Nothing to do with RPMs....it is a torque issue. Also has a lot to do with the weight of the car.

 

I don't think you understand how reductions this question sounds. It's like asking "how high do I need to jump so the Southwest flight to Chicago arrives 10 minutes early?"

The answer is 42.

 

18. Your variable in the invert is simply A*(t*2), not A*(b+(t*2)). Simple mistake.

 

You need to go to geeks.com's forum...... we're just building cars and busting knuckles here.

I'm guessing the answer will look more like an algebra problem on an old SAT than "2,455 RPMs".

 

Thank you for all the replies. I am humored by them. None of you k ows the answer so instead you resort to such comments. If you guys look over at the topic in yhis forum "How much torque is needed to lift the front end of a car" Willy made a nice post and that is the reason why I posted here. I am actually waiting for a reply of him and I put it publicly so everyone can read and enjoy it.

 

"None of you k ows the answer so instead you resort to such comments"

That should help

 

Ridiculing answers defenitely did help.

 

You are correct, you deserve a straightforward answer....

Question: "Is it possible to roughly calculate the RPM needed to spin car tires with just the following info?".

Answer: I don't think there is enough info in your postulate as presented, but there may be someone here who will surprise us all. Of course he could just pull a number out of his butt and we'd be clueless as to whether it was correct or not.

 

What's missing are the gear ratios for the transmission and axle drives and the engine’s power curve.

Making assumptions of 2.20 low gear and 3.55 ratio in the axle or drive splitter I’ll solve for the known torque at 2500RPM.

This breaks into four problems:

One is what the coefficient of friction is for a tire.

Second is the force of torque the engine develops, the only known number is 680 NM at 2500 RPM. I'm converting this and the vehicle weight to English measures for my ease.

Third is the torque available at the axle through the gear ratios at that engine RPM.

Fourth is the effect of tire size on lever arm between the axle center and the tire perimeter. Surprisingly tire width has little to do with longitudinal spin up.

1) Roughly the coefficient of friction for a tire is about .9. The vehicle weighs 1665 Kg which is 3670 pounds US. Divided by 4 that puts 917 pounds on each tire assuming a 50-50 weight distribution front to rear and side to side. 917 pounds times .9 is 825 pounds of available traction per tire. For 2 tires that's 1650 per axle, for 2 axles or 4 tires that's 3300 pounds.

2) 680 NM is 501 pound feet of torque at 2500 RPM at the crankshaft. Given this is the only know data point for torque I'll solve for this.

3) Torque is multiplied by the gear ratios; the low of 2.20 in the transmission increases the torque to 1102 pound feet on the driveshaft. The 3.55 axle gear reduction increases the torque to 3912 pound feet.

4) For a two wheel drive on an axle the entire 3912 pound feet are available and are divided by the powered tires or 1956 pound feet per tire. A 20 inch diameter tire has a lever arm of 1.66 ft/2 equals .833 ft. This times the 825 pounds of available traction equals 687.2 pound feet per tire or 1374.4 pound feet for the axle. This would spin the tires quite easily.

5) The All Wheel Drive divides the axle ratio torque across two axles and 4 tires. That is 3912 pound feet divided by two axles or 1956 pound feet of torque per axle. The tire traction doesn't change so with driving torque of 978 pound feet per tire and 687.2 pound feet of traction available the AWD solution still says the tires can be spun with the engine at 2500 RPM but not with the vigor of a 2 wheel drive.

Phew my brain is tired! Hopefully I got the math right, I'm better at the concept than the calculations.

Bogie

 

Wow! Bogie! That was some amazing input! I'm still reading through this and digesting it.

 

They will spin at 1 RPM in a vacuum.
Did you really expect an answer to this. There are so many variables that come into play that are not even related to the vehicle.
As a side note, I don't think RPMs would be one of the variables.

 

I've always thought that oldbogie was one of the smartest people on the site. Now I'm wondering if he's just been fooling me all along. LOL!

 

Bogie has the correct answer. It isn't a matter of rpm, it's again a matter of torque to overcome the friction between the pavement and the tire patch on the road. No matter how fast your car is going, the speed of the tire patch in contact with the road surface is zero - unless you apply enough torque that overpowers the friction locking the tire to the road. if the engine can generate the proper torque at that speed, you can bust the tires loose and spin them. Top fuel dragsters lay down a huge tire print so can transfer huge amounts of torque throgh the tires to the road without breaking traction. Even then those monsters can generate enough torque at 300mph to light up the tires. Amazing machines!

Similarly if the coefficient of friction is lower, you can bust the tires loose with less torque. Best place to guarantee lighting up your tires is on ice! But I will never know because I refuse to live where, if you don't do eveything right, you die from the weather.

The other parameter you can work with is gear ratio. An engine in a car with rear end gears @ 2.5:1 must put out twice the torque as a car with 5:1 gears to break the tires free.

In transportation power, it's all about torque.

 

So, in order to get the RPM, which I was looking for, I would apply this formula?
(Power and Torque: Understanding the Relationship Between the Two, by EPI Inc.)

HP = Torque x RPM ÷ 5252

And solve it for RPM:

RPM = (HP / Torque) * 5252


Breaking it down again:

1) the force needed to break traction is:
- 825 pounds per tire
- 1650 pounds for RWD
- 3300 pounds for AWD

2) 3912 pound feet are available, but we don't necessarily need all. So I'll just use the pounds above as the force/torque necessary (if that would be correct)

Applying the formula above:

RWD RPM: 550 HP / 1650 pounds feet * 5252 = 1750 RPM? Looks a bit low to be honest.
AWD RPM: 550 HP / 3300 pounds feet * 5252 = 875 RPM??

Did I calculate that correctly?

 

Doesn't work like that. You are using peak engine HP numbers the occur at very high rpm and peak torque numbers that occur somewhere high-mid rpm range. The math may work out but the physics don't since you are making some wrong assumptions. At those low rpm your torque is going to be WAY lower than the numbers you show. And you must consider the torque at each tire, not total torque. Your calculation says a 4-wheel drive car can break the tires loose easier than a 2-wheel drive car which is intuitively and actually incorrect. Think about it. The total engine torque is being distributed to 4 tires in the former case as opposed to two tires in the latter, thus less chance of breaking traction at any one tire.

Again you shouldn't be looking for a magic rpm to spin the tires. You need to be looking at the torque curve to see where along the increasing rpm of an engine you might have enough torque to do the deed; for the avilable traction (foot print, coefficient of friction, etc.) at the time. What you need in general is an engine too powerful at low rpm for the weight of the car, and/or low gear reduction to multiply the vailble engine torque to overpower the tire traction at one or more rpm values, and/or a camshaft that yields low speed torque which necessarily reduces the sexy top hprsepower rating of the engine which occurs at the top end of the RPM range.

The top fuel dragsters I mentioned above takes all of these parameters to ridiculous extremes; way too light a car; WAY too powerful an engine, etc. Its power curve shows it is capable of frying its tires at any conceivable speed. The only way they avoid it is to cover the starting line with glue to artificially increase the coefficient of friction then feather the application of the clutch (i.e., burn the clutch to a crisp every run) to be just below the break-loose torque application all the way down the track. They are successful some times, go up in smoke a lot of the time. They must be on that ragged edge all the time or they will not be competitive. A perfect drag racing run would be to apply as much of the available 10,000hp to the pavement through a slick that was not deformed, not shaking, and had maximum contact footprint with the track, and front tires that never touched the track.

 

Thanks Willys. I was hoping not to look at torque curves and the like getting into the nitty gritty details and was hoping to mark RPM "lines" in the cockpit.

There is a complete car database from the german ADAC, which would give me just about most information of each car. It would have been interesting for me to compare RWD Diesel engines, which have lower RPM, but sometimes much higher torque with 700 nm, if you look at a factory BMW Alpina D3 or D4, which would outperform factory BMW M's by now in acceleration 0-60 mph.

 

In the event that this isn't just an exercise on paper, barometric pressure, temperature and humidity will factor in. Not only to the physics of traction but also affecting the engine's peak performance. Also, are all four wheels routing through lock-up diffs and won't they contribute their own variables to the equation as they slip and stick?

 

"I would like to calculate (very roughly!) how much RPM I would need to start spinning tires in a RWD or AWD car

Assume the following: the car is stationary, in 1st gear, the tires are cold, neither flat or overpressurised ie "normal", and assume I have enough torque. Neither suspension, nor mixture of the tires nor their profiles should be taken into consideration."

The answer to a question like this is much more easily found by trial and error, as there are just too many variables.

Sounds like maybe a manual trans? If so, i don't see anyone considering the biggest variable (which actually does have something to do with "how much RPM")...the amount of stored energy in the already spinning crankshaft/flywheel assy. With a heavy enough flywheel and aggressive enough clutch, you could have enough stored energy to be able to start spinning the tires at under 1000rpm even if the engine died when the clutch was dumped

 

"A 20 inch diameter tire has a lever arm of 1.66 ft/2 equals .833 ft."

I don't see where the 20" "fit in". What would this show for a 21" 19" or 18" tire?



I just realized I pulled the wrong numbers and Bogie nicely provided them:

"4) For a two wheel drive on an axle the entire 3912 pound feet are available and are divided by the powered tires or 1956 pound feet per tire. A 20 inch diameter tire has a lever arm of 1.66 ft/2 equals .833 ft. This times the 825 pounds of available traction equals 687.2 pound feet per tire or 1374.4 pound feet for the axle. This would spin the tires quite easily.

5) The All Wheel Drive divides the axle ratio torque across two axles and 4 tires. That is 3912 pound feet divided by two axles or 1956 pound feet of torque per axle. The tire traction doesn't change so with driving torque of 978 pound feet per tire and 687.2 pound feet of traction available the AWD solution still says the tires can be spun with the engine at 2500 RPM but not with the vigor of a 2 wheel drive."


RWD RPM: 550 HP / 1374.4 pound feet * 5252 = 2101 RPM
AWD RPM: 550 HP / 687.2 pound feet * 5252 = 4203 RPM

This looks better in my opinion. To be honest I testdrove this RWD beast to which I provided the numbers and the tires spun quite easily. Hence I was curious to find out numbers, because I was more focused balancing the car than looking at the RPM

 

Second approach would be:

RWD:
2500 RPM / 3912 pound feet * 1374.4 pound feet = 878 RPM
2500 RPM / 1956 pound feet * 687.2 pound feet = 878 RPM


AWD:
2500 RPM / 978 pound feet * 687.2 pound feet = 1756 RPM

Don't know which approach is correct. The 1st one is more realistic in terms of RPM than the 2nd.

 

How much torque does your engine make at 2500 RPM?

 

My 85 horse power Case skid steer can break all 4 tires loose when I push something thats not wanting to move,,,but if I full power just launch the machine it will wheelie for a foot or two then its at top speed.Once the tires lose grip the formula changes,or at least the coeficient of traction changes

 

This thread is starting to sound like that swallows and coconut conversation in Monty Python and the Holy Grail.


King Arthur: The swallow may fly south with the sun or the house martin or the plover may seek warmer climes in winter, yet these are not strangers to our land?

1st soldier with a keen interest in birds: Are you suggesting coconuts migrate?

King Arthur: Not at all. They could be carried.

1st soldier with a keen interest in birds: What? A swallow carrying a coconut?

King Arthur: It could grip it by the husk!

1st soldier with a keen interest in birds: It's not a question of where he grips it! It's a simple question of weight ratios! A five ounce bird could not carry a one pound coconut.

King Arthur: Well, it doesn't matter. Will you go and tell your master that Arthur from the Court of Camelot is here?

1st soldier with a keen interest in birds: Listen. In order to maintain air-speed velocity, a swallow needs to beat its wings forty-three times every second, right?

King Arthur: Please!

1st soldier with a keen interest in birds: Am I right?

King Arthur: I'm not interested!

Second Swallow-Savvy Guard: It could be carried by an African swallow.

King Arthur: Will you ask your master if he wants to join my court at Camelot?

1st soldier with a keen interest in birds: Oh yeah, an African swallow, maybe, but not a European swallow. That's my point.

Second Swallow-Savvy Guard: But then the African swallow's not migratory...

Narrator: Meanwhile, not more than two swallow's flights away, Arthur and Bedivere had discovered something. Oh, that's an unladen swallow's flight away, obviously. There were more than two laden swallow's flights away, four really, if they had the coconut on a line between them. I mean, if the birds were walking, and dragging the coconut...

 

Yes, I detect a little Rod Serling influence myself!!

 

Have some of William Shakespears "As you like it". Whatever you see in this thread keep your coconuts to yourself

 

OK here the nitty gritty some asked for:

Power & Torque curve:

===


Gear Ratio & Drive Axle:

First Gear Ratio 1) 4.71 Second Gear Ratio 1) 3.14 Third Gear Ratio 1) 2.11 Fourth Gear Ratio 1) 1.67 Fifth Gear Ratio 1) 1.28 Sixth Gear Ratio 1) 1.00 Reverse Ratio 1) 3.32 Clutch Size N/A in Final Drive Axle Ratio 1) 2.56 Seventh Gear Ratio 1) 0.84 Eighth Gear Ratio 1) 0.67
====

Tire friction coefficient can be calculated from stopping distance, if I am correct:

Braking, 70-0 mph: 151 ft

Braking Distance

d = V2/(2g(f + G))

CF: let's just take 0.9 for now, since I am in a hurry - I'll resolve the formula for f and the get the number lateron.


===

With regards to Bogie's lever arm: He used just the rims. I want to ask if this would be better. At the bottom you have a calculator for diameter:

Calulators from www.4Lo.com

===


I hope I provided the needed nitty gritty some were asking for. I am still looking at calculating the RPM needed on paper that will start to spin tires.

Thanks to all for your input.

And with regards to those with coconuts and spinning wheels in space, I leave you with this video, which shows there are countless other ways monkeys came up with to spin wheels, not only including dropping the clutch, tying your car up, or driving against a wall.. Enjoy the video!

10 Awesome Ways To Get To Minimum Tread Depth - YouTube

 

OK here the nitty gritty some asked for:

I hope I provided the needed nitty gritty some were asking for. I am still looking at calculating the RPM needed on paper that will start to spin tires.


805.82rpm based on this equation

 

From an engineering position......Postulating a question as such makes this exercise pointless. The always false plug in numbers, thus false calculations would be too great a dithering amplitude for any use.

 

Yes, but if we tied a string between two swallows......