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Having troubles quantifying inertia

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LAN_403

The Zander

Tue Apr 10 2007, 03:45AM

Registered Member #295 Joined: Tue Mar 07 2006, 06:59PM
Location:
Posts: 23

Ok, the title may be technically incorrect as inertia is more a concept than a quantity, but it perfectly illustrates my dilemna. i have recently taken an interest in designing PIC-based preformance car electronics (boost controllers mostly) and there is one piece missing in my understanding of automobile mechanics. simply put, why is it easier to break tires loose (over come rubber/asphalt friction with prime mover's torque) when the car is at a standstill than when it is moving. assume that the vehicle has a single fixed gear ratio, and a perfectly flat torque curve.

ex. clutch drop from 5000rpm results in lose of traction, while simply accelerating the vehicle smoothly until revs reach 5000rpm results in maintained traction. the force of friction between the tires and road has not changed, and the engine always outputs the same torque when @ 5000rpm and maximum load.

any help is greatly appreciated.

...

Tue Apr 10 2007, 05:28AM

Registered Member #56 Joined: Thu Feb 09 2006, 05:02AM
Location: Southern Califorina, USA
Posts: 2445

well, with theclutch you would pop the tires loose when you had the whole mass of the engine spinning at 1500rpm, then suddenly hook it up to the tires.

With newer cars, you have a torque converter which can conver the difference in speed between the engine and transmission (tires) into torque, so when you have the engine spinning at 5500 rpm and the tires at 0, you get more torque.

I wouldn't be suprised if it really is just that you can get more torque to the weels at low speed...

Coronafix

Tue Apr 10 2007, 06:52AM

Registered Member #160 Joined: Mon Feb 13 2006, 02:07AM
Location: Melbourne, Australia
Posts: 938

Power = torque x rotational velocity
Applying power slowly will increase torque slowly.
By revving the engine to 5000rpm and then dumping that power into the wheels,
it wants to continue rotating and hence the burning rubber.
Don't know if that's any clearer?

Steve Conner

Tue Apr 10 2007, 09:44AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Yes, it's the energy stored in the rotating parts of the engine that breaks the tyres loose when you dump the clutch. Once they are broken loose, the dynamic coefficient of friction is smaller than the static one, so provided you have a decent amount of power, the wheels are likely to keep spinning until you take your foot off the gas and let them hook up again. Unless you're Sam Barros, in which case they fall off.

On rear-wheel drive cars and motorcycles, you have another effect. As you accelerate, the weight of the car/bike/whatever is transferred back over the rear wheel(s), which makes them grip better. If you dump the clutch suddenly, there is not enough time for this to happen, so the odds of breaking traction are higher still. Drag racers spend a lot of time practicing the perfect launch, with almost all of the vehicle's weight on the rear wheels and the rear tyres right at the limit of grip. (Or these days, get launch control firmware to do it for them...)

Dr. Slack

Tue Apr 10 2007, 01:34PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Your engine's torque rating is a maximum continuous rating. It does not output a constant torque, it outputs whatever torque is demanded until something gives (much as a variac outputs as much current as the load demands, until something gives).

If the engine is at a steady 5000rpm, and you steadily increase the load, into (say) a water-cooled brake, then as you exceed the maximum torque, the engine will slow or stall from lack of bang behind the pistons.

If the engine is at 5000rpm, and you rapidly connect it to a stationary vehicle, via wheels, clutch, gearbox drivetrain etc, then the effect will be to try to bring the engine and the vehicle to the same speed. It requires a huge amount of torque to bring the engine down to road speed, or fling the vehicle up to engine speed, and that torque generally exceeds the wheel/road friction interface, which slips. However, the clutch might be the lowest torque withstanding link, which would then slip. It turned out that in my Rover 1400 of a few years ago, the lowest torque-capable part of the drive chain was the teeth on my first gear pinion. The shock to the wallet changed my driving style, and I now use the clutch more as a linear control than a switch.

Simon

Wed Apr 11 2007, 01:41AM

Registered Member #32 Joined: Sat Feb 04 2006, 08:58AM
Location: Australia
Posts: 549

The_Zander wrote ...

the force of friction between the tires and road has not changed

It has.

The Zander

Thu Apr 12 2007, 09:25PM

Registered Member #295 Joined: Tue Mar 07 2006, 06:59PM
Location:
Posts: 23

RE: ...

i agree, except that torque converters so far as i've seen appear only in automatic cars..

RE: Goldsphere

What your getting at it that the rotational inertia of the engine resists the change in velocity applied by the sudden application of the clutch and thus excerts more power @ the wheels. concrete, but i still don't know how to quantify that added energy.

RE: Steve C

thank you for clarifying goldspheres response, and adding the mass-component redistribution comment

RE: Neil thomas

Firstly, i dont entirely conccur with your assertation that the engine outputs as much torque as the load requires. that would imply that, for example i am cruising along the highway and i depress the gas pedal fully, the engine under this condition outputs "X" lb.ft of torque. Now imagine i am climbing a steep hill and i depress the gas fully, the engine outputs "Y" torque. by your analysis X does not equal Y. when in fact it must. the gas pedal is mechanically (or at least virtually linked in drive-by-wire cars) to the throttle-body, the device that controls airflow into the engine. conversly if the pedal is in the same position in the two above scenarios, so must the throttle body, and thus the same amount of gasoline is being burnt, resulting in the same power. so really the engine outputs as much torque as the pedal controlling airflow requires, or @least that is how i see it.

Secondly you said "It requires a huge amount of torque to bring the engine down to road speed", that is exactly the torque that i am trying to quantify!, i presume the effective torque at this point would be engine torque + "inertial torque". or somthing like that ....

RE: Simon

i appologize for the poor wording, i will clarify

BEFORE the tires slip the force of friction (static coefficient) in both the clutch drop, and gentle clutch are the same. imagine the free-body diagram of the wheel road interface, in both scenarios there would be an arrow indicating the force of the engines torque pointing backwards in relation to the cars movment. and an arrow representing the force of friction opposing it. it would seem that i then have to identical FBDs with different results. an impossibility, solved by the addition of the force of "interial torque" augmenting the force of normal torque in the clutch drop scenario.

--------------------------------------- -------------------
now that's out of the way, i can pose the remainder of my question, how do i quantify the magnitude of the additional torque imposed by the inertia of the spinning engine components. the force that causes wheel spin when dropping the clutch but not when easing it out.

Thanx
-TheZander

...

Thu Apr 12 2007, 10:45PM

Registered Member #56 Joined: Thu Feb 09 2006, 05:02AM
Location: Southern Califorina, USA
Posts: 2445

The extra energy comes from all of the mass in the engine spinning at 5k rpm. To quantify that you would need to do some pretty ugly math, add up the energy stored in all of the moving components in your engine... The torque you would get is (at first) limited by whatever the clutch will give you before it slips. You could probably measure this experimentally by like spinning the engine up to full speed, then take your foot off the gas and engage the clutch--and see how far the car roles...

Also, just because a throttle body is full open does not mean that you are getting the max power out of an engine. If you have soo much load on the engine so it can only spin at 1krpm you will only get a tiny fraction of the power you could get out if you geared it down so it was spinning at 5krpm (or whatever it runs at for max power). I am going to assume you can figure out why this is...

Dr. Slack

Fri Apr 13 2007, 07:45AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Firstly, i dont entirely conccur with your assertation that the engine outputs as much torque as the load requires.

That's your perogative, but, until you do, you won't understand what is going on. Read what I said carefully, "as much torque as the load requires UNTIL SOMETHING GIVES". I explained what gave in the situation you outlined. If both X and Y are below the maximum torque that the engine can produce BY BURNING GAS, then what you say is true. If the hill is steep enough and Y>max gas burning torque, then the engine will output Y for a while, part from the pressure of the burn on the pistons, part from the torque required to slow the engine down.

As this is (mainly) an electrical forum, I am going to use quite an electrical analogy here, it will allow you to use a simulator such spice, simmetrix, LtCAD to simulate what is going on. For rotational speed w read voltage, torque TQ is current, moment of inertia MOI is capactitance, spring constant is inductance, angular momentum is charge, energy is energy (good old energy!) = .5CV2 = .5MOIw2 = TQ x angle, all the quantities and their conversion are exact. For a clutch which can slip at a defined torque, use a constant current biassed bipolar or a constant voltage biassed FET

Consider very carefully what happens when you drop the clutch at 5000rpm. Assume by dropping the clutch that the full spring pressure goes instantaneously onto the plates. The fast engine (charged capacitor, rotational speed w = voltage, energy = .5CV2 = .5Iw2 to represent the spinning mass) connected to a current limited power supply to represent the torque available from burning gas, is then connected to stationary wheels (uncharged capacitor) by two principle components, the torque (current) limiting clutch, drop the clutch by applying base current or drain volts, and the springyness (inductance) of all the transmission components. For a simpler simulation, leave out the inductance

I'm not going to describe the whole thing here, you can set up the simulation and play with it. Just notice that the current flowing between a charged and uncharged capacitor can reach very high levels, just as the torque available from the engine can be very large when required to slow down rapadily. The clutch limits the torque to the wheels. The wheels limit the toque to the road. If you could fit a really serious clutch and drop it instantly, with the transmission lock on, you could slow the engine down a lot more rapdily. Eventually, you could try to slow the engine down so rapidly that the crankshaft sheared while the pistons continued to move. As ever, the torque the engine will supply is a) whatever it is asked for by the load until b) something gives.

The "huge amount of torque you are looking for" is the limiting torque of the clutch or the road wheels, and has nothing to do with the engine. It doesn't matter whether you have a 2 liter commuter or a 10 liter Hummer, and leave the throttle down or blip the throttle off as you drop the clutch, once the clutch slips or the wheels break away, the torque is limited by the mechanism that "gives".

stop4stuff

Fri Apr 13 2007, 09:46AM

Registered Member #64 Joined: Thu Feb 09 2006, 06:25AM
Location: Southampton, UK
Posts: 68

Energy from the engine is transferred more directly to the tyres through the drivetrain when the clutch is dumped causing the tyres to spin... & the vehicle doesn't go anywhere.

When the clutch is eased out the clutch slips, the gearbox side of the clutch assembly spins up more slowly & energy is lost as heat in the clutch assembly... & the vehicle pulls away.

So I think the answer would be that although the engine developes the same amount of torque in both situations, in the latter case, energy is converted to heat & energy going through the drive train is less than the engine is putting out.

Engine_torque - clutch_heat_loss = torque_available _to_drivetrain

paul

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