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Post by Stiggy on Sept 18, 2014 13:25:24 GMT
I felt this may be an interesting topic for conversation. I have asked a few pro's in the game of mapping/dyno's and an ecu manufacturer but not found a precise the answer to; "how much power can we put down before we lose traction" Obviously it depends how much load is on the tyres, the tyre width, weight transfer, road conditions etc. If you take an average 7 type car as an example. Call it 600kgs with fuel and driver. 25% weight distribution to make calcs easier so 150kg on each tyre. Call it a dry road with road tyres that are say 200mm wide. To make it easier lets assume there is no loss in the drive due to transmission/bearings etc. If you dial in weight transfer this may confuse things. Ratio wise, lets call it 5-1 via first gear and 4-1 via the open (not LSD) diff. If we assume we have just 100 ft lbs at 6000 rpm, x by 5 in first gear =500, x4 for the diff and that gives 2000 ft lbs of torque at the tyre contact point assuming the wheel is 2 feet dia. That would be a one foot lever from hub centre to the road. We then need to divide this by the 2 wheels. I select LAUNCH control or just hit the gas up to 6000 rpm, flick my foot off the clutch and hold tight. Question is, how much torque can we apply before we lose traction. Opinions very welcome. Clutch slip will soften the hit on the tyres and rpm/torque will not be constant, but say at the maximum, how many ft lbs will those tyres transmit? Please let me know if I have missed something. Maybe modern traction control cars have a memory showing how much torque in which gear was available at the point of slip.
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Post by tomaff on Sept 18, 2014 13:41:37 GMT
I cant answer an exact torque figure, or equation.
I can say that my 240bhp exocet does lose traction easily in 1st 2nd and 3rd if your not carefully. But it does go like stink. I am fitting some wider stickyer rubber this winter that will hopefully aid my endeavours.
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Post by mawdo81 on Sept 18, 2014 16:19:01 GMT
Why is it torque v power? Torque is a measure of force at a standard (1 ft or meter depending on the unit) distance from the fulcrum. Power is a measure of force per second.
You need force to go and maintain after you add rolling resistance and drag, so don't you want both? Sent from my GT-I8190N using proboards
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Post by tomaff on Sept 18, 2014 16:45:30 GMT
I took it as stuart asking if there is a way to work out the amount of torque needed to loose traction and if there is an equation for this taking into account weight etc?
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Post by mawdo81 on Sept 18, 2014 17:20:37 GMT
Sorry, too much looking at the title!
Well traction is all about static friction and the force needed to make 2 surfaces slide.
The force is proportional to the downwards force of the tyre on the road, and also the area of the tyre on the road. So tyre pressures are going to have an impact on this.
Additionally the force is proportional to the coefficient of static friction between the 2 surfaces. Now I'd expect that these are incredibly hard to get, due to the shear amount of variables concerned, but I'd also expect there to be some minimum standards somewhere to base some calculations on.
Tyre pressure would give contact patch area so all you need is the coefficient of friction & the actual equations...
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Post by bingo on Sept 18, 2014 20:06:33 GMT
I can give u a very precise answer....
It's all a load of monkey nuts
Sent from my SM-N9005 using proboards
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Post by leahy on Sept 18, 2014 20:09:16 GMT
I sense electric power in mind
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Post by mawdo81 on Sept 18, 2014 20:32:13 GMT
Ignore the bit re tyre pressure, I'm forgetting the second law!
Amontons' First Law: The force of friction is directly proportional to the applied load. Amontons' Second Law: The force of friction is independent of the apparent area of contact. Coulomb's Law of Friction: Kinetic friction is independent of the sliding velocity.
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Post by Stiggy on Sept 18, 2014 21:49:01 GMT
I sense electric power in mind Electric motors can provide full torque from zero RPM but here I am interested to hear from owners with ICE as to how much torque they feel they are applying before break away. I am doing some calcs right now. I will post my results in the morning for a 600kg car inc driver. Not all drivers would be included in this calc. I have changed the name of the thread as we are considering the co efficient of friction, I remember it as the symbol "u". If I can conclude a set of criteria tonight then we can compare applied loads in relation to acceleration and G forces. This info should give a better understanding of the importance of weight saving, although traction improves with increased loads. I hope to produce an abridged set of figures to help you make the most of your MEV.
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Post by Stiggy on Sept 19, 2014 10:53:18 GMT
Calculations of this nature can be very complex, we can however abbreviate the factors to give us an indication of what we can achieve in our cars. Firstly we must consider the weight. Call it 600kgs with a driver on board and assume we have 150kgs on each corner. That equates to 1500N per wheel or 3000N in total on the 2 driven wheels. To convert that to torque we times by .3 as that is the radius of the driven wheels and we arrive at 450Nm I prefer to adjust that to ft lbs so we go online and use a calculator to find that 450Nm = 331 ft lbs per driven wheel or 662 in total. Now if we factor in weight transfer we can see a dramatic change. As you launch you may see your rear suspension squat by 1/2" and if you have 300lbs per inch springs then you have just increased the load on the rear driven wheels by 300 lbs or actually nearly 100%! Other factors play an important role such as tyre temp/pressures/road surface/ambient temp/tyre compound etc. But we need a guesstimate, I am going for a factor of 1 given that in poor conditions that maybe .5 and with soft compounds on a warm dry track it could be 2. So we take the 662 ft lbs times 1 = well the same! Compare this to your engine output of say 120 ft lbs @4500 rpm times 5 when in first gear and times 4 as it goes through the diff ratio and you can multiply the available torque by 20 which is a whopping 2400 ft lbs. That is why we slip the rear tyres if we try too hard even if we only have a non tuned 1.6 engine. We can only put down about 255 of that available power i.e 662 of the 2400 available, How sad! Now we could add extra weight but then you need extra torque, we could increase tyre width or use softer compounds or just stay home until conditions are perfect. What I do enjoy is watching the front wheels spin on hot hatch boy's car with his 2.0 supercharged whatever, I can out accelerate him in my 1.6 naturally aspirated MEV. Weight transfer is not on his side, he drives the front wheels! His over zealous use of his available power probably due to lack of experience is not helping either. Smooth delivery is often key. What is interesting is the G force we can achieve. We say we have 600kgs to move, divided by that 3000N and we have 5.0 m/s squared which is equal to .5 G. Now when we check out those online calculators again we see that we should get to 62 mph from stand still in just 5.5 seconds. Not too shoddy. But we do slow down when changing gear unless 62 is achieved in first.
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Post by carbonbob on Sept 20, 2014 17:18:20 GMT
Personally doesn't compute lol right foot possibly to heavy to calculate hehe
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Post by lukiez on Oct 3, 2014 23:23:52 GMT
I select LAUNCH control or just hit the gas up to 6000 rpm, flick my foot off the clutch and hold tight. Question is, how much torque can we apply before we lose traction. Opinions very welcome. Clutch slip will soften the hit on the tyres and rpm/torque will not be constant, but say at the maximum, how many ft lbs will those tyres transmit? Please let me know if I have missed something. may i ask to what this all may lead to? are you trying to factor in the reduced mass/higher power outputs to help reduce wheel spin compared to a production car or...? apart from giving yourself a mild headache if its for adding tracion control to and aftermarket ECU it shouldn't be necessary... Maybe modern traction control cars have a memory showing how much torque in which gear was available at the point of slip. in a word no, unless things have changed recently but tracion control systems use the ABS monitoring (wheel speed sensors and rings) whereby instead of picking up when a wheel locks up (sensor signal flatines), when a wheel starts to spin up more compared to the other wheels the tracion control sees the increased signal frequency and acts this on until all wheels are comparitively even. depending on the device used, it can close a secondary throttle valve unless it has fly-by-wire, another is to alter ingition timing and/or fuel injection - some cars have been known to apply the brakes. all this is done on the fly hence its not required to store torque or power values with weight, gear ratios etc. in the memory since it can sense when the tyre is breaking traction. Note: - ABS & tracion is often a conbined system on manufacturer's ECUs or one effects the other when a fault occurs.
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Post by coverdad on Oct 4, 2014 11:35:36 GMT
Road surface is a really important factor of to how much friction you can achieve. Then there are the vehicle dynamics just after the power starts to bite, those pushing down the back and lifting the front and others causing twisting torsion affecting the frame and suspension Finally there are the aerodynamic forces of the bodywork, air density and prevailing wind. No matter how much or little power I usually manage to stall or whelk spin every seven I have driven in equal measure.
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Post by coverdad on Oct 4, 2014 11:36:44 GMT
Whelk spin! I love predictive text
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