Share Posted November 28, 2007 And my next reply: MASS IS THE ONLY VARIBLE, your freaking nuts man, Look up what a variable means PLEASE, DAMN. If "MASS IS THE ONLY VARIBLE", and we "Look up what a variable means", then by definition the 1000lb car and 10,000lb car have to accelerate down the track at the same rate. Are you saying they do? Quote Link to comment Share on other sites More sharing options...
Challenger Posted November 28, 2007 Share Posted November 28, 2007 Another way to think of it is how hard would it be to break internal parts If you were to launch a car on jacks. (wheels off ground) compared to on its wheels. When jacked up it weighs nothing. Quote Link to comment Share on other sites More sharing options...
rudypoochris Posted November 28, 2007 Share Posted November 28, 2007 Pretty sure things on heavier cars break easier since there is: 1. More traction 2. Less loss of power through inertia (due to slower acceleration), but that doesn't really matter when dropping the clutch. F=MA simply proves that acceleration will be less with a greater mass. Less acceleration does mean less loss of force through inertia in the drivetrain, but as stated earlier, not significant if your talking about breakage right when you drop the clutch. Breakage after the clutch drop though would be higher. Quote Link to comment Share on other sites More sharing options...
Pop N Wood Posted November 28, 2007 Share Posted November 28, 2007 Every body is talking past each other. WTF are you trying to argue? State a contention first. Quote Link to comment Share on other sites More sharing options...
ihatejoefitz Posted November 28, 2007 Share Posted November 28, 2007 Every body is talking past each other. WTF are you trying to argue? State a contention first. And break the first, most important law of internet arguing? Quote Link to comment Share on other sites More sharing options...
johnc Posted November 29, 2007 Share Posted November 29, 2007 F= MA has time in the calculation which is not appropriate for this discussion. You should be discussing Loads, not Force. Quote Link to comment Share on other sites More sharing options...
woldson Posted November 29, 2007 Share Posted November 29, 2007 The way i see it as long as you keep tracktion the entire drive train will see the full force of the torque being applied. The only diff will be accelleration and the amount of time it is being applied based on a static distance. If this was not true than the same drive train in a lighter rig will not have the same amount of torque being applied down a 1/4 mile track. Might ask yourself where did the force go? After reading over the years here, it seems that the drive train parts need to match the amount of hp and tq no matter what veihical it gose in. !If hooking up at all times! So i go up stairs and take a shower and got to thinking more. The law of phyics stats that a object in for every action there is an equal and opposite reaction. So in essence if an object 10000 tons or an ounce they will accerate and move at the same rate. When opposing forces are applied, i.e. gravity, accelleration and rate of speed will very. In both seniros the amout of force applied was constance, where as the they differed was the recation of the object's, acceration and speed. If iam all washed up, please advise!!!!hehe. Quote Link to comment Share on other sites More sharing options...
(goldfish) Posted November 30, 2007 Share Posted November 30, 2007 There is shock loading to think about also. Totally different that the load of say a mustang dyno. Quote Link to comment Share on other sites More sharing options...
Tony D Posted November 30, 2007 Share Posted November 30, 2007 There is shock loading to think about also. Totally different that the load of say a mustang dyno. How so? Set up a 100% Grade, and do a 7000 rpm holeshot with the car strapped to the Mustang, and I lay money something will give, and my thought is it will not be the drum or 5" Driveshaft on the Dyno's Braking Mechanisim! Quote Link to comment Share on other sites More sharing options...
Administrators BRAAP Posted November 30, 2007 Administrators Share Posted November 30, 2007 How so? Set up a 100% Grade, and do a 7000 rpm holeshot with the car strapped to the Mustang, and I lay money something will give, and my thought is it will not be the drum or 5" Driveshaft on the Dyno's Braking Mechanisim! Yeup! Exactly! Good point Tony. In that example, my money is on car parts scattered across the shop floor… It seems that most of you guys are on the right track and saying the same thing yet getting your panties in a wad over specifics, (myself included)… Power. The rate at which energy is being released vs the ability of the parts in question to absorb the release of said power. If a part fails, then that busted part had obviously experienced more “force” than it was designed to absorb. Shock loading is just a fancy term for, “a lot force being exerted rather quickly”. A part will fail at a given “force” load. If that force load is exceeded, that part will fail. Generally speaking, the drive train in our cars are designed to absorb and transfer considerably more force than our engines produce via the combustion events alone. Now lets make an attempt to exert more force to the tires from a dead stop, i.e. accelerate quickly. In the drag racing community, this scenario is referred to as “launch”. When you launch your car from rest, engine revved up, the engine and all its rotating and reciprocating components are about to release a great deal of stored up energy above and beyond that being produced by the combustion events. The sum of that energy varies with flywheel/clutch mass, diameter, (polar moment), rate at which you side-stepped the clutch pedal, etc. Ever notice how hard your stock L-28 Z car will snap your head back when you dump the clutch at high RPM in first gear from a dead stop? But only for a split second till the energy that was stored in the rotating mass has been absorbed, then the car accelerates with less authority solely on the force of the combustion events. Grab second gear, lots of RPM, dump the clutch again, neck snaps back for a split second,... Do this all over again and you hear a loud BANG, and the car doesn't move any more! The designed torque rating of a part was exceeded, tow bill and parts bill exceeds wifes allowable "stupid tax" limit. Quote Link to comment Share on other sites More sharing options...
Tony D Posted November 30, 2007 Share Posted November 30, 2007 ..."tow bill and parts bill exceeds wifes allowable "stupid tax" limit." Hello' date=' Triple A? I need a tow from Tustin to Riverside. Yeah, my number is blah blah blah... Uh huh, what do you mean I don't have Platinum Plus Service any more? WHO cancelled it? WHEN? No, I'll just drive it home, I was hoping to save the turbo...but maybe I can get it home without using boost much... ______________________________________ Next Call: Honey, why did you cancel the Platinum Plus AAA Service? Oh, it was $100 a year instead of $68 and it was 'wasted money'? You do realize now I will have to drive my car home with a bad turbo, and it might burn up the bearings, it's howling like mad if I boost it. You saving $32 a year in tow service 'we never use' by cancelling it the month BEFORE MSA, will likely cost us $752 for a new turbo! No, I'm not paying for it! YOU cancelled the service, you are going to buy my new turbo, honey! There goes your Emerald Ring this June! (And no lie, that is EXACTLY where the money came from: Emerald Ring Fund! Took her 6 years to get that ring...) Every time she wears that ring, I remind her about how a Turbo as more metal than the ring, so it means more to [i']me[/i]! I digress.... "Stupid Tax" I turned it around on her that year. Take a woman's jewlery out of her grasp and make it clear it was because of something she did...and they will start thinking about 'saving $32 a year'! What I should have done was say "Well, at $32 a year, it's going to be another 20 years before I buy that ring!" But that would have had other consequences.... Quote Link to comment Share on other sites More sharing options...
woldson Posted November 30, 2007 Share Posted November 30, 2007 I understand baap's post and i think most do concerning launch, but what about after. I believe that as you accelerate and change gears stresses on the drive train will change and more power will be lost into different parts of the drivetaian. Soo when and what other parts will break down the track and why? Is there more loose of hp and tq due to increse rotations of parts due to friction? I do know 4th gear dose not actually exist, hence more power to the tires. I know ive seen viedos of failures after launch. Feel free to flame:) Quote Link to comment Share on other sites More sharing options...
Gollum Posted November 30, 2007 Share Posted November 30, 2007 Watching one of the latest top gears from the current season (series 10) I got a really good visual extreme of what we're talking about. Jeremy was testing a car designed by mclaren engineers (can't remember the name of the car) and it's basically a road going F1 car. Weight is measured in negative pounds and it has something like 575 hp out a 3.5 liter V8. He kept having problems driving it at low speeds because the handling was SHOT from the lack of downforce from the air speed. Basically without serious downforce the car doesn't have the mass to give the tires any traction. If you take a block of anything you have near you, (looking at a postit pad here at work) set it on a flat surface and push it. Now put a heavy object on it and push it... harder? duh. The lighter you make a car the stickier and wide the tires need to be to be able to create the same extreme loads on the drivetrain. There simply isn't the traction avaible. As mass changes traction changes. And simply hooking a car up to tow something wouldn't simulate this at all, the weight would have to be ON the tires in order to push DOWN towards the ground. So draw whatever conclusions you want out of this, but I think I've explained my idea of the physics involved in a very communicative and understandable manner. Quote Link to comment Share on other sites More sharing options...
Daeron Posted November 30, 2007 Share Posted November 30, 2007 In my experience, this is the place where shawn just smiles and nods at the tranny guy as he rebuilds my buddies T5 in his camaro, and finally gets him a transmission that doesn't blow up from the way my buddy drives. Thanks for all the illumination on a confusing topic; in the end my knowledge is no more concrete than it was, but I feel confident that I understand the "big picture" of the situation, and it all boils down to what the tranny guy said. Basically, in the real world, dumping the clutch breaks things The way I have always thought of it, to simplify it, is that torque is rated in foot-pounds... and those pounds are intimately linked to the pounds that the power is moving. The specific physics underlying it are beyond my ability to explain, (damn i need School.) but this thread has made me certain that I understand it myself. Tony, Great story. Might not be the best marital advice for all people, but it was amusing to read. Quote Link to comment Share on other sites More sharing options...
Sideways Posted November 30, 2007 Share Posted November 30, 2007 F= MA has time in the calculation which is not appropriate for this discussion.F= MA has time in the calculation which is not appropriate for this discussion Hope you guys dont mind if i chime in with my thoughts- First theres a LOT of things going on, and people need to look at the larger picture as a whole- and not just single details. To elaborate a little theres metal/material fatigue, stress, force, mass, and time- they ALL come into play when were talking about increased mass and why it does what it does to the drivetrain. If you increase the mass, but do not increase the force- The acceleration time lowers. Its now taking longer to accelerate from one rpm to the next- So youre now asking the same gears to take the force of the engine for a longer duration. This is how force, time, and material fatigue come into play on some levels quite easily id imagine. Also with added weight, comes added stress (also a physics term worth looking in to) on all the parts at use. Its not right to say it makes the parts in question stronger when subtracting weight or weaker when adding it, youre just altering how much youre physicaly demding or asking of the materials potential strength. (If you take into consideration the impact-strength of materials, that last bit on the importance of added stress with added weight becomes clearly important in how it effects the risk of breaking youre drivetrain on a clean launch with a heavier car) Small simple example lets compare 2 basic machines, two lever systems. Each uses a 10 foot board, with a 2-inch high fulcrum in the middle, only one is lifting 10 pounds- and one is lifting 100 pounds (both of which are placed on the immediate end of the board). Neither system is phsyicaly stronger then the other, you are however demanding more of one systems arm then the other and run a higher risk of going beyond its yeild strength. Very similar situation when comparing cars with equal engine/drivetrain specifications except the other is heavier- Its not that ones stronger, youre just exerting more stress on the components at work If anyone has any further thoughts of corrections please feel free to do so, this is based merely on my small understanding of the forces at play Quote Link to comment Share on other sites More sharing options...
Top Secret Posted December 5, 2007 Share Posted December 5, 2007 Setup an engine and transmission on a dyno, rev it to 6k and drop the clutch. Setup an engine in a 4000 pound Chevelle with drag radials, and 4 link, load it up to 6k and drop the clutch. Watch all the parts fly all over the place. People need to keep reverse forces in mind. A motor can only push as hard as what is being pushes against it. It is the same idea as why a car moves when the tire pushes against the earth, it is because the earth is pushing back. If the motor puts out 600 foot pounds of torque, the drivetrain is only going to be stressed with 600 foot pounds of torque if there is enough grip to push it back. Increasing the weight of the car is one way of many to increase the "blowback force". Getting a better drag suspension is another. Wider, stickier tires is again another. The reason a 600 foot pound motor in a go-kart does not require as strong of a drive train is because the pavement cannot push back on the drive train with 600 foot pounds of torque. Quote Link to comment Share on other sites More sharing options...
johnc Posted December 5, 2007 Share Posted December 5, 2007 If the motor puts out 600 foot pounds of torque, the drivetrain is only going to be stressed with 600 foot pounds of torque if there is enough grip to push it back. You forgot about gearing. 600 ft. lbs. of torque at the crank. Multiply that by a 3.00 first gear and now you're at 1,800 ft. lbs. at the driveshaft. Multiple that by a 4.11 rear gear and now you're at 4,200+ ft. lbs. at the end of the axle. Quote Link to comment Share on other sites More sharing options...
(goldfish) Posted December 6, 2007 Share Posted December 6, 2007 Huh, maybe I should check up on this more How so? Set up a 100% Grade, and do a 7000 rpm holeshot with the car strapped to the Mustang, and I lay money something will give, and my thought is it will not be the drum or 5" Driveshaft on the Dyno's Braking Mechanisim! True. I was vague in my post. But I hope most people got the point that shock loads from a clutch drop is different than a slow build up. Say driving up an increasing grade road? better Quote Link to comment Share on other sites More sharing options...
Top Secret Posted December 6, 2007 Share Posted December 6, 2007 You forgot about gearing. 600 ft. lbs. of torque at the crank. Multiply that by a 3.00 first gear and now you're at 1,800 ft. lbs. at the driveshaft. Multiple that by a 4.11 rear gear and now you're at 4,200+ ft. lbs. at the end of the axle. A good point. 600 ft pounds x 3.0 first gear x 4.11 rear would net you just shy of 7400 foot pounds of torque! Ofcourse, good luck hooking up 600 ft pounds with a final drive ratio of 12.33 That is a lot of power for the earth to push back on. It is a commonly known fact that the earth is rotated with the torque of just two Chevy big blocks. Quote Link to comment Share on other sites More sharing options...
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