Taylor Race CV joints

[Phil883]

Hi,

Do you know how much torque a Taylor Race CV joint could resist.

It will be for an formula hybrid system and with the electric motor and ICE, we have a torque of 820 Nm to a wheel.

Thank you

[thewoundedsoldier]

Their FSAE Tripod should be sufficient. We used them and custom made the housings from 7075. The stubaxle is a pain, though.

Why so much torque? Trade that for some !SPEED!

[thewoundedsoldier]

BTW...does all of it go to one wheel?? http://fsae.com/groupee_common/emoticons/icon_eek.gif http://fsae.com/groupee_common/emoticons/icon_eek.gif

[Drew Price]

Do you think your tires will be able to react 820NM from a standstill?

Best,
Drew

[thewoundedsoldier]

Originally posted by Drew Price:
Do you think your tires will be able to react 820NM from a standstill?


Formula Hybrid cars are heavy. I saw a couple teams with cars over 800 lbs.!

Assuming a *light* 600 lb. hybrid, this torque would only be about 1.2g.

[Hector]

Originally posted by Drew Price:
Do you think your tires will be able to react 820NM from a standstill?

Does torque exceeding what your tires can handle magically disappear? Anything the tires can't handle goes into the inertia of the wheel and drivetrain components. If your tires can only handle half the torque you're trying to give them, the other half goes to inertia. Still transmitting 100% of your torque.

I talked to a GM engineer/judge at comp this past year and he told me that the difference in torque they measured when you do a clutch drop with tires on the ground vs. tires on wet ice is only 20%, meaning that roughly 80% of your torque is going to inertia, not tires. Granted, their wheels/tires are bigger than ours but still...

[Thrainer]

When I was testing our carbon axles, I broke a TRE FSAE tripod (center part split open; crack started from the spline) at just 500 Nm. It was after about 100'000 cycles, though.
In the static test, the tripods survived 1100 Nm.

So I would strongy advise against using these for 800 Nm.

Regards,
Thomas

[Phil883]

Thank you to all

So, our ICE could give 190 Nm and the electric motor 82 Nm at peak. Also, we have a speed ratio of 4:1 between the motor shatf and the differential. It gives 1088 Nm at the differential and a Torsen T1 could put 75 % of the total torque to one wheel. So a wheel could take 816 Nm.

[scotty young Taylor Race]

Thanks Hector.....

The newest version of the 02002011 tri-pod has a forged center. These dudes are rated at 1700Nm.
I see wheel spin way before this....I would have to say at least half of the hybrid teams have been running them since the beginning of the program.

scotty
Taylor Race.

[Phil883]

Originally posted by scotty young Taylor Race:
Thanks Hector.....

The newest version of the 02002011 tri-pod has a forged center. These dudes are rated at 1700Nm.
I see wheel spin way before this....I would have to say at least half of the hybrid teams have been running them since the beginning of the program.

scotty


Taylor Race.
Do you also have a solution for the halfshaft because the one you sold could only resist to 600 Nm.

[Drew Price]

Originally posted by Hector:

Does torque exceeding what your tires can handle magically disappear?


It doesn't disappear, it just isn't generated.



Best,
Drew

[scotty young Taylor Race]

The axles are 4340 and are rated conservatively
at 625Nm (460Ftlbs).

They where tested to static failier at 1871Nm
( 1380ftlbs)

There is always the option of going to the next larger size axle and tri-pod or CV joint.

scotty
Taylor-race.com

[oz_olly]

Hey Scotty,

What are your next sized tripods and axles rated to? Are they aimed Formula Ford sized cars?

[Kirby]

Originally posted by oz_olly:
Hey Scotty,

What are your next sized tripods and axles rated to? Are they aimed Formula Ford sized cars?

Hey Mate,

They are the formula ford ones. I have one sitting around in the workshop if you want to have a look, I can send you the tripod and a housing.

there is the added benefit that they make replaceable steel inserts for the housings.

[oz_olly]

Hey Kirby,

Hope you got back from Melbourne ok and the Monarchy didn't tax you too hard.

You wouldn't happen to know the diameter of the tripod balls, PCD and diameter of the centre piece would you?

[scotty young Taylor Race]

Close Kirby....

Yes they are the Aluminum housings (94mm ) with the steel liners

But we have made our own gun drilled 30 spline axles.This is the new system we designedfor the DSR/CSR cars.

The package per side is around 6 pounds or so.

The tri pods are rated at 1500 Nm ... Eric form UTA who works for us is doing some axle data.

scotty
Taylor Race Engineering

[rink2563]

oz_olly, check out http://www.taylor-race.com/pdf/ACF319.pdf there are some drawings of the tripods a few pages in.

[Tim.Wright]

Originally posted by Drew Price:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Hector:

Does torque exceeding what your tires can handle magically disappear?

It doesn't disappear, it just isn't generated.

Best,
Drew </div></BLOCKQUOTE>

I'm going to disagree with both of these (though Hector is practically correct). If there is 1000Nm on the rear axle at the moment the tyre looses traction, then you can't argue it isn't generated. The 1000Nm has to be reacted somewhere.

When the tyre torque drops, the complete driveline will start to accelerate. Some torque will go into reacting the contact patch force, the rest will go into accelerating the rest of the driveline. This agrees with what Hector said.

Where my point of view is different is that the residual torque used to accelerate the driveline (e.g. 20% in a low friction condition) is not applied to the complete driveline. The first component in line (e.e. clutch) will see all of the residual torque. The second component (eg gearbox) will see the residual torque minus the torque used to accelerate the first component. The third will see the residual torque minus the acceleration torque of the first and the second components etc until you reach the wheels. The split of the torque depends on the split of the inertias in the system of which the final rotating assembly of wheels, brakes, axles is undoubtedly the biggest so it will see most of the residual torque. Additionally, the final rotating assembly is usually through a substantial gear reduction, so the effecive inertia (if transformed back to the flywheel) will be massive so it will see probably 90% of the torque once you loose traction.

Tim

[Jimmy01]

Hey Tim, I'm going to have to disagree with you on this one:

Additionally, the final rotating assembly is usually through a substantial gear reduction, so the effecive inertia (if transformed back to the flywheel) will be massive so it will see probably 90% of the torque once you loose traction.
The wheel assembly will be spinning much slower than the crankshaft so its effective inertia will be reduced compared to faster spinning components as the acceleration is also scaled down. If you had a 10:1 gear reduction from engine to wheels and the wheel inertia was 10 times that of the crankshaft(it will obviously be significantly more than that - anyone got rough numbers?) then the torque required to accelerate each would be equal.

[Tim.Wright]

The effective inertia does not depend on the speed, only the gear ratio (and it is multiplied by the square of the reduction), but I do agree the torque involved will be less because of the lower speeds.

The more i think about this, the more I realise you dont want to calculate effective inertias back to the flywheel anyway because you loose the important speed part of the equation.

Tim

[JasperC]

Well, it's not THAT hard in my humble opinion...

T = Ia or torque equals inertia times angular acceleration. You can determine I for each component in the driveline, and the angular acceleration of one component is equal to that of another component times the drive ratio between them (you don't need the square of reduction for calculating torques, only when calculating the kinetic energy of a component).

So basically, what Jimmy said is right:


Originally posted by Jimmy01:
If you had a 10:1 gear reduction from engine to wheels and the wheel inertia was 10 times that of the crankshaft(it will obviously be significantly more than that - anyone got rough numbers?) then the torque required to accelerate each would be equal.


Kind regards,

Jasper Coosemans
Chief Drivetrain 2009-2010
DUT Racing Team
Delft University of Technology

[JasperC]

Oh, I realized that I forgot to mention one important thing, which is that of course the torque available at the crankshaft is (in Jimmy's example) only 1 tenth of that at the wheels.

So while the torque required to accelerate the crankshaft may be in the same order of magnitude as the torque required to accelerate the wheels (I don't have the exact figures for inertia of wheels and driveline components, so I'm not entirely sure about this), the larger part of the excess torque produced by the engine will indeed be used to accelerate the crankshaft and other engine parts.

Bottom line is that probably half shafts and tripods will not see much more torque than the tyre can deliver to the track, but it may be a BIT more. http://fsae.com/groupee_common/emoticons/icon_smile.gif Would be very interesting to do some calculations on this to get a decent idea of the figures instead of just guessing at the order of magnitude.

Cheers,
Jasper