Mr.Integra

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Lsos &raquo;</TD></TR><TR><TD CLASS="quote">Friction = Weight * coefficient of friction.

Weight: It's right there in the formula. Look it up in any physics book.

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wrong wrong WRONG again..

kinetic frictional force = the NORMAL FORCE * coefficient of friction....

dood.. stop spreading false information.. seriuosly..

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Lsos &raquo;</TD></TR><TR><TD CLASS="quote">

All else equal, that is the ONLY reason it takes more power to move a heavier car at 50mph: the fact that weight affects friction. </TD></TR></TABLE>

WTF.. it takes MORE energy to move a heavier mass.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Lsos &raquo;</TD></TR><TR><TD CLASS="quote">
There is no other reason. There's air resistance slowing it down, rolling resistance, and friction in the drivetrain. If it wasn't for those, it would take NO power to move a car at 50mph, no matter how much it weights.
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WTF

i'm not even gonna comment on anything else.

you need to take physics 101 again and get anything but a D or F.

GSpeedR

Let's start off by making one thing clear: what makes the car move?

Answer, the rotational momentum/kinetic energy of the spinning drive wheels is translated into linear (translational) momentum/energy. The angle (rotational position) is multiplied by the radius to obtain linear position. Linear velocity? v = rw, where w is angular velocity, and v is linear velocity. It is with the friction between the tire and the road that this is possible. And without sliding tires, this is pretty straightforward.

I'm betting that there is a definite issue of semantics here, as people are arguing about "power output", whp, etc. A dyno only measures how fast the engine can accelerate the wheels.

The total weight of the vehicle is not a parameter for an inertial dyno (such as a DynoJet). However, the only reason that it is a parameter for the Mustang dyno is because the "brake resistance" must be proportional to the resistance the vehicle experiences due to its viscous drag (air resistance) and its mass (F/m = a). The Mustang dyno simulates the frictional forces that the vehicle experiences while in motion. This is the only place where the vehicle weight is a consideration, besides providing enough weight for the tires to grip the drums. Yet, it still requires energy to spin the heavy 2700lbs drums on both dynos. And this energy is still supplied by the engine. Obviously, it is easier to accelerate a drum (essentially a large flywheel) that does not experience a large resistive force. Remember that whatever dyno you are using, must take the mass/radius of the drum into account, because it is like adding another flywheel that requires kinetic energy(which they all do). It would be possible to use drums massive enough that the vehicle weight could be adequately simulated on a dyno without using the a resistive brake, but there is a small amount of math to convert moments of inertia into translational inertia (mass). Here is a thread that has a decent approach to this, if anyone is interested: http://forums.audiworld.com/s4/msgs/1032814.phtml.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">Friction = Weight * coefficient of friction. </TD></TR></TABLE>

This is fine. What's the normal force? The force of gravity on the vehicle aka weight.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">All else equal, that is the ONLY reason it takes more power to move a heavier car at 50mph: the fact that weight affects friction. There is no other reason. There's air resistance slowing it down, rolling resistance, and friction in the drivetrain. If it wasn't for those, it would take NO power to move a car at 50mph, no matter how much it weights. </TD></TR></TABLE>

F = ma, or F/m = a. To accelerate a heavier mass, the force must increase. This is regardless of frictional forces that the car will also experience. The more mass an object has, the more energy/force is required to accelerate it. So long that the tires remain under static (rolling) friction with the road, the linear motion of the car is directly connected to the rotational torque seen at the crankshaft. So long as you don't disengage the clutch, the lighter car will be easier to accelerate, decelerate and keep at a constant speed.

Mr.Integra

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">

This is fine. What's the normal force? The force of gravity on the vehicle aka weight.

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the normal force DOES NOT always equal to weight

Lsos

I agree.

I agree.

I'm not going to comment on this because a walking human is very different than a moving car. For one, a human doesn't have wheels and he has to constantly accelerate and decelerate his feet to keep moving.

You guys ever heard of Newton? Here's his first law:

Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.

That external force is friction (or gravity if you're pushing it uphill). That is why you need to work to keep a car moving. No other reason. How much force/ work does it take to keep a 200000lb space shuttle flying at 14000mph in space? 0.
You need work to get it up to speed, but to keep it going all it takes is overcoming friction. Nothing else.

That may be so, but you cannot make a statement like this:

Ultimately what I'm saying is yes, a heavier flywheel will make your car accelerate slower. But it's not because it robs power, it's because it takes energy to accelerate. The effect of it is very dependant on how fast the engine is accelerating. A heavy flywheel would have almost zero effect on how fast a car with a CVT accelerates, because the engine would be at a constant rpm.

Saying that a flywheel robs power is like saying that that a passenger robs power.


Modified by Lsos at 10:15 AM 8/29/2003

GSpeedR

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Mr.Integra &raquo;</TD></TR><TR><TD CLASS="quote">the normal force DOES NOT always equal to weight</TD></TR></TABLE>

We both know that the vehicle weight is the normal force in this case, there's no need to nitpick over it.

GSpeedR

The only time that an external force is NOT applied to the car (neglecting air resistance, and tire friction) is when the engine is at a constant speed.

This is where the confusion between everyone is. The rollers of the dyno, just like a car begin at rest. Just like Newton says, they are going to stay at rest until an external force is applied to them. So the wheels deliver a periodic rotational force, that the friction between the tire and road can transmit to make the car accelerate. A big *** truck still requires more force to accelerate the same amount as a small car, in every instance. Even in space. In space, I can't push a 200ton ship 2000mph because I can't deliver that much force, even though there is nothing slowing it down.

It will take a tremendous amount of force to accelerate that huge spaceship to 14005 mph, however.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">You need work to get it up to speed, but to keep it going all it takes is overcoming friction. Nothing else.</TD></TR></TABLE>

When the clutch is engaged and the tires are NOT slipping the linear position of the vehicle is directly depenedent upon the rotational position of the crankshaft because they are directly connected. The car will accelerate when the engine accelerates, it will decelrate when the engine decelerates, and it will remain at a constant velocity only when the engine remains at a constant velocity (rpm). If you accelerate the vehicle to 50mph and diconnect the engine from the wheels (disengage the clutch), then the heavier car will indeed remain at a constant velocity for a longer period of time. However, cars don't work this way, and certainly not on a dyno.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">Ultimately what I'm saying is yes, a heavier flywheel will make your car accelerate slower. But it's not because it robs power, it's because it takes energy to accelerate. The effect of it is very dependant on how fast the engine is accelerating. A heavy flywheel would have almost zero effect on how fast a car with a CVT accelerates, because the engine would be at a constant rpm.

Saying that a flywheel robs power is like saying that that a passenger robs power. </TD></TR></TABLE>

Since we are talking about a DynoJet and a Mustang dyno, both of which do not measure the power output at the crank, but rather at the wheels. A flywheel requires energy to be accelerated that would otherwise be transmitted to the wheels, decreasing the net power output seen at the wheels (if there was no flywheel we would see more power at the wheels), so yes, I say a flywheel "robs" power. Vehicle weight, however, comes into play at a later stage not seen by a dyno, as the rotational energy of the wheels is translated into linear energy, that must overcome a large mass. If there was a "dyno" that measured how fast an entire car moves, and determined the torque/power from such data, then a passenger could "rob" power. I wouldn't say that on the DynoJet or Mustang Dyno, however.

Read the flywheel thread from before, the flywheel is directly connected to the engine. So even at a constant engine speed, the heavy flywheel does not release its extra stored energy to the rest of the drivetrain as it is directly connected to the clutch, and a lighter flywheel will allow more torque to reach the drivewheels at that constant speed.


Modified by GSpeedR at 1:35 PM 8/29/2003

Lsos

I agree with most of everything you said, and in my other posts I have said the exact same things in different words.... except a couple things:

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">If there was a "dyno" that measured how fast an entire car moves, and determined the torque/power from such data, then a passenger could "rob" power. I wouldn't say that on the DynoJet or Mustang Dyno, however.</TD></TR></TABLE>

The only way a passenger could rob power on this hypothetical dyno is if you didn't add his weight to the vehicle weight when making calculations. If you input a car's curb weight into a G-tech accelerometer and then place 10 passengers into it, yeah, it will show that your car has less power than without the passengers. That is wrong though! It has the same power, just more weight that wasn't accounted for! That's what an inertial dyno does, it doesn't take flywheel weight into consideration when making power calculations.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">Read the flywheel thread from before, the flywheel is directly connected to the engine. So even at a constant engine speed, the heavy flywheel does not release its extra stored energy to the rest of the drivetrain as it is directly connected to the clutch, and a lighter flywheel will allow more torque to reach the drivewheels at that constant speed.</TD></TR></TABLE>

I don't understand the logic here. At a constant speed the flywheel doesn't absorb nor release any energy. That means that all the energy (torque) should reach the wheels.

raene

GSpeedR is the voice of sanity in this thread.

He's doing an excellent job explaining, I just want to make sure that anyone reading this thread in the future listens to him... that's why I'm posting

Lsos, the flywheel at a constant speed DOES absorb some energy to keep moving... 'tis the real world my friend, and all moving objects have a tendency to slow when faced with friction, esp. large rotation objects spinning in an air-filled chamber when they have little bumps all over their outer edges. The heavier flywheel will slow less rapidly but will take more energy to stay at a constant speed. The lighter flywheel will require less energy to be fed to it to maintain that constant speed and overcome wind resistance. The lighter flywheel will thus slow down more rapidly, but will require less energy than the larger to keep moving.

GSpeedR

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Lsos &raquo;</TD></TR><TR><TD CLASS="quote">The only way a passenger could rob power on this hypothetical dyno is if you didn't add his weight to the vehicle weight when making calculations.</TD></TR></TABLE>

Yep. The vehcile weight is not a necessary parameter on a dyno. With the resistance brake dyno, you would enter vehicle to determine a good frictional force to simulate the weight of the vehicle.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">If you input a car's curb weight into a G-tech accelerometer and then place 10 passengers into it, yeah, it will show that your car has less power than without the passengers. That is wrong though!</TD></TR></TABLE>

I figured part of the problem is semantics. The torque or power output from an engine is not dependent upon flywheels, transmission gears, wheels, etc. It is only dependent upon the force of the piston stokes spinning the crank. I know you have been saying this, and it is completely true. For other people the thread Where Torque Comes From has some good discussion regarding this.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">That's what an inertial dyno does, it doesn't take flywheel weight into consideration when making power calculations.</TD></TR></TABLE>

I don't have personal experience with an inertial dyno (or any dyno for that matter), but I understand that the inertial (and the Mustang style) measure the acceleration of two drums connected to the wheels. Anything moving part that has mass and is connected somewhere between the crankshaft and these drums will "absorb" energy/torque (require energy to spin). So the flywheel, the driveaxles, and the wheels and tires will have some affect on the "hp at the wheels" as measured by the dyno.

Example: pretend you have a 30 ton flywheel. If the engine cannot spin this, then it will certainly not be able to spin the dyno drums. The dyno will measure a zero torque output at the wheels.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">I don't understand the logic here. At a constant speed the flywheel doesn't absorb nor release any energy. That means that all the energy (torque) should reach the wheels.</TD></TR></TABLE>

All moving things have energy. They were supplied this energy to begin moving, and release this energy when stopped (a heavy flywheel is harder to stop because it has more energy that must be released). With the engine at a constant speed, the flywheel is accelerated to whatever corresponding speed the engine is at. The flywheel is not able to slow down however, because the engine will not let it. The flywheel required a certain energy to stay at a speed and it will continually require that energy.

Lsos

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">All moving things have energy. They were supplied this energy to begin moving, and release this energy when stopped (a heavy flywheel is harder to stop because it has more energy that must be released). With the engine at a constant speed, the flywheel is accelerated to whatever corresponding speed the engine is at. The flywheel is not able to slow down however, because the engine will not let it. The flywheel required a certain energy to stay at a speed and it will continually require that energy. </TD></TR></TABLE>

I still don't understnad why the flywheel would require energy to keep it spinning at constant speed. It doesn't want to slow down, it doesn't want to speed up...it just wants to remain at whatever speed it's at. Is there a formula to figure out how much energy is required to keep a flywheel spinning at constant speed?

We have raene supposedly agreeing with you on this...but yet he's not.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">Lsos, the flywheel at a constant speed DOES absorb some energy to keep moving... 'tis the real world my friend, and all moving objects have a tendency to slow when faced with friction, esp. large rotation objects spinning in an air-filled chamber when they have little bumps all over their outer edges. The heavier flywheel will slow less rapidly but will take more energy to stay at a constant speed. The lighter flywheel will require less energy to be fed to it to maintain that constant speed and overcome wind resistance. The lighter flywheel will thus slow down more rapidly, but will require less energy than the larger to keep moving.</TD></TR></TABLE>

You say here that ultimately it's the friction that needs to be overcome, nothing else....exactly what I've been saying all along. A heavier flywheel will have the same air resistance as a lighter flywheel. Coefficiend of drag is not determined by weight. Therefore the same amount of energy should be required to keep a heavy flywheel spinning as a light one. The light flywheel might seem like it needs less energy to bring it up to its nominal speed after it's been slowed down by air resistance....but it's been slowed down much more than the heavy flywheel, so in the end it all evens out and both require the same power to keep moving. Ultimately it's just friction you're trying to overcome, and there's not that much more with an 18lb flywheel as an 8lb flywheel. I wouldn't even say there's a quarter horsepower's worth of friction in there.

raene

I SO didn't want to get dragged into this

OK, I've reread the entire thread very slowly... Lsos, I agree with you a lot more now than I did before; simple semantics... I was getting confused in what you were saying

OK, now the question is: You have two wheels, each stuck on a stick. Both are identical diameters, both are identical thicknesses, both are identical in every aspect except one is made of wood weighing 5 lbs and the other is made of lead weighing 50 lbs. Both are spinning with the axis of rotation in the X plane, Z being vertical, gravity being applied downward through the Z plane, normal force upward through the contact points on the sticks. They have identical bearings on the sticks to allow the stick to rotate.

It doesn't really matter which will stop rotating first. To maintain speed, constant force must be applied; since it's rotational, constant torque. That is the key - constant torque, not torque being applied after each has slowed. Which requires less torque? The lighter. hp = (tq * rpm) / 5252, so hp is measured via torque. Imagining the stick as the crank, the lighter flywheel takes less torque to maintain rotational velocity at all engine speeds, even constant. The rest of the torque is fed to the drivetrain which spins the drums. I'm not saying it's a huge difference but it is indeed a difference.

AFAIK a Mustang dyno is not a load dyno, but an inertia dyno. And FWIW, all inertia dynos measure torque, not hp. HP is inferred by torque by repeated application of the above equation.


Modified by raene at 4:26 PM 8/29/2003


Modified by raene at 4:26 PM 8/29/2003


Modified by raene at 4:27 PM 8/29/2003

GSpeedR

The angular momentum of the flywheel remains constant under constant angular velocity (L = I w). So there is no net torque upon the flywheel (because the engine is at a constant speed), remember that torque is the time-derivative of angular momentum (derivative of a constant = zero).

K = 1/2 I w^2, such that K = 1/4 MR^2 w^2

E = E' (conservation of energy)

K(crankshaft) = K(flywheel) + K(clutch) + K(axles) + K(wheels) + ...

The only change we have made is the mass of the flywheel.

The Kinetic energy of the crank (initial energy) is the same, so to conserve energy, the kinetic energy of the clutch, axles, wheels, etc. must decrease. Conservation of energy works without acceleration, and I can produce the same argument using angular momentum, which is also conserved.

However, we must first understand that there is zero net torque at a constant engine speed.

Proof:
F = ma
Torque = r x F
Torque = r*F(tangential)
Torque = r*[m*a(tangential)]
Torque = r*m*r*(alpha), where r*alpha = a(tan) alpha is the angular acceleration (the rotational equivalent to a)
Torque = I*(alpha)

The engine (crank) is rotating at a constant speed, so the alpha = 0, and hence there is no torque. This requires one assumption however, that we consider the individual power pulses of the pistons to be a smooth linear event unstead of a sinusoidal torque "curve". Essentially, the engine is always producing a torque (because the power stokes of the pistons produce a tangential force upon the crank), but the net effect is that the flywheel rotates at a constant speed.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">We have raene supposedly agreeing with you on this...but yet he's not.
You say here that ultimately it's the friction that needs to be overcome, nothing else....exactly what I've been saying all along.</TD></TR></TABLE>

I believe the purpose of raene's discussion of friction is to show that the both flywheels are trying to slow down. However, there is a constant supply of kinetic energy from the crankshaft and so the flywheels are never given a chance to slow down. However, I'm not sure that we are arguing the same thing here. I am discreetely discussing the requirement of energy

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">Ultimately it's just friction you're trying to overcome, and there's not that much more with an 18lb flywheel as an 8lb flywheel. I wouldn't even say there's a quarter horsepower's worth of friction in there. </TD></TR></TABLE>

Friction is not the issue at hand here. Both flywheels experience the same amount of friction. You are arguing that the large inertia of the heavy flywheel places a smaller load on the engine because the flywheel will keep spinning for a longer period than the lightened version. This evens out the pulsation of torque (smoothes out the sinusoidal torque curve) making the car run more smoothly, yet it will also allow the engine to spin with less load. A heavy flywheel will require less energy input from the engine to continue spinning at a constant speed, yet requires more energy at each input, future acceleration notwithstanding.

At constant engine speed, the rotational inertia of the flywheel does not affect the ability of the engine to turn the wheels, because they are already turning.

Edit: I am not satisfied by my response at this point, but I have other stuff I need to do. I'll will do a bit of research and hopefully I can find something more definitive.

Edit 2: I got lazy and just started classes again, so I suppose I'll just leave it at that.


Modified by GSpeedR at 9:44 PM 9/5/2003

Del_Slowest

so which ones' better? lol