Torque calc. at rear shaft

[geniusanurag]

I was trying to find out the torque that would be acting on the rear shaft but got confused that what affect would torsen differential would bring on to it. If the maximum engine's torque is x at the differential(after multiplying with gear ratios), would it become 2.8*x/3.8 (2.8 is the TBR of torsen) on the shaft whose wheel has more traction.
Or, as torsen redistributes torque only during cornering and maximum torque rpm would not be reached during cornering, so the rear shaft can be designed with x/2 torque.

[geniusanurag]

I was trying to find out the torque that would be acting on the rear shaft but got confused that what affect would torsen differential would bring on to it. If the maximum engine's torque is x at the differential(after multiplying with gear ratios), would it become 2.8*x/3.8 (2.8 is the TBR of torsen) on the shaft whose wheel has more traction.
Or, as torsen redistributes torque only during cornering and maximum torque rpm would not be reached during cornering, so the rear shaft can be designed with x/2 torque.

[absolutepressure]

I'm not exactly sure what you're asking. Are you saying that since you have two half-shafts, each half-shaft only carries half of the engine's load? If that's what you are thinking, sorry, but it doesn't work that way. Let's talk about an open diff for right now, I'll let someone else explain how the torsen applies more useful force to one side than the other, because they can probably explain it better than I can. But anyway, I know in an open diff setup, the diff applies equal amounts of force to each tire at all times.

That is to say, if one wheel is on ice, and only providing a useful 5lb/ft of torque, the wheel on dry pavement is also getting supplied with only 5lb/ft of torque, even though it has traction.

More along the lines of what you're asking though, let's say you just had a sprocket bolted up to a solid shaft. The forces on either end of the shaft are obviously going to be the same, since it's solid. Now, just because you put a couple gears and shit inbetween them, that doesn't change anything since the gears are still solid objects. It just lets one side spin faster than the other since the gears can rotate with respect to eachother, but like I said, each side has the same amount of torque, which is equal to the engine output at that point in time.

Also, picture a torsen with one wheel on the ice. Let's say the ice is perfectly smooth, and has no friction. The torsen is going to lock itself up (somehow) and apply all of the engine's torque to the gripping wheel, which would mean each half-shaft needs be able to handle the entire engine load. (since the roles can be reversed and the other side will get all of the engine's torque)

I hope I answered your question, and I hope I'm not wrong, I'm the engine guy on our team. What I just said is pretty much stuff I remember from "howstuffworks.com"

[FryGuy]

Technically, the tires are the limiting factor here. The rear drive shafts should be designed based on the maximum grip of the tires rather than the the engine torque.

However, if you don't have tire data then the engine torque is probably a close approzimation but be aware that you are not really getting the whole picture.

Also be aware that if you are braking off of the differential (or some other type of in-board brakes), it is possible that your braking torque might exceed you engine torque and should be used instead.

With that being said do like absolutepressure said and assume each driveshaft is capable of taking on the full torque and design it accordingly. Its also better to overdesign it than underdesign it and have it break on you.

[Conor]

Calculate the torque as a result of the friction force on the tire using the highest coefficient of friction available for that tire, apply an appropriate safety factor and you should have a pretty good approximation. Don't over think it with the differential, you want the absolute worse case scenario for that shaft, so not considering a reduction from the differential will add another appropriate safety factor. Also, don't forget that weight shifts when you accelerate a vehicle so that has to be considered in the normal force on each tire that you're looking at. These are only my suggestions, but I think they'll give you a good approximation with the easiest math and shortest calculations.

[SR-Mike]

Fry Guy and Connor,

Yes, you are correct to say that the tyres do effect the magnitude of the torque in the drivetrain, be that the driveshafts, diff, whatever. But the Maximum frictional force is definitely not the factor that governs the maximum forces seen by these components.

As i have said on this forum before, you cannot assume the drivetrain has no mass and hence no moment of inertia. Think about it, if you lift the car off the ground and accelerate the driven wheels there is still some torque being transferred through the drivetrain. Where as if the limiting factor was tyre grip as you've said the torque should be zero, shouldn't it?? you tell me.

And Connor, Nice pic ;-)

Cheers
Mike

Strathclyde Uni Alumni 02-06

[Rex]

I agree with Mike - while the tire friction estimate should get you in the right ballpark, spinning up your wheels, hubs, etc shouldn't be ignored unless proven insignificant through calculations.

Also when considering that weight shifts when you accelerate (per Connor's post), consider also that you could be turning and accelerating. Many half shaft failures happen during straight-ahead launches, but many more happen at corner exits.

Connor, the Lotus icon is great - I am always amazed how many of ACBC's ideas are words to live by in the FSAE environment. Do you drive one? Or have some other type of affiliation with them?

[geniusanurag]

[QUOTE]Originally posted by Rex:
I agree with Mike - while the tire friction estimate should get you in the right ballpark, spinning up your wheels, hubs, etc shouldn't be ignored unless proven insignificant through calculations.

Also when considering that weight shifts when you accelerate (per Connor's post), consider also that you could be turning and accelerating. Many half shaft failures happen during straight-ahead launches, but many more happen at corner exits.

QUOTE]

I have calculated the torque taking into acount the frictional grip of the tyre and the rear weight shift during accleration, but still they are less than the maximum amount of torque that the engine is giving at its peak even after taking into account the friction losses in the drivetrain. Where would this extra torque go??

[FryGuy]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content"> I have calculated the torque taking into acount the frictional grip of the tyre and the rear weight shift during accleration, but still they are less than the maximum amount of torque that the engine is giving at its peak even after taking into account the friction losses in the drivetrain. Where would this extra torque go?? </div></BLOCKQUOTE>

Where does the extra torque go?
Wheel spin. If your engine torque exceeds the friction force of the tire you will go into excess slip and you'll sit there spinning your tires.

Therefore the maximum torque available is limited by the frictional force. I disagree with SR-Mike on this. While i agree that inertia should be taken into account, i don't think the inertia contributes enough of an extra load to be of concern. The longitudinal force of the tire falls off greatly after a certain point (wheelspin) and i highly doubt the extra inertia as a result of the increased rpm is anywhere near the amount of torque lost to slippage.

Now i havent done the inertia calculations because im unsure of the rpm and angular acceleration at that point. But this might start off some good conversation as im trying to avoid studying for finals http://fsae.com/groupee_common/emoticons/icon_smile.gif

[Steve Yao]

FryGuy, do the calculations.

Absolutepressure, A torsen with one wheel on ice will behave no differently than an open diff. The torsen requires *some* amount of opposing torque from the unlaiden wheel to generate its torque bias.

[Boston]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by FryGuy:
I disagree with SR-Mike on this. While i agree that inertia should be taken into account, i don't think the inertia contributes enough of an extra load to be of concern. </div></BLOCKQUOTE>

Impact loading under shifting and clutch drops (due to inertia) are absolutly significant. Unless you have the ability to perform strain gauge tests on the shafts, you will need to apply a generous saftey factor on top of the maximum frictional force.

[The Bunker]

I agree with Boston, The sudden impact from jamming down a gear on the track or a hard launch assuming decent pavement conditions will be the most likeley time of shaft failure. Even if your tires slip some, the down shifting can sometimes cause a hopping/chattering in the drivetrain that wreck components! Whether its a chain or a shaft, it will find the weekest link in your driveline. Just my 2 cents from what I've seen.

[Welfares]

SEY, you are asolutely correct about the torsen, if you lift a wheel you lose drive, we were having trouble with this on the skid pad, when we tried to explain this phenomenon to the judges they commented that we "did not understand the operation of a torsen differential"

I like to describe our new "diff" as "infinately biasing".

On the subject at hand, we used a fat car and excessive weight transfer in the calculations without including inertia and we havn't bent anything yet, well, nothing we did calcs for anyway.

[SR-Mike]

I think some people are almost there but most are missing some important things.

Yes FryGuy,
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content"> If your engine torque exceeds the friction force of the tire you will go into excess slip and you'll sit there spinning your tires. </div></BLOCKQUOTE>
But this is exactly why forces due to the inertia of the system are vitally important. The engine just doesn't just stop producing torque when the wheel starts to spin, instead what happens is that where before this engine torque had been used to accelerate both the drivetrain (inc wheels tyres etc) and the car, the majority of the torque is now being used to accelerate only the drivetrain. So the torque is always there acting somewhere throughout the drivtrain system. Where exactly most of it acts is dependant on the ratio's of inertias between components and gear ratio's etc but it seems a logical assumption that the wheels/tyres have a relatively high moment of inertia compared to the other components in the system; hence the majority of the torque will still be transferred through the chain, diff, driveshafts and hubs.

So without even considering shock loading all components should at least be designed to withstand the forced generated by the maximum torque output of your engine.

Shock loads again can be explained by remembering that all components in the system have mass. Say you've got a very loose chain which allows the engine sprocket to turn marginally before the chain is taught. the engine will in this case have a a revolution or so to build up some angular momentum while under minimal load, then as the chain becomes tense you basically get a collision between the spinning engine/gearbox and the stationary diff which also has mass and the result is a rapid transfer of angular momentum in the form of a big force down the chain. This example can be likened to many different cases including bad gear changes or even bumps seen by the driven wheels.

If you have values for all the moments of inertia of all the rotating components on your car, it should be possible to calculate the lot.

Hope this helps.
Mike

[Conor]

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Rex:
I agree with Mike - while the tire friction estimate should get you in the right ballpark, spinning up your wheels, hubs, etc shouldn't be ignored unless proven insignificant through calculations.

Also when considering that weight shifts when you accelerate (per Connor's post), consider also that you could be turning and accelerating. Many half shaft failures happen during straight-ahead launches, but many more happen at corner exits.

Connor, the Lotus icon is great - I am always amazed how many of ACBC's ideas are words to live by in the FSAE environment. Do you drive one? Or have some other type of affiliation with them? </div></BLOCKQUOTE>

Thank you, my one big dream in life is to own an Exige, and I'll be damned if I never do.

As for the rest of the thread:

I'm not disagreeing at all with what you guys have to say about calculations, but sometimes the best thing is to do your calculations the best you can, and then spend a ton of time testing. You can calculate until you're blue in the face, but I don't care what you say, no calculation can garuantee the failure or survival of a part in real world situations. There's too many variables to consider. Do your best, may an extra pair of shafts if need be, and then beat the piss out of what you got. That's the best way to figure it out.

[Rex]

The Exiges are great, though a bit small for my taste. My wife wants one. There's nothing that drives quite like a Lotus - I drive a 94 Esprit. Just be sure to budget in some maintenance expense when you get one!

Point well made about making an extra set of half shafts. Given the setup time and effort involved, I would always make a spare set. If the first set doesn't break you've got a proven set for use next year.

[Steve Yao]

...but if the first set does break then you have an extra set of shafts you know will fail. So it becomes necessary to identify the mode of failure.
If it was improperly sized/designed then you don't want to put the same part right back on the car and have the next failure possibly be catastrophic.

[John Valerio]

...why would you make your backup drive shafts (or any backup/replacement part for that matter) the same dimensions as the original? as you said that kind of defeats the purpose.

[Rex]

I was thinking along the lines of waiting to heat treat the second set, and trying a different heat treatment if the first set broke.

[SR-Mike]

Sufficient testing is the most significant step in the production of any FSAE car from my point of view, but it can't replace design completely. And if you have don't know what loads your components are required to take it doesn't rank as engineering design to me, and it will be treated as such by the design judges.

And the likelihood is that if you make 2 you'll only end up breaking 2 if they haven't been designed properly.

And Connor, Its worse not owning one when you have to drive past hundreds of them on the way into work!!!

Mike

[Conor]

I was implying to make an extra set after the first broke, if it did, but I worded the sentence very poorly and the construction misguided you, as the reader, to interpret my idea as making two sets of the same shaft. I apologize and you guys are correct, making two sets of the same shaft so they both would fail would be pretty useless unless, once the failure occurred, the other set could be easily modified to prevent another failure (like Rex said). Either way, test, test, test... That way you'll really know if the shaft will fail once it stands up to 40 repetitive clutch drops. And as for the Lotus Elise, I've always been very compelled by the vehicle because of the philosphy used to design it. The car is all about amazing handling through weight savings and simplicity. It's a complete testament to the KISS rule. And the fact that such performance comes in an afforable package also makes it very appealing. The pleasure of driving shouldn't be a thing that's reserved for people with very deep pockets.