Race cars - Torque vs Hp - The Undiscovered Country (for many)

[zanick]

Thanks Chis, I think I am back on track now. (literally, going racing tomorrow too! )

Anyway, as far as the intgration of the power curve, yes, I think 75% would probably do it for sportscar and closer to 80% for factor race cars or spec open wheel type cars.

Sorry about the confusion of the HP curves. I mean they are the same shape, with the x-axis scaled. So, when both A car and B car are coming off a turn at 55% of max rpm, they both have the same HP created by :) engine torque thorugh the gears, making the same rear wheel force to the ground. (it might not even be same wheel torque, if the tires are different diameters. (all part of the gearing part of the equation, right?)

as far as the broader HP curve, in the example, the HP peaks were identical, but one curve just had more HP over the operational range. Generally, this is more true for a high torque engines, but occasionally, you see one that has a slightly broader HP curve. generally, it means the other high torque engine has much upside potential for mods.

Ive enjoyed the talk as well . Trust me, very few, but the motorsport top engineers get this stuff, even at this basic level. Its nice to get grounded here with the right thinking and terminology. I worked in the industrial controls arena for 12 years as well, so I have a great handle on the basic stuff. (torque, gear ratios and efficiency, basic motion control profiles,etc). Delt with all sorts of flavors of small electric motors and their associated electronics, including servos and stepmotors. So, I have the high level understanding, but the devil is in the details. :) . Obviously for racing, its an avantage to know how to optimize your equipment performance on a given race day!

Thanks,

mk

That sounds exactly on the button for the physics terminoligy for what is going on. .

As a performance indicator are you thinking of inegrating for an acceptable power band. So say find the area between 80% max power?

I'm still a little confused as why what you mean when you say the horse power curves are the same. As that to me means the torque must be the same as the graphs overlay perfectly. Do you mean that when the x-axis is scaled, the curves have the same trend?


Also you mention getting a broader HP curve, so you sacrifice peak HP for gains along more of the rev range. (in effect maximising the area under the power curve). This is essentially the same as altering the torque curve to sustain at higher rpm (I think). My friends is acutally better than me at this, i'll ask him about it tonight.

I'm really enjoying this discussion, its making me think quite hard and polish up on stuff I've not read in quite a while.

 

[Gnosis]

I get it, its getting to be less and less of a chicken and egg thing. force is what causes the acceleration. got it. Hp is a measure of of the rate the force does the work. HP is a vision into what the potential force is though right?

Mk

Just when I think you're maybe finally getting it, you go right back to that association you have with horsepower and force as per your last sentence above.

Horsepower isn't a "vision" or "insight" into the potential force. All force in the combustion engine is strictly produced by torque and torque alone. Torque IS measureable on a dyno. Horsepower is merely a rating of how much "WORK" (not force) can potentially be accomplished per a given RPM via the available torque of the engine. "Work" and "force" are two entirely different quantities just as horsepower and torque are entirely different.

Horsepower should only be thought of as a rating, as in a "horsepower rating", a work rating.

Try to grasp the concept of “horsepower rating” from this next example:

YOU could potentially twist a ratchet (thereby apply some measure of torque) all day long. This also means we could actually give you a horsepower rating of some kind rated per an entire day’s work. Let’s assume that the absolute torque limit that you can apply per your ratchet is 200 ft pounds and the conveyor belt you’ve been manually cranking with your ratchet has only required 20 ft lbs (piece of cake). This conveyor belt is carrying buckets of material to the top of a silo, then dumping the contents. Suddenly, someone accidentally overloads one of the buckets on the conveyor belt with 400 pounds of material. Doh!

At this point, you HAVEN’T lost your willingness to work neither have you lost your “horsepower rating” (that work which you are capable of accomplishing per day). You simply lack the strength to apply any more than 200 ft pounds of torque to lift the excessive 400 pound load. Since torque is the ONLY factor causing the conveyor belt’s movement and you lack the torque to lift the 400 pound load, all work has ceased. If you stepped over to an identical conveyor belt that didn’t have the excessive 400 pound load, you’d be able to continue at your rated horsepower rating for the rest of the day. Insufficient torque is the issue here, NOT insufficient horsepower.

That's why I stated it WASN’T lack of horsepower that prevented my 1.5 HP mini-bike from climbing that steep incline; it was the engine’s sheer lack of torque, as torque is the ONLY force being produced by a combustion engine and I cannot state this enough. Horsepower is a work rating, torque actually causes the rotation and acceleration of the crankshaft.

 

[zanick]

I understand your push for clarifying that it is the force that does the work. But, it is the HP "rating" as you call it, that can indicate the potential force at any speed. (if you don't know the force, as power=fv. In your analogy, clearly if 1.5HP doesn't give the force you need to climb the hill, more power will give more force. (and I know you are going to say, if I give I more force, it will have a higher power rating as a result ;) ). But, I don't understand why HP isn't an indication of force at any vehicle speed. I am talking about HP and its indication of a force's ability to do work. This is why I was leaning toward watt-seconds (or HP-seconds) as a determinant factor and useful term on the subject of automobile comparisons. If you have a battery , its energy potential is rated in KW-Hours. (same thing as Hp-seconds, watts-seconds etc) That battery can give you the power to be able to lift 550lbs in 1 second over 1 foot. (if this 10v battery is 74amp/second, 1hp-second battery, 746watt-seconds) 746w-seconds, 74amp-seconds, 20ma-hour, etc) . This power rating tells me how much work I can do and how fast I can do it. I can lift 550lbs in 1 foot in 1 second or 1100lbs 1 foot in 2 seconds, or 225lbs, 1 foot in 1/2 second. Its a power limit of how fast you can accelerate a mass at a given velocity, or the rate of doing 'work' . Again, I understand that the force does the work, but you don't really need to know it to get an acceleration of a mass at a velocity, right? Rate of change of kinetic energy is one example I can think of where if I know this, the answer can be power, without the knowledge of force or torque. A chassis dyno has drums and can measure hp without using a torque value, even though a torque value is easily produced from the change of sped of the drums. Without an engine speed input, engine torque could be any value, so the output is engine HP and rear wheel MPH.

Now, we might have glazed over the main question here, but when folks look at HP ratings of vehicles, or torque, which is more meaningful as an indication of potential rear wheel forces at any speed? Certainly engine torque is only an indication, if we are talking about comparing two equal HP cars, with grossly different engine torque and RPM values. If you don't look at the gearing and vehicle speeds, HP will be able to be compared by only plotting both HP curves on one another. if they are same shape for an given range of vehicle speed, the cars will accelerated the same (assuming the same car and gear boxes adjusted for same range of speeds). In other words, both identical cars have gear boxes that allow for the same MPH speed of the cars in each gear. Average power, but more importantly, HP-seconds being a determinant factor on which car will accelerate faster. In other words, if two curves have the same area under the usuable rpm range, but one has more area under the top rpm are vs the lower rpm area in the usable range, the one with he area in the higher rpm range will have more "Hp-seconds" and will accelerate an equal car faster through a wide range of vehicle speed

The point I was trying to make earlier, was that even an integration of the HP curve doesn't exactly find the answer, as this is due to the varying time spent at the higher rev areas. Area under the engine torque curve doesn't work, as most engines make most their torque below an area where they would even be operating in. The most rear wheel torque (after the gear box reductions and torque multiplications) would be found at the maximum amount of HP available at any vehicle speed.

Getting back to your mini bike analogy, if you don't have over 1.5hp or some force at the rear wheels , you are not going to be able to climb that hill. the rate of change of kinetic energy or adding potential energy (mini bike climbing hill) can not be met by the power source and resultant forces from the rear wheel on the road. If we were using the battery, we could turn up the power setting and climb the hill. (using more amps) the capacity would be reduced, but the system would be using a faster rate of doing work, and use more power. !nstead of 1100watt-seconds of power, you could be using 2200watt-seconds (or near 3Hp) to keep that current rate of speed, while climbing the hill. 2 x the amp draw, 2x the power and 2x the torque to the drive wheels. Clearly, the force has been increased by increasing the current draw, which is by definition a higher rate of doing work (higher power). How is this logic not correct?

These are the questions many are looking to have answered in the right terms.

Thanks,

Mk

Just when I think you're maybe finally getting it, you go right back to that association you have with horsepower and force as per your last sentence above.

Horsepower isn't a "vision" or "insight" into the potential force. All force in the combustion engine is strictly produced by torque and torque alone. Torque IS measureable on a dyno. Horsepower is merely a rating of how much "WORK" (not force) can potentially be accomplished per a given RPM via the available torque of the engine. "Work" and "force" are two entirely different quantities just as horsepower and torque are entirely different.

Horsepower should only be thought of as a rating, as in a "horsepower rating", a work rating.

Try to grasp the concept of “horsepower rating” from this next example:

YOU could potentially twist a ratchet (thereby apply some measure of torque) all day long. This also means we could actually give you a horsepower rating of some kind rated per an entire day’s work. Let’s assume that the absolute torque limit that you can apply per your ratchet is 200 ft pounds and the conveyor belt you’ve been manually cranking with your ratchet has only required 20 ft lbs (piece of cake). This conveyor belt is carrying buckets of material to the top of a silo, then dumping the contents. Suddenly, someone accidentally overloads one of the buckets on the conveyor belt with 400 pounds of material. Doh!

At this point, you HAVEN’T lost your willingness to work neither have you lost your “horsepower rating” (that work which you are capable of accomplishing per day). You simply lack the strength to apply any more than 200 ft pounds of torque to lift the excessive 400 pound load. Since torque is the ONLY factor causing the conveyor belt’s movement and you lack the torque to lift the 400 pound load, all work has ceased. If you stepped over to an identical conveyor belt that didn’t have the excessive 400 pound load, you’d be able to continue at your rated horsepower rating for the rest of the day. Insufficient torque is the issue here, NOT insufficient horsepower.

That's why I stated it WASN’T lack of horsepower that prevented my 1.5 HP mini-bike from climbing that steep incline; it was the engine’s sheer lack of torque, as torque is the ONLY force being produced by a combustion engine and I cannot state this enough. Horsepower is a work rating, torque actually causes the rotation and acceleration of the crankshaft.

 

[rcgldr]

I used a 1.5 HP lawnmower engine to power a mini-bike and I geared it to yield a top-speed of 35 MPH while on flat ground with my body weight of 140 pounds. However, while riding at 35 MPH on flat ground, I came to one of the steepest streets in my neighborhood. My mini-bike steadily lost speed and would not make it all the way up to the top of the steep hill.

Assume mini-bike weighs 60 pounds. Weight of rider and mini-bike is 200 lbs. To climb a 20 degree hill, it will take sin(20) x 200 = 68.4 pounds of force to climb the hill (ignoring drag factors here, assuming rear wheel horsepower is 1.5).

If the rear wheel horsepower is 1.5 hp, then you need to solve:

1.5 = 68.4 x speed / 375
speed = 8.22 mph.

In order to climb a 20 degree hill, the mini bike would have to be geared down so it's makes 1.5 rear wheel horsepower at 8.22 mph.

If the hill was vertical, with the mini-bike attached to a chain going up the side of the hill, then it would lift 200 lbs at 2.81 mph

 

[xxChrisxx]

Your thinking is correct but you have the wording wrong zanick.

Rate of change of kinetic energy won't tell you the force nor acceleration. K.E = .5mv^2
Just because they use the same terms doesn't make them mathematically compatble. The dyno doesn't acutally measue energy, as energy can't be directly measured.The thing you are describign for the dyno is rate of change of momentum of the drum. momentum = mv
Dy differentiating this you get the magic formula of F = ma.

 

[zanick]

Thanks Chris,

But if acceleration = power/(mass x velocity or a=P/p , the rate of change of momentium would be proportionally higher from 50-60mph as it would from 60 to 80mph and so would the force for the two speed ranges. However, the power would be much higher at the higher speed range. Thats why I seemed to remember power equaling rate of change of kinetic energy, not momentium. I agree that the rate of change of momentium would get you to "Force".

so I guess what I'm saying is that the rate of change of kinetic energy gets you power and thus acceleration at any given velocity, and from power you could get force. (working backward )right?
mk

Your thinking is correct but you have the wording wrong zanick.

Rate of change of kinetic energy won't tell you the force nor acceleration. K.E = .5mv^2

The thing you want is rate of change of momentum. momentum = mv
Dy differentiating this you get the magic formula of F = ma.

 

[xxChrisxx]

This is where are going wrong. you are correct that Power is not the rate of change on momentum. But what the dyno actually measures, and what it outputs on a screen is not the same thing. From what it measures it does some mathematical operators. to give out a nice convenient power figure. To same you messing around calcualting it yourself.

The dyno:
1) measures the rate of change of anguar momentum of the drum in a given time.
2) this gives the acceleration of the drum.
3) From f=ma you can find the force at the wheel giving said acceleration.
4) deterimes the speed at the tire = surface of drum = radius times angular velocity.
5) determines the power by multplying force time speed (at the tire contact patch).

This is that's going on from what the dyno measures, what the computer then calcualtes and operates. Then produces the power figures.

This cannot be done the other way round as energy cannot be directly measured (see this link:http://en.wikipedia.org/wiki/Energy#Measurement) to get acceleration directly.

We know that as energy is a concept that is inferred, so is power. You can mathematically operate the equations once you konw evreything to give you what you want, but that doesn't mean you can do it in reality.Also i'll discuss the climbing a hill thing with you in a bit, as that's a tricky one to talk about over a forum. Its another time that figurees can be misleading. you look at the figures and see one thing, but reality tells you something different.

 

[rcgldr]

The dyno:
1) measures the rate of change of anguar momentum of the drum in a given time.
2) this gives the acceleration of the drum.
3) From f=ma you can find the force at the wheel giving said acceleration.
4) From this rear wheel force and the distance traveled you get the work done.
5) The change of work done in a given times tells you the power.

Not quite:

4) deterimes the speed at the tire = surface of drum = radius times angular velocity.
5) determines the power by multplying force time speed (at the tire contact patch).

 

[xxChrisxx]

Not quite:

4) deterimes the speed at the tire = surface of drum = radius times angular velocity.
5) determines the power by multplying force time speed (at the tire contact patch).

Both are actally correct :P, the last two steps manipulate the units in the same way. That just inculdes a time function at step 4 and not 5.

I prefer the way you said it though, I think its slightly more clear. So i'll steal your answer and edit my other post :D

 

[zanick]

Ok, we got a little side tracked on how power is measured on a dyno. I still think if you have the rate of change of the KE, can't you solve for acceleration and then work backward to find the force (and then easily convert to torque)? Going back to the linear world, if a car is accelerating at a certain rate, and we know two velocities, the mass of the car, doesn't average power=dw/dt ? It seems that we would have to calculate the change in work from the rate of change of KE. Kind of digging a hole here, I know. :)

Anyway, getting back to the original point and question, what would the terms used in comparing two same vehicles with equal HP "rated" engines, one being higher torque than the other, at different rpm ranges each? I am pointing at watt-seconds or HP-seconds to determine which one would yield the greatest rear wheel forces at any vehicle speed.
(gearing proportioal to each engines rpm range to achieve the same speed in any gear).
Integration of the rear wheel torque curves would obviously do this, but a succession of them would have to be used due to overlap. integrating the HP curves wouldn't do it either, as time spent in higher rpm ranges is different and would give incorrect weighted values for the results. I keep on falling back to HP-seconds or watt-secons as the answer, and I am really asking, in the physics world, what would be the best way to address this with the correct terminology. The car is an interesting subject, with varying force even if power was constant, but its not due to gearing and power varies as well. What I have found, is the easiest way to determine optimal acceleration is to look at the HP curves adjusted to the same vehicle speed operating ranges. Or, just look at gearing spacing, and use same percentage drop of each engine to determine the area that will be maximized and used under the HP curves. However, this doesn't adress the time factor, that's why it seems like HP-seconds is the right way to look at this to compare performance.
Same could be done for rear wheel torque figures, but would be a little more cumbersome to calculate.

It seems that comparing the shape of two HP curves, with the same proportioal engine rpm ranges is the easiest way to compare vehicle performance.*

Thoughts?

mk

*two engines, one with a 10,000rpm redline and the other 5,000rpm redline and shift points.
Both would have 25% rpm drops per shift per gear. Both engines have the same peak HP values, but would have different shaped curves or the same shaped curves, but either way, grossly different engine torque values.

 

[xxChrisxx]

You are refusing to budge from incorrect thinking and its very difficult to give a techincally correct explination until you do. You need to stop thinking about KE now, and start thinking about momentum if you want a technically correct explanation.

By simply using KE you are ignoring what the dyno is actually measuring. Therefore any further premise from that is based on false mathematics. You are approaching the problem backwards.

Just because you can maniplulate the equations the correct way on paper doesn't mean it is done like that in reality. Without a context energy means very little, and its the context that gives it importance.

What you are saying is correct for the car acceleration, but you are butching the way this is acutally calculated. Its also leading you to incorrect thinking. That more power is more force. This isn't correct and is also why you are getting the 1.5hp motor climing a hill messed up a bit.

More torque is more force, more power is force more quickly. (you cannot do this the other way around)
More power does not = more force.

Now for car acceleration, the rear wheel force (and by extansion engine torque) is larger than the drag forces. So for each torque operation you have a net positive force, so if you apply that more quickly you will accelerate more quickly.

For climbing a hill, the forces stopping the motion are much much higher. Its concevable that for a low powered (read low torque) motor, the torque will not provide the force to overcome the 'drag' forces at the rear wheel. You have a net negative force. Now no matter how often you apply this net negative force it will NEVER become positive and you will never climb the hill.

By your reasoning if you increase the power with the same engine (it spin it faster) you will be developing more power, but you will still never get up the hill.

This is the reason why trucks are so good at climbing hills with a full load, tons of torque. So they can climb steeper hills, but the low power means they won't do it quickly. If you hook up a formula 1 engine (same power or even lots more power rating but much lower torque) to the truck it won't have the pulling power to climb with several tons attached.

your battery/electric motor example. You say doubling the power will allow you to lift something heavier, this is true. But as they are operating at the same rpm, the way it doubles power is by doubling the torque outrput.

I'm going to come up with a worked example to show this.

 

[zanick]

ll concede that calcuating dyno force is really based on acceleration of the load and power is calculated from that. Again, not really the point of the discussion, but thanks, as I do want to stay on the accepted side of the explanations.

HOWEVER, your truck and the climbing hill example does not seem right. first of all , if you add more power to a mini bike climbing a hill, but failing, more hp will result in a higher force, as seen by power=fs. by adding NOS for example to that engine along with Fuel, we increase the power rating of the engine. Its rate of doing work is increased proportionally. If you just increase rpm, as you suggest, the mini bike would have to go faster up the hill, greatly increasing the force demands. however, if the engine rpm increased and the gearing allowed for the same equivilant speed, (and the engine HP curve was flat, or engine torque went down in proportion to rpm going up) then , you would be utilzing more of the available engine power and would be able to climb the hill. torque and force at the rear wheels would go up. By saying that just increasing the frequency or RPM a negative net force, makes no sense, as you would be changing the vehicle speed higher.
This is exactly why cars with gear boxes, downshift to climb hills at the same rate of speed. to take advantage of more available HP of the engine . rpm goes up, torque goes up at the rear wheels due to gearing and more hp is utilized.

With the truck vs the formula one engine, both can climb the hill at exactly the same rate if they have the same HP. Rate of doing work, right? the F1 engine does it by high rpm and low torque, the truck the opposite. in the end, if both are using the same HP, both will have the same rear wheel forces at an speed . The real reason that trucks use high torque, low rpm engines, is wear and efficiency. How long would a 15,000rpm engine last pulling a truck up a 100mile incline? The truck could do it for 100s of thousands of miles due to operating at very low rpm. The truck makes decent HP at very low rpm. very broad HP curve. the F1 engine is peaky and has a narrow max HP range.

I can easily give you resultant rear wheel torque to prove that any engine rated at the same hp, even at much less engine torque, will create the exact same rear wheel forces to the ground. Thats a simple set of equations. This seems to be in direct conflict it with your last sentence. This would be consistant with, "power=force x speed".

mk

Now for car acceleration, the rear wheel force (and by extansion engine torque) is larger than the drag forces. So for each torque operation you have a net positive force, so if you apply that more quickly you will accelerate more quickly.

For climbing a hill, the forces stopping the motion are much much higher. Its concevable that for a low powered (read low torque) motor, the torque will not provide the force to overcome the 'drag' forces at the rear wheel. You have a net negative force. Now no matter how often you apply this net negative force it will NEVER become positive and you will never climb the hill.

By your reasoning if you increase the power with the same engine (it spin it faster) you will be developing more power, but you will still never get up the hill.

This is the reason why trucks are so good at climbing hills with a full load, tons of torque. So they can climb steeper hills, but the low power means they won't do it quickly. If you hook up a formula 1 engine (same power or even lots more power rating but much lower torque) to the truck it won't have the pulling power to climb with several tons attached.

I'm going to come up with a worked example to show this.

 

[xxChrisxx]

ll concede that calcuating dyno force is really based on acceleration of the load and power is calculated from that. Again, not really the point of the discussion, but thanks, as I do want to stay on the accepted side of the explanations.

HOWEVER, your truck and the climbing hill example does not seem right. first of all , if you add more power to a mini bike climbing a hill, but failing, more hp will result in a higher force, as seen by power=fs. by adding NOS for example to that engine along with Fuel, we increase the power rating of the engine. Its rate of doing work is increased proportionally. If you just increase rpm, as you suggest, the mini bike would have to go faster up the hill, greatly increasing the force demands. however, if the engine rpm increased and the gearing allowed for the same equivilant speed, (and the engine HP curve was flat, or engine torque went down in proportion to rpm going up) then , you would be utilzing more of the available engine power and would be able to climb the hill. torque and force at the rear wheels would go up. By saying that just increasing the frequency or RPM a negative net force, makes no sense, as you would be changing the vehicle speed higher.
This is exactly why cars with gear boxes, downshift to climb hills at the same rate of speed. to take advantage of more available HP of the engine . rpm goes up, torque goes up at the rear wheels due to gearing and more hp is utilized.

With the truck vs the formula one engine, both can climb the hill at exactly the same rate if they have the same HP. Rate of doing work, right? the F1 engine does it by high rpm and low torque, the truck the opposite. in the end, if both are using the same HP, both will have the same rear wheel forces at an speed . The real reason that trucks use high torque, low rpm engines, is wear and efficiency. How long would a 15,000rpm engine last pulling a truck up a 100mile incline? The truck could do it for 100s of thousands of miles due to operating at very low rpm. The truck makes decent HP at very low rpm. very broad HP curve. the F1 engine is peaky and has a narrow max HP range.

I can easily give you resultant rear wheel torque to prove that any engine rated at the same hp, even at much less engine torque, will create the exact same rear wheel forces to the ground. Thats a simple set of equations. This seems to be in direct conflict it with your last sentence. This would be consistant with, "power=force x speed".

mk

Jesus.Power = force * speed is NOT based in reality.

It is a mathematical relation devised purely for convenience,

 

[xxChrisxx]

ll concede that calcuating dyno force is really based on acceleration of the load and power is calculated from that. Again, not really the point of the discussion, but thanks, as I do want to stay on the accepted side of the explanations.

HOWEVER, your truck and the climbing hill example does not seem right. first of all , if you add more power to a mini bike climbing a hill, but failing, more hp will result in a higher force, as seen by power=fs. by adding NOS for example to that engine along with Fuel, we increase the power rating of the engine. Its rate of doing work is increased proportionally. If you just increase rpm, as you suggest, the mini bike would have to go faster up the hill, greatly increasing the force demands. however, if the engine rpm increased and the gearing allowed for the same equivilant speed, (and the engine HP curve was flat, or engine torque went down in proportion to rpm going up) then , you would be utilzing more of the available engine power and would be able to climb the hill. torque and force at the rear wheels would go up. By saying that just increasing the frequency or RPM a negative net force, makes no sense, as you would be changing the vehicle speed higher.
This is exactly why cars with gear boxes, downshift to climb hills at the same rate of speed. to take advantage of more available HP of the engine . rpm goes up, torque goes up at the rear wheels due to gearing and more hp is utilized.

With the truck vs the formula one engine, both can climb the hill at exactly the same rate if they have the same HP. Rate of doing work, right? the F1 engine does it by high rpm and low torque, the truck the opposite. in the end, if both are using the same HP, both will have the same rear wheel forces at an speed . The real reason that trucks use high torque, low rpm engines, is wear and efficiency. How long would a 15,000rpm engine last pulling a truck up a 100mile incline? The truck could do it for 100s of thousands of miles due to operating at very low rpm. The truck makes decent HP at very low rpm. very broad HP curve. the F1 engine is peaky and has a narrow max HP range.

I can easily give you resultant rear wheel torque to prove that any engine rated at the same hp, even at much less engine torque, will create the exact same rear wheel forces to the ground. Thats a simple set of equations. This seems to be in direct conflict it with your last sentence. This would be consistant with, "power=force x speed".

mk

Jesus.Power = force * speed is NOT based in reality.

It is a mathematical relation devised purely for convenience, it shas no physical basis. I am talking about the power part, not the force * speed.

By injecting nos you are increasing the engine TORQUE at a given rpm because you are increasing the force on the piston and by extension the power. NOT THE OTHER WAY ROUND.

To talk of gearing, why when you are coming to a hill do you downshift? If power is the key factor in rear wheel force it shoudlnt matter about the gear you are in as the engien is pumping out the same power. They downshift to take advantage of the TORQUE multiplication of the lower gear ratio. Seriesly do the calculations for the F1 and truck engine. I gaurantee you will make the mistake that power = force. IT DOESNT!

See how it works : http://auto.howstuffworks.com/question381.htm

^^ read it^^

You are really confusing yourself by flitting between the rear wheel horsepower and rear wheel torque. give all the equations you can.

you seem to clearly have problems thinking in terms of torque and power.

Seriously please go and buy a book on this. i'll have a look through my library for the best one.

 

[zanick]

ll concede that calcuating dyno force is really based on acceleration of the loads and power is calculated from that. again, not really the point of the discussion, but thanks, as I do want to stay on the accepted side of the explanations.

HOWEVER, you truck and climbing hill example does not seem right. first of all , if you add more power to a mini bike climbing a hill, but failing, more hp will relate in higher force, as seen by power=fs. by adding NOS for example to that engine along with Fuel, we increase the power rating of the engine. Its rate of doing work is increased proportionally. If you just increase rpm, as you suggest, the mini bike would have to go faster up the hill, greatly increasing the force demands. however, if the engine rpm increased and the gearing allowed for the same equivilant speed, (and the engine HP curve was flat, or engine torque went down in proportion to rpm going up) then , you would be utilzing more of the available engine power and would be able to climb the hill. torque and force at the rear wheels would go up. By saying that just increasing the frequency or RPM a negative net force, makes no sense, as you would be changing the vehicle speed higher.

you also say: "More torque is more force, more power is force more quickly. (you cannot do this the other way around)
More power does not = more force. But with power=fv, it does indicate that more power would create more force. (or to keep on the force side of thinking, a higher power rating would indicate that more force would have to be produced at the same vehicle speed to climb that hill.


Now for car acceleration, the rear wheel force (and by extansion engine torque) is larger than the drag forces. So for each torque operation you have a net positive force, so if you apply that more quickly you will accelerate more quickly."


With the truck vs the formula one engine, both can climb the hill at exactly the same rate if they have the same HP. Rate of doing work, right? the F1 engine does it by high rpm and low torque, the truck the opposite. in the end, if both are using the same HP, both will have the same rear wheel forces at an speed . The real reason that trucks use high torque, low rpm engines, is wear and efficiency. How long would a 15,000rpm engine last pulling a truck up a 100mile incline? The truck could do it for 100s of thousands of miles due to operating at very low rpm. The truck makes decent HP at very low rpm. very broad HP curve. the F1 engine is peaky and has a narrow max HP range.

I can easily give you resultant rear wheel torque to prove that any engine rated at the same hp, even at much less engine torque, will create the exact same rear wheel forces to the ground. Thats a simple set of equations. This seems to be in direct conflict it with your last sentence. This would be consistant with, "power=force x speed".

mk

Now for car acceleration, the rear wheel force (and by extansion engine torque) is larger than the drag forces. So for each torque operation you have a net positive force, so if you apply that more quickly you will accelerate more quickly.

For climbing a hill, the forces stopping the motion are much much higher. Its concevable that for a low powered (read low torque) motor, the torque will not provide the force to overcome the 'drag' forces at the rear wheel. You have a net negative force. Now no matter how often you apply this net negative force it will NEVER become positive and you will never climb the hill.

By your reasoning if you increase the power with the same engine (it spin it faster) you will be developing more power, but you will still never get up the hill.

This is the reason why trucks are so good at climbing hills with a full load, tons of torque. So they can climb steeper hills, but the low power means they won't do it quickly. If you hook up a formula 1 engine (same power or even lots more power rating but much lower torque) to the truck it won't have the pulling power to climb with several tons attached.

I'm going to come up with a worked example to show this.

 

[xxChrisxx]

You are pretty much wrong in everything you've just said regarding power.

Read the links on how stuff works, or go to your amazon and buy a book on this subject.

Until you realize your fundamental error in thinking you will never learn why you are going wrong. You are clinging to P=F*s, but if you'd acutrally read up on this you know that that relation is not what is heppening in reality. you will discorver its a mathemetical relation devised by people AFTER pissing about with torque and rpm and then doing lots of calculations to it for so long. You will also discover why its the torque that creates the force at the wheel NOT POWER.

Until then there is nothing I can do to increase your inderstanding of this.

 

[zanick]

I get what you are saying, especially with the NOS example. we get more torque, and subsequently more force at the rear wheels. I just have to change my direction of thinking. (coming from car land.)

But, when you talk of gearing below, you lose me. power is the key. you ask the question below. power is key and as you are crusing along at 50mph with your truck, you are using 100hp of your 800hp rating. Here comes the hill, you need much more force to climp the hill, you downshift to take advantage of the engines power available at the higher rpm and higher fuel requirements. At that prior power setting, you might have only been at 10% of the power available. now, you take advantage of higher gear reduction by downshifting, you now are in the higher rpm ranges and at the higher hp range at full throttle. You also are maximizing you rear wheel forces at the max HP range.

Now, you asked me to compare a high torque, low rpm truck engine and F1 engine at the same power, so here it is. Let's keep it simple. 1000hp 1000ft-lbs of peak torque for the truck and 1000hp and 250ft-lbs of peak torque for the F1 engine. The both go to climp a hill. The Big diesel engine is in the truck. At 50mph, the truck is at 4000rpm and producing 8000ft-lbs of torque at the rear wheels. (1000ft-lbs with 8:1 gearing) At the same 50mph, the truck with the F1 engine is running at 16,000rpm and 250ft-lbs of torque, but the rear wheel torque is the same 8000ft-lbs. (250ft-lbs with 32:1 gearing). vehicle speed is the same, HP is the same, engine torque is way different, yet, rear wheel forces are identical. Where is the mistake there?

mk

Jesus.Power = force * speed is NOT based in reality.

It is a mathematical relation devised purely for convenience, it shas no physical basis. I am talking about the power part, not the force * speed.

By injecting nos you are increasing the engine TORQUE at a given rpm because you are increasing the force on the piston and by extension the power. NOT THE OTHER WAY ROUND.

To talk of gearing, why when you are coming to a hill do you downshift? If power is the key factor in rear wheel force it shoudlnt matter about the gear you are in as the engien is pumping out the same power. They downshift to take advantage of the TORQUE multiplication of the lower gear ratio. Seriesly do the calculations for the F1 and truck engine. I gaurantee you will make the mistake that power = force. IT DOESNT!

See how it works : I am finding the link now

You are really confusing yourself by flitting between the rear wheel horsepower and rear wheel torque. give all the equations you can.

you seem to clearly have problems thinking in terms of torque and power.

Seriously please go and buy a book on this. i'll have a look through my library for the best one.

 

[zanick]

Im hearing you about that its really the force that determines the power, and power is just the power rating. But, you asked me to provide the example, and I did. Your truck powered by F1 or its normal engine, doesn't make your point, as i proved, both can yeild the same rear wheel torque at the same HP ratings, and grossly different engine torque values.
Power ratings is what you and I were comparing here and they seem to yield the same rear wheel torque at any vehicle speed. (if they are running at the same power level)

You mention i was wrong in "everything " i had a said regarding power. can you be more specific? you make the comparison of the truck powered by F1 engine vs its normal low reving high torque engine. that was actually incorrect. If power ratings are the same, the same rate of work can be done by either, and this means the same rear wheel torque can be produced!

In the electric motor and battery rating world, you get things rated in KW or KW-hours. unit measures of work. This indicates the potential rates of doing work. Is this wrong as well?

mk

You are pretty much wrong in everything you've just said regarding power.

Read the links on how stuff works, or go to your amazon and buy a book on this subject.

Until you realize your fundamental error in thinking you will never learn why you are going wrong. You are clinging to P=F*s, but if you'd acutrally read up on this you know that that relation is not what is heppening in reality. you will discorver its a mathemetical relation devised by people AFTER pissing about with torque and rpm and then doing lots of calculations to it for so long. You will also discover why its the torque that creates the force at the wheel NOT POWER.

Until then there is nothing I can do to increase your inderstanding of this.

 

[xxChrisxx]

You keep moving the goal posts by changing gearing. Gearing is a torque multiplyer.

He cannot change the gearing he has on his bike when he comes to a hill as he is geared for a certian road speed. For that gearing he doesn't have the RW torque available. When you said that he doesn't have the power available to do it, this is incorrect as the engine always produces the same power output, the phenomenon you are talking about is torque. The determination of whether you can climb the hill is torque, the determination of how fast you can climb is power. A rwhp value doesn't tell you if you can climb the hill or not, a rw force value would.

So to find if you can climb, you use:

RWTorque * wheel radius = Force

The RWHP tells you how fast you can climb it. from P = F * S

If you were theoretically detmining gearing before you set off to go this ans you had a RWHP curve, It would be correct to use P = F* S to determine the gearing for a set speed.

If you have set gearing available (ie you've already set off) The P = F* S does not tell you anything useful.
In this case you have to use RWT*wheel radius = F to determine if you have sufficient torque at the rear wheel to do the job. If you find that you don't you either have to go away and get the gearing that will give the correct RWT or you could slap bigger wheels on the bike. You can then find how fast you can climb the hill form P=T*angular velocity.

As jeff pointed out, he could climb that hill if he was geared for it. Then again with gears you can get any mortor to lift any amount. It'll just do it very slowly. So you see you can use the RWHP formula in some situations, but you can use the torque calculation in all situations.

What you were doing was the former, but applying it to the latter situation. So you were not calcualting anything useful. He stated he was geared for something already, and that it was his lack of torque that was causing the problem. Stating that he can do becuase P = F * S tells him he can is pointless as he can't chgane the gearing to get to the poitput that that equations states.

If you cannot climb a hill at the peak engine torque value, you cannot climb it at peak engine power. (as torque is lower at peak power). This transfers down through the transmission so that. If you can't climb a hill at peak RWT you cannot climb the hill at peak RWHP. It means you need to fiddle with the gearing to be able to do so. If you can't fiddle with the gearing and can't fit a larger wheel, you can't get up the hill. This is something that is easier to visualise using a torque calculation as that is the acutal physicla driving force.

I've got a question that I'm just finishing up that I'm going to post after your next reply. It should demonstrate that power is not the critical factor.

 

[zanick]

Not really, I think we are having the classis problem of rear wheel torque vs engine torque adding to the cross lines of the discussion. you tell me.

I think I see a potential problem. So, getting back to the mini bike. If he is running 1.5 hp and running some top speed say, 30mph, you are right. he has a certain amount of force being generated to sustain that speed. Stilll, power=force x speed. he is not at max torque of the engine, he is at max HP which as you say is higher than max torque of the engine. this is generating the maximum amount of force at that particular speed, 30mph. Now comes the hill. he can't choose a lower gear, because that would slow him down, so you are right. limied by max HP and rear wheel torque. if he adds NOS, it increases force by x% so he can climb the hill at that same speed.

You keep on saying that the bike cannont climb the hill if it doesn't have the torque, what you mean to say, is that if it was at max HP prior, and you wanted to keep the speed constant, yes that would be true. with more HP, you would have more force and then you would. you say its more force so then you can then determine the hp has increased as well. Thats fine. In the electrical /mechanical world, if i dropped in a higher output battery, to the elecrict motor powered bike, it could cause the motor to run with more torque and thus at a higher power level.

truck analogy with its stock engine vs F1 engine. If both engines are installed with appropriate gearing, both will be able to climb the hill at the same speed, using gears appropriate for the trade offs of rpm and speed and mulitplication of torque. One engine would do it at high rpm , low torque and the other would do it with high torque, low rpm. the net forces at the rear wheel would be the same because the rate of work would be the same.

Now , your last paragraph. If I have a transmission, and I can't climb a hill at a certain speed at max engine torque, there is a distinct possibility, and in most cases, that i could climb it at max HP, even though the engine torque would be less at that same vehicle speed, rear wheel multiplied torque would be higher at that same vehicle speed than it would be found at max engine torque. agree? I think you two part sentence is saying that if you have a fixed gear, and you can't climb the hill at max torque, certanly at max HP you wouldn't be able to climb it, because the speed would be much higher. and your second part of the sentence, is saying that if you can't climb the hill with max rear wheel torque, as multipled thorugh the gear box at that speed, certainly you wouldn't be able to climb it at max hp, because those two points are the same engine rpm. max HP would be the point at which you would have max torque at any given speed. just like if we had an infinitely variable gear box, the engine would operate at max hp and not max torque to achieve max torque levels at the rear wheels at any given vehicle speed.






mk

You keep moving the goal posts by changing gearing. Gearing is a torque multiplyer.

He cannot change the gearing he has on his bike when he comes to a hill as he is geared for a certian road speed. For that gearing he doesn't have the RW torque available. When you said that he doesn't have the power available to do it, this is incorrect as the engine always produces the same power output, the phenomenon you are talking about is torque. The determination of whether you can climb the hill is torque, the determination of how fast you can climb is power. A rwhp value doesn't tell you if you can climb the hill or not, a rw force value would.

So to find if you can climb, you use:

RWTorque * wheel radius = Force

The RWHP tells you how fast you can climb it. from P = F * S

If you were theoretically detmining gearing before you set off to go this ans you had a RWHP curve, It would be correct to use P = F* S to determine the gearing for a set speed.

If you have set gearing available (ie you've already set off) The P = F* S does not tell you anything useful.
In this case you have to use RWT*wheel radius = F to determine if you have sufficient torque at the rear wheel to do the job. If you find that you don't you either have to go away and get the gearing that will give the correct RWT or you could slap bigger wheels on the bike. You can then find how fast you can climb the hill form P=T*angular velocity.

As jeff pointed out, he could climb that hill if he was geared for it. Then again with gears you can get any mortor to lift any amount. It'll just do it very slowly. So you see you can use the RWHP formula in some situations, but you can use the torque calculation in all situations.

What you were doing was the former, but applying it to the latter situation. So you were not calcualting anything useful. He stated he was geared for something already, and that it was his lack of torque that was causing the problem. Stating that he can do becuase P = F * S tells him he can is pointless as he can't chgane the gearing to get to the poitput that that equations states.

This comes down to:

If you have a given gearing - use torque eq.
If you have a given speed (and can vary gearing) - can use torque or power eq.

If you cannot climb a hill at the peak engine torque value, you cannot climb it at peak engine power. (as torque is lower at peak power). This transfers down through the transmission so that. If you can't climb a hill at peak RWT you cannot climb the hill at peak RWHP. It means you need to fiddle with the gearing to be able to do so. If you can't fiddle with the gearing and can't fit a larger wheel, you can't get up the hill. This is something that is easier to visualise using a torque calculation as that is the acutal physicla driving force.

 

[xxChrisxx]

Will repost q in a second.

Ok Q is:

14.4Kg block on a 45% slope

The motor used is:

100 Watt at 200rpm
4.77Nm at 200rpm

167.2 Watt at 400 rpm
4Nm at 400 rpm.

The reduction gearing is 15:1.
The rolling radius is 1m

Can the motor move the block up the hill?

EDIT: you can assume no losses through the drivetrain, so RWHP = engine HP.

Now after ansering this I'll give you the second part and address the truck and F1 engine thing.

Also buy this book :https://www.amazon.com/dp/1844253147/?tag=pfamazon01-20
and read the last chapter. It's not a very technical book, but is perfect for what we are talking about.

 

[zanick]

Well first off you have two motor speeds and thus driving speeds, fixed by a 15:1 gear box. If I can't change that, then I can't get speeds that utilize the available power of the motor.
the power range for the two speeds is 1:1.65, but the speed difference is 2x. so, I would need much more power to achieve 2x the speed of the 400rpm motor setting. If the slower motor setting works and uses all available power to lift the mass, even with optimal gearing, 1.67x greater power setting could only allow for a small increase in drive speed.

Thats at first glance. it looks like yes for the 100watt setting, but no for the 167watt setting due to the amount of power needed to raise the mass at 2x the speed.

Will repost q in a second.

Ok Q is:

14.4Kg block on a 45% slope

The motor used is:

100 Watt at 200rpm
4.77Nm at 200rpm

167.2 Watt at 400 rpm
4Nm at 400 rpm.

The reduction gearing is 15:1.
The rolling radius is 1m

Can the motor move the block up the hill?

EDIT: you can assume no losses through the drivetrain, so RWHP = engine HP.

Now after ansering this I'll give you the second part and address the truck and F1 engine thing.

 

[xxChrisxx]

First of all buy this book:
https://www.amazon.com/dp/1844253147/?tag=pfamazon01-20

And read the last chapter (hell you can read all of it if you want). It'll tell you exatly wehre your thinking is a bit off.Regarding the question.

The acutal answer is neither will get it moving at that ratio.

Those are the peak figures, so the maximum power the motor can produce is 167.2 W. The maximum torque it can produce is 4.77 Nm.

The point of the above was to show that for a given gearing its the torque that creates the moving force. A minimum moving force requites a minimum torque at the wheels. The power figure has no bearing on this force. If it did the power figure would cause larger rear wheel forces and get it moving.

The next thing I was going to ask was try it with 21:1 (ive misplaces my notes on this i think that was the ratio). You'll find that the force at the wheel is then enough to move the block at the lower power figure but not the higher.

If you did rev it up (as more power apparently creates more force at the rear wheel) you should move faster. In fact it does the opposite, it you did rev it up you would no longer be providing enough force to overcome the resistance.

Do the acutal calucaltions and you'll see this is the case.Here is my working for the 1st part:

14.4*9.81*cos(45)= 100N
You need at least 100N to get the block moving up the slope.

It can be shown that the rear wheel forces for each settign are:

100W setting.

Engine_Torque*gearing = Wheel torque.
4.77*15= 71.55 Nm at the wheel.
As the wheel is 1 m Rolling radius.
T=F*d
Force at wheel is 71.55 N

This gives a net negative force of 28.45N

for the higher power setting the wheel force (using the same method above is)
Force =60N

Using the power method: we know the power of the motor and the force necessary.

P=100W
F=100N

using P = F * S

100/100 = 1 m/s So to get the thing moving we need a wheel speed of 1m/s or below.

Wheel speed at lower power = circumfrence of wheel * angular velocity
for 1 second.

S = (2*pi)*(200/(15*60)) = 1.39 m/s
S = (2*pi)*(400/(15*60)) = 2.79 m/s

The wheel speeds are too high to provide the adequate forces to start the block moving.

So more power doesn't = more rear wheel force. And the torque method is fer better at this (its more clear that it won't work) as it is a snapshot in time.

 

[rcgldr]

100 Watt at 200rpm, 4.77Nm at 200rpm
167.2 Watt at 400 rpm , 4Nm at 400 rpm.
The reduction gearing is 15:1.

It's not a fair comparason, setting the gearing for the same speed = 1.3963 m/s, with a radius of 1 meter:

100.00 watt at 200 rpm, 4.77465 Nm at 200 rpm, reduction gearing 15:1, force = 71.620 N
167.55 watt at 400 rpm, 4.00000 Nm at 400 rpm, reduction gearing 30:1, force = 120.00 N

As expected, 120.00 / 71.62 = 1.6755 = 167.55 / 100.00

 

[xxChrisxx]

You've totally missed the point of the arguement.

It was to show that the instantaneous force was dictated by torque only (not power). To do that they must have the same gearing. If you could change the gearing willy nilly, then the question becomes moot.

Thats why I told him to use 21:1 next. after that 25:1 and after that 30:1.
It shows that power output of the engine is the same, but the gearing changes to torque at the wheels.

I was then going to move on to what if you wanted them to move at the same rate. Now this is a power over time.

So with the 15:1 reduction at the higher power, you have a lack of torque to get it moving. But not when you use a different gearing.

I need him to break the thinking that its power dominating the size of the force (thats torque). and get him to understand that power essentially shows the frequency of the force. So although at the higher power, you have a smaller force. that force acts more often.

HOWEVER: The instantaneous force MUST be large engouh to move the block, otherwise the frequency is irrelevent. Think about trying to hammer in a nail with a feather. You can hit it as often as you want, but you can provide enogh linear force to get it to move. (thats a bit of an over simplified comparison, but it'll do for the purposes of explaining what I want)

 

[rcgldr]

It was to show that the instantaneous force was dictated by torque only (not power).

I understand that torque is the angular equlivalent to (linear) force.

To do that they must have the same gearing.

Why? If you have gearing, then the gearing should be matched to the expected demands of the motor. This is why cars have transmissions and differentials. This is why a P51D Mustang (WW2 airplane) gears it's prop down to 1437 rpm while it's engine spins at 3000 rpm. Since the title of this thread is race cars, I would assume that those cars would be geared appropriately.

In your example, you specify a torque, a gearing and a wheel radius, allowing the linear force at the wheel to be calculated, but without knowing that this torque occurs at non-zero rpm, it's not known if the motor can move the block.

Take the case of an electronic stepper motor. Dynamic torque (moving) is about 70% of holding torque (not moving), so for example.

reduction gearing 15:1, wheel radius 1 meter, 60N load (required force)

torque == 5.0Nm at rpm = 0, force = 75.0N
torque <= 3.5Nm at rpm > 0, force = 52.5N

The motor has enough torque to hold the load in place, but it can't turn against the load.

Knowing only the torque isn't sufficient to know if it will move unless it's also known that that torque can be generated when the motor is turning. Some knowledge of angular velocity (the fact that it is non-zero) in addition to torque (so therefore power) is required to know if the motor can perform any work.

 

[zanick]

you really missed the point of the entire discussion with that one. I had similar values at my quick glance, using .707 x the 14Nm of the load up the hill. Even still, you missed the part that if you didnt use gears to utilize the available power, of course the load can't be lifted up the hill. there is not enough power available to move those two loads. However, if you slowed things down a bit, the power could be used to move the mass up the hill.

geaering doesn't create HP , it gives you the ability to utilize it. It also multiplies torque, by sacraficing rpm. Because at 200rpm, and the resultant 2m/second requires more torque than the motor can provide. P=fv. however, at a slower speed, the hp can be utilized though deeper gearing and would move/lift the load. I you have more power at 400rpm on the motor, 100watts going to 167watts, then obviously, double the gearing again, for the same output speed , and you will have increased the final torque output proportionally.
You are using more electrical power, creating more mechanical power, and thus have more force generated at any given output speed.

A stepping motor, like Jeff brought up, has its greatest torque at the first signal to move. It has something called a torque vs angle of displacment. however if you want to get a load to move, you can easily gear it, and take advantage of its maximum power capabilities. If the load is above its holding torque, it can't move it Gear it down, and now it has more torque and is better utilizing the available power capacity.

go back to your original idea of the truck powered by its engine or a F1 engine. you said that at the same hp, the F1 engine doesn't do a good job, and that is just not true with half the engine torque and double the gearing with its double the engine speed, the same rear wheel torque is achieved. This is the point of the discussion as I see it.

HP (or HP rating) will determine the torque at the wheels at any given speed, and is proved by the combination of the identities showing:

Acceleration =power/mass x velocity.

why don't you summarize what the last chapter in the book says. I've been racing awhile and always have plotted out torque curves in each gear. I now use HP curves with geared % rpm drops. I can compare other powerplants as well, without looking at engine torque curves and get accurate assesement of which engine will perform best over a given speed range. basically, he who uses more of the available HP, if two cars have the same hp, will win the race (all other things being equal). even if they are different in HP , a lower hp can win if it can be used in such a way where it has more average HP over the operational speed range. Conversely, and more commonly, the broader HP engines can yeild more available torque at the rear wheels, even if its engine torque is lower compared to a high torque same HP engine.

The point of this really was, to confer with the physics community to talk about what terms would be appropriate to describe the charactistics of engines in race cars . Is it HP-seconds that we want to optimize? Becauase as I see it, the amount of time spent at the highest engine hp levels available, will yield the max amount of rear wheel torque, and acceleration at any given vehicle speed.



mk

First of all buy this book:
https://www.amazon.com/dp/1844253147/?tag=pfamazon01-20

And read the last chapter (hell you can read all of it if you want). It'll tell you exatly wehre your thinking is a bit off.


Regarding the question.

The acutal answer is neither will get it moving at that ratio.

Those are the peak figures, so the maximum power the motor can produce is 167.2 W. The maximum torque it can produce is 4.77 Nm.

The point of the above was to show that for a given gearing its the torque that creates the moving force. A minimum moving force requites a minimum torque at the wheels. The power figure has no bearing on this force. If it did the power figure would cause larger rear wheel forces and get it moving.

The next thing I was going to ask was try it with 21:1 (ive misplaces my notes on this i think that was the ratio). You'll find that the force at the wheel is then enough to move the block at the lower power figure but not the higher.

If you did rev it up (as more power apparently creates more force at the rear wheel) you should move faster. In fact it does the opposite, it you did rev it up you would no longer be providing enough force to overcome the resistance.

Do the acutal calucaltions and you'll see this is the case.


Here is my working for the 1st part:

14.4*9.81*cos(45)= 100N
You need at least 100N to get the block moving up the slope.

It can be shown that the rear wheel forces for each settign are:

100W setting.

Engine_Torque*gearing = Wheel torque.
4.77*15= 71.55 Nm at the wheel.
As the wheel is 1 m Rolling radius.
T=F*d
Force at wheel is 71.55 N

This gives a net negative force of 28.45N

for the higher power setting the wheel force (using the same method above is)
Force =60N

Using the power method: we know the power of the motor and the force necessary.

P=100W
F=100N

using P = F * S

100/100 = 1 m/s So to get the thing moving we need a wheel speed of 1m/s or below.

Wheel speed at lower power = circumfrence of wheel * angular velocity
for 1 second.

S = (2*pi)*(200/(15*60)) = 1.39 m/s
S = (2*pi)*(400/(15*60)) = 2.79 m/s

The wheel speeds are too high to provide the adequate forces to start the block moving.

So more power doesn't = more rear wheel force. And the torque method is fer better at this (its more clear that it won't work) as it is a snapshot in time.

 

[xxChrisxx]

Look if you care in the slightest about being correct about the physics of this.

Buy/borrow/steal and read the following books.

John Heywood - Internal Combustion Engine Fundamentals
Richard Stone - ICE
A.G Bell - Four-Stroke Performance Tuning
Paul Van Valkenburgh- Race Car Engineering & Mechanics

1. is probably least relevant but is pretty much the engine bible.
2. stone is good for practical workings.
3. Good non technical book reguarding practical tuning
4. a very good all round car setup book. especially the section on gearing and performance.

I have read all of these at some point over the last 3 years. They are all thorough and will show you where you've gone wrong.

Read Bell first. then any of the others. I am in no way inclined to type out whole sections from the book.

 

[xxChrisxx]

I understand that torque is the angular equlivalent to (linear) force.

Why? If you have gearing, then the gearing should be matched to the expected demands of the motor. This is why cars have transmissions and differentials. This is why a P51D Mustang (WW2 airplane) gears it's prop down to 1437 rpm while it's engine spins at 3000 rpm. Since the title of this thread is race cars, I would assume that those cars would be geared appropriately.

In your example, you specify a torque, a gearing and a wheel radius, allowing the linear force at the wheel to be calculated, but without knowing that this torque occurs at non-zero rpm, it's not known if the motor can move the block.

Take the case of an electronic stepper motor. Dynamic torque (moving) is about 70% of holding torque (not moving), so for example.

reduction gearing 15:1, wheel radius 1 meter, 60N load (required force)

torque == 5.0Nm at rpm = 0, force = 75.0N
torque <= 3.5Nm at rpm > 0, force = 52.5N

The motor has enough torque to hold the load in place, but it can't turn against the load.

Knowing only the torque isn't sufficient to know if it will move unless it's also known that that torque can be generated when the motor is turning. Some knowledge of angular velocity (the fact that it is non-zero) in addition to torque (so therefore power) is required to know if the motor can perform any work.

jesus.

im going to answer this point 1 last time.

this was a demonstation about a key physical concept to the op. for the purposes of demonstating that physical concept, the gears had to be kept the same.

Gears are torque multipliers. The point was to show that the force at the wheel is given by torque only. not power. if it was given my the power figure, an increase in power should give an increase in wheel force. it doesnt.

in my exapmle i specify a torque AT AN RPM. however this point is moot as:

torque defines the force at the wheel. So a huge torque at 0 rpm would give a huge force that had the POTENTIAL! (READ THIS 50 TIMES) POTENTIAL! to move the block.

if you applied this 0 times per second. which is where power comes in. you would not move the block. as the potential is not being used.

likewise if you had a force that ws not large enough DID NOT HAVE THE POTENITAL due to incorrect gearing. you could have all the power available in the gear but you wouldn't shift he block.
THIS IS THE POINT YOU NEED TO KNOW BOTH! POWER AND TORQUE TO KNOW IF IT'LL MOVE. YOU CAN KNOW POWER and RPM or torque abd RPM. or anything else, but in the end you'll always be boiling down the equations for a torque and power.

 

[xxChrisxx]

thats me done for this now. Read the books I posted.