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You can't say tractive force is based on Power... it is not. Power is based on force, distance, and time. Engines make torque and Power is calculated, not the other way around. While in a basic sense you are right, you just got it the wrong way.
That's wrong, and we've already been over that. Engines don't just make torque: they output a force over a distance in a certain amount of time, or torque*rpm. If they don't make power (i.e. if they only make torque), then they don't get any work done: it's like the electric motor that is stalled against a load. Again, we've already gone over this.
I am going to say it one more time... Power tells you what you can do. Torque (Force) tells you what you will do.
And you're wrong one more time. Power literally, BY DEFINITION, is a measure of what is being done or what an engine can do: it is a measure of how much force can be exerted over a distance within a time frame. Force (torque) alone tells us jack **** about what an engine will do. I made that blatantly obvious when I told you the torque of a car's engine and asked you how fast it could go. You correctly replied that that wasn't enough information, because torque does not tell us what an engine will do. If all you know is the engine's torque, you can't know: 1) the acceleration force it can apply at a given road speed, 2) the top speed it can achieve in a car of a given drag/friction, or 3) how fast it can accelerate a certain mass.
You know the power curve, minus small inertial loss changes between gears, is the same profile at the wheel for all gears in the transmission. So, the same power appears at the wheels based on engine RPM through all gears. Torque curve however is not the same and neither is wheel speed.
I don't know what torque curve you're talking about - you're confusing yourself again. The tractive force to accelerate a car is always, always equal to power at the tire divided by road speed. It is always determined by power, never crankshaft torque. And as I've noted previously, tractive force is not a torque but rather it's a linear force.
Last edited by MatthewMiller; 02-06-2019 at 12:31 PM.
That's wrong, and we've already been over that. Engines don't just make torque: they output a force over a distance in a certain amount of time, or torque*rpm. If they don't make power (i.e. if they only make torque), then they don't get any work done: it's like the electric motor that is stalled against a load. Again, we've already gone over this.
And you're wrong one more time. Power literally, BY DEFINITION, is a measure of what is being done or what an engine can do: it is a measure of how much force can be exerted over a distance within a time frame. Force (torque) alone tells us jack **** about what an engine will do. I made that blatantly obvious when I told you the torque of a car's engine and asked you how fast it could go. You correctly replied that that wasn't enough information, because torque does not tell us what an engine will do. If all you know is the engine's torque, you can't know: 1) the acceleration force it can apply at a given road speed, 2) the top speed it can achieve in a car of a given drag/friction, or 3) how fast it can accelerate a certain mass.
I don't know what torque curve you're talking about - you're confusing yourself again. The tractive force to accelerate a car is always, always equal to power at the tire divided by road speed. It is always determined by power, never crankshaft torque. And as I've noted previously, tractive force is not a torque but rather it's a linear force.
For us (cars), "e" is the driveline efficiency. Typically, people use 85%, but it doesn't matter what driveline loss you use as long as you keep it equal in your comparisons. Or, just leave it out altogether and just work with power at the tires: TE=375(P/V). To be uber-clear, Tractive Effort (TE) is the "F" in F=MA. Notice that there are zero ***** given about crankshaft torque. You don't need to know it. Power is the only engine parameter that matters.
Last edited by MatthewMiller; 02-06-2019 at 03:35 PM.
For us (cars), n is the driveline efficiency. Typically, people use 85%, but it doesn't matter what driveline loss you use as long as you keep it equal in your comparisons. Or, just leave it out altogether and just work with power at the tires: T=2650(P/V). I'm too lazy to convert to mph and pounds right now. Maybe later. To be uber-clear, Tractive Effort (T) is the "F" in F=MA. Notice that there are zero ***** given about crankshaft torque. You don't need to know it. Power is the only engine parameter that matters.
So, if you are only calculating based on Power... why can't you go 180mph in 1st gear... or why can't you do it in 6th?
Where does the F come from?
- It comes from Power delivered to the tires.
Where does Power at the tires come from?
- It comes from Power delivered from drive train
Where does the power come from in the drive train
- It comes from the engine
Where does the power from the engine come from?
- It comes from the ability of the engine to produce Torque at RPM
Hmmm.... Engines produce torque... end of story. You can't produce a formula with P(power) where power is not defined by torque as it relates to a cars drive train using an internal combustion engine. The motor produces a twisting force... by definition torque.
You can have One million Horsepower at your tires and not be able to go over 30mph. Resistance forces don't care about power, and neither does a free body diagram in physics. On the other hand, if you are producing One million pounds of force (which here is torque divided by the diameter of the wheel/tire) you better hold on to something because there won't be enough force to hold you back.
Let me try to get this through another way. Force, Distance, and time.... all the units we need to define power.
A human has a capacity to do work and we would call this a unit of power.
Said human lays on a bench press and 100lbs are loaded. This human has a reach and that will be the distance work is done.
This human produces 4 reps in 10 seconds. Thus, utilizing his power to do work.
Now, we double the weight and half the reps... same reach but this time the person completes the exercise in 5 seconds.
The equation balances, same power is used, same work is done.
Now let's double the weight again. Same power Capacity
For the equation to balance, we would need to do one rep of 400lbs in 2.5 seconds.
Except, not everyone can move 400lbs on a bench press. Not sure I can anymore... been some time. So, now the weight doesn't move and no work gets done because there is not enough force to move the weight(resistance force).
The resistance force doesn't car how much Power capacity is there. It only matters if you have enough force to overcome it. Your equation only solves if it is actually being done. Force can be present absent of motion.
Let me try again to say this differently. If you have a car going 180mph, you can calculate the Power it is consuming based off the equations you are using. If you want the car to go that fast, you can use your equations to see how much power you will need. I don't disagree with this. It will not tell you if the car sitting in your garage with that amount of power available will do it, only that it has the potential to do it.
As I said above. If you take the Torque/HP curve, multiply it through the drive train, and move the tires. The inverse relationship between torque and speed through gears produces an almost identical horsepower curve for every gear at the wheel. Though, we know acceleration rates change as gear changes are made. Because while the speed goes up, force goes down... Power is the same. There are significant steps down in force production, and significant steps up in wheel speed. What changes between gears is wheel speed and torque output, not the power.
My point from the beginning is from a design perspective, you have to know the force output to know what you can do. Just having a 400hp motor doesn't tell you a given car will actually achieve a certain speed.
Possibly, based off the equations you are putting up, if you are arguing that if a car is traveling a certain speed against a known resistance we know how much power is being utilized. I agree with this and the torque/force doesn't matter because you are already doing it. The equation is describing what is happening and will hold true.
Yikes. Is this question for real? You've asked it twice, so unfortunately, I'm guessing that you're serious.
I was literally about to post the same question. I have the feeling that we're all being punked.
Originally Posted by KyleF
Uh huh... and then if you have a car with 400hp, why can't it go 180mph in 1st gear? Why can't it do it in 6th?
It can't do it in either gear because the engine would not be anywhere close to its peak 400hp output at 180mph in either gear. 1st gear would have the added feature of blowing the engine up well before 180mph was ever reached.
I think the whole issue of transmissions with discrete gear ratios gets some people confused. This is why I have said several times that it's easier to understand the concept of power if you consider a car with a CVT, which can keep its engine at its peak-power rpm at all road speeds. The whole point of a regular transmission with several discrete ratios is to approximate the CVT: to keep an engine as close to its peak-power rpm as possible at all times during max acceleration.
Last edited by MatthewMiller; 02-06-2019 at 04:03 PM.
My point from the beginning is from a design perspective, you have to know the force output to know what you can do. Just having a 400hp motor doesn't tell you a given car will actually achieve a certain speed.
I can't make this any more plain: this is literally, exactly what power tells you. Again, this is by definition! The entire concept of power exists so that we have one single number that will tell us how much work an engine can do and how fast it can do it.
Possibly, based off the equations you are putting up, if you are arguing that if a car is traveling a certain speed against a known resistance we know how much power is being utilized. I agree with this and the torque/force doesn't matter because you are already doing it. The equation is describing what is happening and will hold true.
You and I have one thing in common: neither of us know what you're talking about. I just gave you the formula to determine the accelerative force based off of power, and only power. You don't need to know the engine's torque or rpm or gear ratios or any other damn thing. It doesn't matter if the engine is making 400hp and 400lb/ft at 5252rpm, 400hp and 800lb/ft at 2626rpm, or 400hp and 200lb/ft at 10504rpm. You just need to know power and road speed to know how much force is available at the contact patches to accelerate the car.
There is no way to make this any more simple or obvious. If you don't understand why power is the engine parameter that matters (and why torque doesn't matter), then I can't help you. If you're too obtuse to understand that the engine obviously has to actually be outputting its peak power at the targeted speed (i.e. geared appropriately for the task at hand), then I also can't help you with that.
Last edited by MatthewMiller; 02-06-2019 at 04:06 PM.
I was literally about to post the same question. I have the feeling that we're all being punked.
It can't do it in either gear because the engine would not be anywhere close to its peak 400hp output at 180mph in either gear.
It can't do it in either gear because the engine would not be anywhere close to its peak 400hp output at 180mph in either gear. 1st gear would have the added feature of blowing the engine up well before 180mph was ever reached.
This is my point. You have the same 400hp motor. It doesn't achieve a given speed on hp alone, other consideration have to be made.
This is my point. You have the same 400hp motor. It doesn't achieve a given speed on hp alone, other consideration have to be made.
It's not your point, although I've no doubt you think it is. If the motor isn't putting out 400hp, then it isn't putting out 400hp. It doesn't matter what the sticker under the hood says or the magazine article says: it only matters what amount of power the engine is putting out at the moment. You gear the car to keep it as close to its peak power speed as possible for whatever you're trying to do. Again, for the sake of making this understandable to you, imagine the car has a CVT and can always be kept at peak-power rpm regardless of road speed.
Uh huh... and then if you have a car with 400hp, why can't it go 180mph in 1st gear? Why can't it do it in 6th?
It could, it just needs to be geared correctly, gearing just allows you to put the engine in the correct power band to utilize the power that is available. If you had a 3000 stall converter, a good oil cooler, 2.59 rear gears and 1:1 first gear you could get the car to 180 MPH, it would take a while but could get there.
Your misunderstanding a basic concept. Power tells you energy that is available to do work, and with that you can figure out what can be done. To set-up the car correctly you do have to know what RPM that the power is avalable at and gear accordingly but that is it. Torque only gives you part of the picture, you must know RPM to figure out power and then power tells you how much work can be done, you gear for the power peak to acheive max speed and acceleration. Torque by itself is meaningless
If you sit down and figure this out you will see that the higher the torque at any given RPM the higher the power. The greatest acceleration of a car will be when you keep the engine operating in the area where it has the highest power under the operating range, not the highest torque. That is because you will gear the car accordingly and that gearing to the tires will actually provide the highest torque and HP to accelerate the car at any given speed.
It's not your point, although I've no doubt you think it is. If the motor isn't putting out 400hp, then it isn't putting out 400hp. It doesn't matter what the sticker under the hood says or the magazine article says: it only matters what amount of power the engine is putting out at the moment. You gear the car to keep it as close to its peak power speed as possible for whatever you're trying to do. Again, for the sake of making this understandable to you, imagine the car has a CVT and can always be kept at peak-power rpm regardless of road speed.
That is funny, you want to tell someone else what the point they are trying to make is? Seriously? That makes no sense...
I think you get too angry to think straight and just want to argue. Somewhere down this rabbit hole I think you lost what I said at first... Back to the beginning of what I said....
Originally Posted by KyleF
Weight is much less of a factor than aero. Air resistance increases exponentially with speed. Mass does not. It's also not just about peak power, but where the peak is made in the RPM range, and where your gearing places you in your power curve at speed. Horsepower is the ability to produce torque at high RPMs, but the amount of torque that is there is important. You have to have enough twisting force to the drive wheels to keep accelerating. This is why you see a lot of land speed cars getting pushed to start. They don't have enough gearing to get going, but once the mass is moving and they have the engine in it's power curve, they can accelerate to insane speeds. These cars are built with top speed, not acceleration in mind.
It seems "twisting force to drive the wheels" really bothers you. Your tractive force that you want to hang your hat on and calculate from power is the twisting force at the wheels (Torque) divided by the wheel diameter. This force will not be the same in 2nd and 4th gear, but the power will.
With your CVT, As the vehicle is accelerating Power output is constant, Velocity is increasing and thus wheel rpm, and torque is going down and thus so is the tractive force. The result is as speed increases, the rate of acceleration decreases Acceleration stops when the torque output divided by wheel diameter is equal to the sum of resistance working against it. You do see how the wheel diameter figures into the overall gear ratio of the car right?
To balance the Free body diagram, you would use forces, not power... but since you have Time, Distance, and Force... you want to solve one side for Power and just damn marry that word. Fine, it works on the conservation of energy principle (Energy in a system cannot be created or destroyed). On the other side you have Speed, Time, Distance of the car moving and a set of forces acting against it. We have already agreed with this information you can calculate power. So, essentially you are converting the resistance forces into a Power and saying the sum of power on one side has to be equal to or greater than the sum of power on the other side. Or should I say the equations you want to use are doing this at a snap shot.
If you don't see it, we agree on this as far as I can tell.
Seems you keep talking about the snap shot in time when the car is actually achieving said speed. Which you said in this post.
What you fail or just don't want to agree on is when I say just because you have 400hp (or any number you choose) you won't achieve a given speed if other considerations aren't made.
I also believe form educational background, when solving problems we always used forces (Physics, Statics, Dynamics, Strength of Materials, etc.). Then if you have force, time, and distance you can calculate power being produced or consumed. You just want to work with power. Which is fine. I prefer to work with force, which is also fine since the resultant forces are working against the same body.
So back to my point about considerations:
Car A: 400hp @6000 RPM, 27" Diamter Tires, 3.07 Ring Gear, ZF6 Transmission
Car B: 400hp @6000 RPM, 27" Diamter Tires, 3.07 Ring Gear, ZF6 Transmission
Car C: 400hp @6000 RPM, 27" Diamter Tires, 3.07 Ring Gear, CVT Transmission
If your point is if all three are going the same speed they are consuming the same amount of power I agree.
That information alone doesn't tell you . As has been discussed in this thread a few other items need to be included. Once being the Cd and frontal area since these two factors will become the greatest resistance force at speed. Some consideration would also need to be put on mass and Cf of the tires... assume the last one to be mainly negligible from one tire to another.
My point is not about an existing car that is currently moving. I am talking about building one that will reach it's potential. What I am trying to convey and can't get across to you is that just because I have a 400hp motor setting on a stand, that alone doesn't tell me what it will do inside the car. I would set up a sheet with my gear ratios (Transmission, Ring, and Tire Diameter) and populate the cells with what output I have Using the engine's curves, either torque or hp, I would use that through gear multiplication calculate the tractive force at the tire and velocity of the car at a given RPM. I would then calculate the resistance force... and then see where they meet. This would tell me where acceleration stops and ultimate velocity is achieved. You equations apply here as at ultimate velocity I have Force, Speed, and Time... and would know what power I am using. Since this is a street car, I find I am not optimally using my Power to produce force.
My peak RPM (point of peak hp for simplicity sake) in 5th says I am not using all 400hp based on velocity, I should be able to go faster but....
My peak tractive force in 6th gear says I can't overcome the resistance forces and if I shift, I will actually start to slow down. And this line is my point about needing to know the force.
If I start playing with gearing(in a perfect world changing transmission, ring gear, and tire diameter being all equally easy to achieve... because it is on a spread sheet.) I would use my calculation to sync in the gearing, but I am going to be looking at the force produced to understand my acceleration rates and ultimate speed. For instance, in this situation you could go to a higher ring gear to extend 5th further until it's peak HP point produces the peak tractive force, would be nice if ring gears came in infinite number of ratios. The other option is to go to a lower ring gear and and turn 6th into a pulling gear beyond 5th, but I am going to need to know the force at the wheel to know if the change will result in a force great enough to overcome the resistance forces.
OK, so commonly going to a higher gear ratio is the way to go, so we can maintain an overdrive ratio
Put in a 2.73, Run the numbers, and still not quite there. I am still not utilizing all of my 400hp at the top of 5th! I am not consuming my 400hp... I can still go faster but I am still RPM limited.
So, next step is up to a 2.59. Run the numbers and now find its too high. At 400Hp and turning 6000RPMs I am not producing enough force to overcome the forces at the target speed.
Now I would put the 2.73 back in and adjust my tire diameter and find the sweet spot where the forces balance. Tires are available in much larger array of diameters than ring gears. So I would save that as the last item to mess with since its easier to find a width to sidewall height that will match the calculations. I am going to be looking to balance my forces, you may look to balance your Powers. Again, either way should be fine since you have force, time, and distance.
The other side of this is, if you know the speed you want to go, the resistance forces at that speed for a given car, and the gearing in the car you can spec the motor and say I need X number of Horsepower at Y RPM. that may get tricky when you start narrowing it down by what it must fit into and what transmission it must fit in as we all know a gear/trie swap is much easier than a motor swap.
In the end, I prefer to run these calculations through as torque since I am looking at forces. Force at the tires being calculated by torque on the hub that is produced at the engine and multiplied through gearing. All of this can be stated in Power since again, you have Force, time, and distance.
If you want to continue to argue about Power versus Torque, just keep in mind Since P is defined by T*RPM/5252... you can always swap it in. Let me borrow someone elses' words..."it’s the reason that anyone telling you that horsepower and torque should be considered equally and separately is significantly off-base. The fact of the matter is that horsepower is the product of torque and another value — RPMs (divided by 5252). It’s not unrelated, separate, or different. In fact, there’s not a single machine in existence that measures a car’s horsepower. It’s a man-made number. When a car’s performance is tested, its torque is measured using a dynamometer. The measure of an engine’s performance is torque. Horsepower is an additional number that’s attained by multiplying the torque by the RPMs." "The mistake most people make when engaging in this debate is considering horsepower and torque independently. Almost everyone argues as if they are separate, unrelated values. They aren’t." "The torque at the wheels is the torque at the engine combined with the torque magnification given by the transmission through gearing. So the transmission only sees what’s coming off the engine, while the wheels see the resulting force combination of the engine plus the transmission. That’s what horsepower represents. Horsepower is the combination of the benefits of the engine’s raw abilities combined with RPMs. And RPMs are what allow us to use gearing effectively, which gives us more torque at the wheels." - D. Miessler 12/5/18
I don't particularly like how he phrase the "plus transmission". I understand what he means because of gear multiplication, but the transmission actually robs some of the engine output.
I think you get too angry to think straight and just want to argue.
He's not the only one.
Originally Posted by KyleF
With your CVT, As the vehicle is accelerating Power output is constant, Velocity is increasing and thus wheel rpm, and torque is going down and thus so is the tractive force. The result is as speed increases, so does the rate of acceleration.
You sure about that? As you go faster, your rate of acceleration increases? IDK about that one.
The rest....too much reading, too poorly articulated to be worth responding to, IMO.
If you want to continue to argue about Power versus Torque, just keep in mind Since P is defined by T*RPM/5252... you can always swap it in. Let me borrow someone elses' words..."it’s the reason that anyone telling you that horsepower and torque should be considered equally and separately is significantly off-base. The fact of the matter is that horsepower is the product of torque and another value — RPMs (divided by 5252). It’s not unrelated, separate, or different. In fact, there’s not a single machine in existence that measures a car’s horsepower. It’s a man-made number. When a car’s performance is tested, its torque is measured using a dynamometer. The measure of an engine’s performance is torque. Horsepower is an additional number that’s attained by multiplying the torque by the RPMs." "The mistake most people make when engaging in this debate is considering horsepower and torque independently. Almost everyone argues as if they are separate, unrelated values. They aren’t." "The torque at the wheels is the torque at the engine combined with the torque magnification given by the transmission through gearing. So the transmission only sees what’s coming off the engine, while the wheels see the resulting force combination of the engine plus the transmission. That’s what horsepower represents. Horsepower is the combination of the benefits of the engine’s raw abilities combined with RPMs. And RPMs are what allow us to use gearing effectively, which gives us more torque at the wheels." - D. Miessler 12/5/18
I can't leave this one alone either;
IDK who "D. Miessler" is, but he's neglecting something incredibly important; RPM. A dyno measures torque..."D" got that part right. What else does a dyno measure? A Torque wrench measures torque. Can I use that to measure power? No I can not, because the torque wrench doesn't measure the other crucial piece of information; RPM/speed. The dyno measures torque and it measures speed; the two variables = power.
"The measure of an engine’s performance is torque and speed". -fixed. A 5000 ft lb engine would "perform" great on good old "D's" dyno....right? 'cause all HIS dyno measures is torque. So 5000 lb-ft is an "awesome performing" engine....right? I wouldn't want it;
Kyle, try to look at it like an ele. motor, I turn on my table saw and before its up to speed I ram a 4x6 into it, it will stall and maybe blow a fuse. If I let it get up to speed it will cut no problem. Ele. motors are known for torque but rated in power. The power in an engine is torque x rpm. Try this torque = amps power = volts Edison lost that fight because he did not want to use high voltage to push the electricity down long lines. If you dump the clutch at too low a rpm you will stall or bog, get the engine up into its powerband and off you go! I hope this helps you to see how torque & horsepower relate to one another.
02-06-2019, 12:29 PM
