Rear bar problems.

[Mike]

Originally Posted by Mark42

Yeah, but right now I don't want to get into welding on new brackets. For one, I don't trust the local shops to get it on square, and the shops I do trust, will charge a bit because of the labor involved. Good is not cheap.

I hope it doesn't come to that.

they make bolt on relocation brackets, and even the weld on ones bolt on first and make it nearly foolproof

[sweetbmxrider]

yes, i had bolt on brackets and never had issues. didn't really wheel hop either with the car lowered a couple inches and 18's with a 295/35 tire. only time it did was in wet weather and a lot of throttle or the one time i had a nail in the tire and one was about 15 psi and the other was 32.

[Slow-V6]

Keep on wheel hoping and your going to break that Eaton rear end of yours!!

The problem is its a F-body with stock suspension and with 3.73's!! You need LCA's with Relocation brackets, Subframe connectors, TQ arm, and maybe even a PHB.

[Slow-V6]

Originally Posted by Mark42

Geez, its just a stock V6 with stock 16" wheels and tires! I don't get why it has this wheel hop problem.

You dont have a stock V6, you have a bolt on V6. My 98 V6 with the same mods as you put down 180rwhp and 210 rwtq. That equates to around 215 hp and 250ish tq! Stock its 200 hp and 225 tq. You have a rear wheel drive car making over 200 hp and 250 ftlbs of tq!! Welcome to life! You didnt have wheel hop with you old rear because it was only 1 wheel peel. Now your rear is trying to grip with both tire but its not getting the power down due to suspension or tires so it hops from both wheels spinning. The more you wheel hop the better chances are that your rear end is going to break!

Have you changed the fluid in the rear since you had it installed? If not I would and make sure you get some posi additive in there and check the wear on the gears!

[Mark42]

Originally Posted by Slow-V6

You dont have a stock V6, you have a bolt on V6. My 98 V6 with the same mods as you put down 180rwhp and 210 rwtq. That equates to around 215 hp and 250ish tq! Stock its 200 hp and 225 tq. You have a rear wheel drive car making over 200 hp and 250 ftlbs of tq!! Welcome to life! You didnt have wheel hop with you old rear because it was only 1 wheel peel. Now your rear is trying to grip with both tire but its not getting the power down due to suspension or tires so it hops from both wheels spinning. The more you wheel hop the better chances are that your rear end is going to break!

Have you changed the fluid in the rear since you had it installed? If not I would and make sure you get some posi additive in there and check the wear on the gears!

You really think the intake mode, exhaust and PCM tune are making that much HP? Cool!

I believe the torque increase caused by the 3.73's and the posi are the real culprits of the wheel hop.

I did change the rear lube (finally). Was surprised at how dirty the lube was after a thousand miles or so. Put in Mobile ! 70-90wt synthetic. Wear pattern on the gears looked good. I cleaned the magnet, and added a drain plug. But did not put in any posi additive because the Eaton TrueTrac does not use clutches, and they say the additive does nothing for it. So its straignt synthetic lube.

[Slow-V6]

Originally Posted by Mark42

You really think the intake mode, exhaust and PCM tune are making that much HP? Cool!

I believe the torque increase caused by the 3.73's and the posi are the real culprits of the wheel hop.

.

My 98 V6 made 168 rwhp bone stock. With a Flowmaster catback , filter ,and a whisper lid I made 181 rwhp. With headers, Aluminum D/S, and those mods I made 188rwhp. On a 125 shot of N20 and those mods I made 313rwhp and 415 rwtq. My car was a 98 with that angled TB and also was a 5-speed.

What tq increase are you talking about? You dont gain TQ from gears!

The real culprit is your suspension and tires! If you were able to get your power down to the ground then you would not be wheel hopping! Going from a open diff to a Posi you should be able to get more traction.

[Mark42]

Originally Posted by Slow-V6

......
What tq increase are you talking about? You dont gain TQ from gears!

With 3.08 gears and an open posi, it was hard to spin one tire. With 3.73 gears and a posi, the tires can be spun at will.

What force increased that now overcomes the resistance of two tires by changing gears?

The horse power did not change, nor did the engines foot lbs of torque rating at the flywheel. But the torque at the back axle has most definitely changed.

So yes, gears change torque.

[V]

Myth #1: Dynamometers only actually measure torque. Power is an abstract quantity that can't be measured, and must be calculated from the torque.
This belief is false and it will not be hard to show that. However, it involves a discussion of how dynamometers actually work, and before we can go there, we have to begin with a brief review of some fundamental physics. I'll try to avoid equations as much as possible, and instead I'll try to instill an intuitive understanding of the basic concepts, which is something that classroom lectures and textbooks often fail to accomplish for many people.
Isaac Newton taught us that any object having a mass M, has an intrinsic resistance to changes in its present momentum when external force is applied. The acceleration of the object is proportional to the strength of the external force, and is inversely proportional to the mass of the object. These observations are summed up succinctly by the infamous equation: F = M x A. The true essence of force, mass and acceleration is embodied in this equation, and this equation tells us how force or mass can be detected and measured. When an object's momentum is not constant, this indicates the existence of an external force, and the strength and direction of that force can be determined by investigating the acceleration. Conversely, if an object is accelerated under the influence of a well-known force, its mass can be determined.
Whenever the aggregate force acting on an object is not directed through the object's center of mass, the object is caused to rotate, and this rotation will continue to accelerate as long as this condition persists. The laws governing rotational motion parallel the laws governing straight-line motion. Torque replaces force, the moment of inertia replaces the mass, and the angular acceleration replaces straight-line acceleration. Whereas velocity is measured in terms of absolute distance in straight-line motion, in angular motion we measure velocity in terms of degrees of rotation per unit of time, which we may call angular distance. Angular velocity is the instantaneous rate at which angular distance is covered. Angular acceleration is the instantaneous rate at which the angular velocity is changing, and is proportional to the applied torque, and inversely proportional to the moment of inertia. The torque that is associated with an applied force is equal to the force (magnitude) multiplied by the distance from the center of mass of the object to the line of force (measured at right angles to the line of force). For an object that is constrained to rotate about a fixed axis of rotation, i.e., a wheel, the distance is measured from the axis of rotation to the line of force. Two objects with identical mass will not in general have the same moment of inertia. For example, if two perfect spheres of identical mass are made of different materials having different densities such that they are not the same size, the larger one will have the greater moment of inertia. In the case of objects having fixed axis of rotation, if two such objects have the same overall diameter and density but differ in the radial distribution of the mass, the one with the mass concentrated closer to the axis of rotation will have a lower moment of inertia.
Work, in the case of linear motion, is the force multiplied by the distance through which the force was exerted and the object was moved. Work and energy are equivalent; it takes a specific quantity of energy to move an object a given distance in opposition to a given resistive force, and that amount of energy is the same no matter how quickly the movement is performed. Power is simply the instantaneous rate at which work is performed (or energy is expended), and is therefore equal to the product of the force and the velocity. If you measure the force required to move an object, the distance that it moves, and the time interval required to move it that distance, then you can calculate the average power that was applied throughout that motion and over that interval of time. Note, though, that average power is no more equivalent to power than average speed is equivalent to speed.
In the case of rotational motion, work is the torque multiplied by the angular distance through which the object is rotated. The parallel with linear motion is again applicable: it takes a specific quantity of energy to rotate an object a given angular distance, and that quantity of energy is the same no matter how quickly the rotation is performed. Each time an object undergoes one complete rotation, a fixed amount of work will have been performed. It follows that the amount of work performed in a given interval of time will depend on the angular distance covered in that time interval, and the rate at which the work is performed will therefore depend on the angular velocity. In rotational motion, power is equal to the product of the torque and the angular velocity, which may be measured in revolutions per minute or rpm.
Let's reiterate the important rules discussed thus far:
Straight-line motion:

• Force = Mass x Acceleration
  Work = Force x Distance
  Power = Force x Velocity

Rotational motion:

• Torque is measured by taking the right-angle distance from the line of force to the axis of rotation, and multiplying the force by that distance. Torque = Moment of Inertia x Angular Acceleration
  Work = Torque x Angular Distance
  Power = Torque x Angular Velocity
  

We should note that these equations are of the type of equation known as "identities". As opposed to stating conditions on the value of a variable, they simply state that the quantities on the two sides of the equal sign are identical quantities.
Torque Multiplication
There is one more important theoretical concept that we have to discuss, that being an important consequence of the relationship among power, torque, and angular velocity as given in the last equation above.
If you use a gearbox to achieve a different rotational rate at the wheel as compared to the engine, the torque at the wheel will not be the same as the torque at the engine crankshaft, even if there are no frictional losses. The reason that this is so is that the rate at which work is done must be constant throughout the system (excepting only for the loss of energy within the system due to friction), and because power is the product of the torque and the angular velocity.
Analogies between mechanical systems and electrical systems are often insightful, so I'll ask you to indulge me for a moment while I digress and talk about electrical transformers. The voltage transformer is the electrical equivalent of the mechanical gearbox. A voltage transformer (which, by the way, only works for alternating current) takes advantage of the inductive coupling between the primary and secondary windings to transform the voltage. In electricity, power is the product of the voltage and the current. Because the rate at which work is performed must be constant throughout the system, the product of the voltage and the current in the secondary must be the same as the product of the voltage and the current in the primary – at least in the case of an ideal transformer where there is no loss to heat. If the voltage is reduced or increased by a factor of N, the current will be inversely changed in that same proportion.
Okay, back to mechanics. Because power is the product of the torque and the rotational speed, and because the power must be constant throughout the system, the transmission increases the torque in the same proportion by which it reduces the rotational speed. If you measure the torque at the rear wheel, you have to use the overall reduction ratio to calculate the engine torque. At a given engine speed, the rear-wheel torque will depend on what gear is selected. Any dynamometer chart that shows torque must be engine torque or else it would be applicable only to a specific gear and the chart would have to specify the gear, which would not be especially useful. Note, however, that if the torque is measured at the rear wheel and then the overall reduction ratio is used to calculate the engine torque, the result will not be the same as the result that you would get if you rigged the dynamometer directly to the crankshaft, because the measured rear-wheel torque is subject to power train losses. Nevertheless, any dynamometer chart that shows torque and that does not specify the gear is most definitely the engine torque, albeit adjusted for drive train losses.
Metrics of torque and power; dynamometer theory
Okay, at last we're ready to talk about the metrics of torque and power, and about how dynamometers work. Contrary to what the urban myth says, there is no fundamental physical reason by which it is impossible to measure work or power directly on a dynamometer, i.e., without deriving the power through the torque measurement. Furthermore, even if it were true that the only way to measure power is by derivation through the measured torque, that fact would have no bearing on the significance of power.
In fact, measuring power on an inertial dynamometer is no more difficult than measuring torque, and is not predicated on the measurement of torque. Whenever the work performed goes to increase the kinetic energy of an object, power is the instantaneous rate of change of the kinetic energy. It is a trivial matter to calculate the kinetic energy of a moving drum if its rotational speed and moment of inertia are known. The rate of change of the kinetic energy can be determined in a manner essentially identical to the traditional method used to determine the angular acceleration of the drum, which is needed in order to calculate the torque.
In an inertial dynamometer, the engine is allowed to accelerate an inertial drum as quickly as it is able. Newer inertial dynamometers use an inertial accelerometer to give the measure of the angular acceleration continuously. With the traditional method, however, each time the drum rotates through a fixed number of degrees, an electrical pulse is generated which triggers the recording of the elapsed time. >From those data points, the average angular velocity can be computed for each of the individual time intervals between adjacent points, by dividing the fixed angular distance by each of the time intervals. Then an average acceleration is calculated for each adjacent pair of average angular velocity values, by dividing the difference in the adjacent average angular velocity values by the corresponding time difference. For each average acceleration value, the average torque over the corresponding time interval is then found by multiplying the average acceleration by the moment of inertia of the drum.
On an inertial dynamometer, the average rate of change of kinetic energy for each time interval can be determined using a method that is essentially identical to the traditional method used to determine the acceleration. The average kinetic energy for a given time interval can be calculated from the average angular velocity over the interval, using the formula: K.E. = ½M x V². (It is necessary to factor in a standard correction to account for the fact that the square of an average is not the same as the average of the squares.) If the difference between the average kinetic energies for two adjacent time intervals is divided by the corresponding time difference, the result will be the average power for that time interval. I want to emphasize that this is not in any way an impractical, far-fetched approach. It is for all intents and purposes identical to the ubiquitous, traditional method used to determine the acceleration and the torque on an inertial dynamometer.
It is also possible to measure power independently of torque on an inertial dynamometer that is equipped with an accelerometer. A computational technique known as numerical integration can be used to derive the angular velocity at many closely spaced points, from the accelerometer readings. The kinetic energy can then be calculated at each of those points. The average rate of change of the kinetic energy between each of those points can then be calculated the same as with a traditional inertial dynamometer, by dividing adjacent pairs of kinetic energy values by the corresponding time differences.
Inertial dynamometers vs. brake dynamometers
With an inertial dynamometer, the engine is allowed to spin up as quickly as it can, accelerating the drum as quickly as possible. Consequently, the measured results reflect the engine's ability to increase its work output rapidly, which ability is more greatly influenced by the engine's own internal inertia than is a measurement of steady-state work output. This factor is treated less rigorously than it could be: there is generally no attempt to quantify the effect that the engine's reluctance to rapidly increase its output has on the measured results, and the measured results are not carefully distinguished from a measurement of steady-state output such as would be obtained on a brake dynamometer. Be that as it may, an inertial dynamometer is a more realistic test of an engine's ability to accelerate a vehicle, whereas a brake dynamometer is at least as realistic when the question at hand is an engine's ability to climb a steep hill at steady speed. Turbine engines have a very high thrust to weight ratio, but due to the combined moments of inertia of all of the turbine blades, they are slow to speed up and slow to slow down. Jay Leno's turbine-powered motorcycle would fare much better on a brake dynamometer than it would on an inertial dynamometer.
With a brake dynamometer, the engine speed is increased in small steps and held steady at each step in order that the braking torque that is required to hold the drum at a steady rotational rate ("dynamic equilibrium") can be read. The throttle is kept fully open, but closed-loop feedback using the drum's motion sensors is used to regulate the braking force in order to hold the drum's angular velocity constant. Several different types of brakes are used, but regardless the principle of using feedback to hold the drum velocity and the engine rpm constant is the same. Depending on the type of brake used, transducers will be used to take measurements of force, or hydraulic pressure, or hydraulic flow, or electric voltage, or current, etc. Calibration factors and formulae will be applied, all in order to determine the opposing torque that is applied via the brake to the drum in order to maintain the drum at dynamic equilibrium.
To summarize, with an inertial dynamometer, the drum is allowed to accelerate as rapidly as the engine can make it accelerate, and the torque that the wheel exerts on the drum is determined using the data from the drum's motion sensors and simple mathematical methods to obtain average rates of change over small time intervals. With a brake dynamometer, the torque that the wheel applies to the drum is found more directly by applying a regulated counter-torque, using transducers to measure the braking force, which must be calibrated and which can be a significant source of error if not done properly with a good understanding of metric processes.
Myth #2: Engine torque is the same as the dynamometer drum torque.
Peculiar though it is that anyone with any real knowledge of the subject could be this misinformed, it is apparent that many of the same people who believe that dynamometers can't measure power, believe this as well. Regardless, the true nature of the relationship between engine torque and dynamometer drum torque is sufficiently important such that I would be remiss if I were to omit that subject from this article.
Regardless of which type of dynamometer is used, the dynamometer measures the torque that is applied to the drum by the wheel, which is applied by way of a longitudinal, frictional force between the wheel and the drum. That longitudinal force can be found by dividing the measured drum torque by the drum's radius. The rear wheel torque can then be found by multiplying the longitudinal force by the wheel's radius. Of course, the longitudinal force is not usually of interest unless perhaps you wish to calculate the theoretical acceleration of your vehicle. To get the rear wheel torque more directly, you divide the drum torque by the drum radius and multiply by the wheel radius. To convert the rear wheel torque to engine torque, you divide the rear wheel torque by the overall reduction ratio. The overall reduction ratio is found by multiplying together the primary reduction ratio between the crankshaft and the transmission's input shaft, the final reduction ratio between the transmission's output shaft and the rear wheel, and the transmission ratio that depends on the specific gear that was used during the dynamometer run.
Myth #3: Acceleration will be greatest when the engine speed matches the engine speed where the torque peak occurs.
Myth #4: Power is an abstract, derived quantity that is meaningless insofar as concerns the goal of selecting shift points that yield the greatest acceleration.
In accordance with the well known relationship among force, mass and acceleration given by Newton's well-known F = M x A, the acceleration of the vehicle is proportional to the longitudinal force at the tire contact patch. (We can ignore the fact that some of the force goes to angular acceleration of the wheels; this will not alter the validity of the analysis or the conclusions.) If the engine torque (adjusted for power train losses) is known for a given engine speed, the corresponding longitudinal force at the contact patch can be calculated by multiplying that torque by the overall reduction ratio and then dividing by the rear wheel radius. So long as the reduction ratio remains fixed at a given gear, the wheel torque, the longitudinal force and the acceleration will all have their maximum values at the engine speed where the torque has its maximum value. That is fine and good, and obvious, but it bears little on the question of how the maximum rear wheel torque and acceleration are determined and maximized when you can use gear selection to alter the engine speed at a given wheel speed.
When you use gear selection to change the relationship between the engine speed and the wheel speed, two things happen. First, the change to the engine speed generally alters the engine torque. Second, the change to the overall reduction ratio changes the torque multiplication between the engine and the rear wheel.
The vehicle will have greatest acceleration at a given wheel speed when the gear selected results in the greatest rear wheel torque. If a gear is selected that puts the engine speed somewhat higher than the engine speed at which the engine torque peak occurs, that numerically lower gear will result in greater torque multiplication and the rear wheel torque will be greater even though the engine torque will be somewhat less than its maximum value. This will be true so long as the torque curve remains reasonably flat above its peak value. Even with engines that have a pronounced peak in the torque curve, the torque curve will be essentially flat for some distance near the peak. If the gear ratios are properly matched to that torque curve and the wheel speed is within the normal operating range, it will always be true that the acceleration will be greatest when the engine speed is higher than the engine speed at which the torque peak occurs.
This begs the question of how the acceleration is related to the power. That relationship is slightly complicated by the coupling of several facts. Power is the rate of change of the kinetic energy, kinetic energy depends on the square of the velocity, and it is not intuitively obvious whether the velocity should change most rapidly at the same engine speed where the square of the velocity changes most rapidly. However, by combining two of the identities (equations) that were presented near the beginning of this article, we can derive an identity that describes acceleration as a function of power, velocity and mass.
Combining the two identities: Power = Force x Velocity; Force = Mass x Acceleration, we get another useful identity: Power = Mass x Acceleration x Velocity. Rearranging the terms to isolate acceleration, we get an identity that describes acceleration as a function of power, velocity and mass:

• Acceleration = Power / (Mass x Velocity)

This formula provides us with a couple of useful facts. For one, it tells us that for a given power and mass, the acceleration decreases as the velocity increases, which is consistent with the fact that kinetic energy increases as the square of the velocity.
Probably of greater interest to most readers is the fact that for a given velocity and mass, the acceleration is directly proportional to the power. There are two distinct and meaningful consequences of the proportionality between acceleration and power. First, at a given speed, the acceleration will be greatest when the gear selected is such that the power at the associated engine speed is the greatest among all the gears. Second, given any two vehicles with identical mass (to include the mass of the rider), the one with the more powerful engine will exhibit the greatest maximum acceleration, regardless of which one produces the most torque. To be sure, the power to weight ratio determines the vehicle's maximum acceleration, which of course is why that ratio is frequently quoted.
The simplest way to assess what the acceleration can be at any given wheel speed, is to convert that wheel speed to the equivalent engine speed for each gear, and then look at the power curve to find which of those engine speeds yields the most power. You can also answer this question from the standpoint of rear wheel torque, but then after looking up the engine torque for each of the engine speeds, you have to turn back around and multiply those engine torque values by their corresponding reduction ratios in order to find the rear wheel torque for each gear. You'll get the same result either way; if you don't, then at least one of the two graphs is in error. However, the torque method requires more computational work as compared to simply looking at the power curve, so why would anyone want to do that?
Just for grins, let's consider an actual example. The March, 2003 issue of Motorcycle Consumer News has a dynamometer chart for the 2003 FJR1300. The Yamaha shop manual gives the primary, secondary, and gear-specific reduction ratios for the five gears. I want to find the optimal road speed for shifting from 1st gear to 2nd gear. I worked out the value of the multiplier for converting the road speed in mph to engine speed in rpm. That multiplier is 60, and the primary and secondary reduction ratios are already factored in to that, but I still have to multiply by the gear-specific reduction ratio. At 50 mph, the engine speed will be 7590 rpm in 1st gear (more or less depending on the accuracy of my measurement of the wheel radius), and looking at the dynamometer chart, I read about 120 hp for that engine speed. That's very close to the peak power, but the peak is located a little higher, just shy of 8000 rpm. Therefore, I expect that I should go beyond 50 mph in 1st gear before shifting to 2nd gear.
At 60 mph, the engine speed will be 9104 rpm in 1st gear, which is just off the chart because when you cross 9000 rpm you're into the red zone, but I can visually extrapolate the curve and estimate 110 hp, well below the maximum hp. (The error in my measurement of the wheel diameter might be the reason for the engine speed being in the red zone at 60 mph in 1st gear, but that error won't effect the essential comparative result, so I will ignore the red zone.) I can't know whether I should shift to 2nd before reaching 60 mph, until I check what the power will be in 2nd gear at 60 mph and confirm that it is more than 110 hp. At 60 mph in 2nd gear, the engine speed will be 6380 rpm, and the chart indicates about 105 hp, which is slightly less than the 110 hp that I'll get at 60 mph if I remain in 1st gear. Therefore, if this chart is correct and subject to the accuracy of my measurement of the rear wheel diameter, the implication is that for purposes of maximum acceleration, I should wait until I reach a speed slightly higher than 60 mph before shifting from 1st to 2nd.
But what happens if I compare the engine torques? At 60 mph in 1st gear, at 9104 rpm, the engine torque has dropped to about 60 lb. ft., a value well below the peak value of about 87 lb. ft. At 60 mph in 2nd gear, at 6380 rpm, the engine torque is about 85 lb. ft., which is very close to the peak. If I were to believe that the acceleration is greatest when the engine torque is greatest, then I would conclude that I should shift well before I reach 60 mph, probably somewhere in the neighborhood of 50 mph. But let's see what happens when we convert those engine torque values to rear wheel torque. When 60 lb. ft. is multiplied by the overall reduction ratio for 1st gear, the rear wheel torque at 60 mph in 1st gear is found to be about 658 lb. ft. When 85 lb. ft. is multiplied by the overall reduction ratio for 2nd gear, the rear wheel torque is found to be about 653 lb. ft. In other words, I have slightly more rear wheel torque at 60 mph in 1st gear than I do in 2nd gear, which suggests that I should wait until I reach a speed slightly higher than 60 mph before shifting from 1st to 2nd, which agrees exactly with the result that I got when I compared the power!

[Slow-V6]

Originally Posted by Mark42

With 3.08 gears and an open posi, it was hard to spin one tire. With 3.73 gears and a posi, the tires can be spun at will.

What force increased that now overcomes the resistance of two tires by changing gears?

The horse power did not change, nor did the engines foot lbs of torque rating at the flywheel. But the torque at the back axle has most definitely changed.

So yes, gears change torque.

So do you gain power from gears like you stated earlier or not?

Originally Posted by Mark42

Its just a posi and gears that are causing:

- More hp/torque to the rear at launch due to the gear change.

- More resistance to wheel spin due to the Eaton Truetrac, which in turn puts more stress on the drive-line components (like LCA bushings and T/A bushing).

- Mild hp increase due to tune and exhaust.

So if you dynoed your car before the gears and dynoed after the gears you would have more tq at the rear wheels after the gear install? If you have more power at the rear end then you have more power at the motor as well! I had 3.23's in my v6 and also had a 3.73 and a 4.10 gear rear end in it and it put down the same tq with all 3 rear ends.

I am thinking that with a lower gear there are more teeth which makes it get up to speed faster with less effort from the motor! I got the best gas miles in my V6 with the 4.10 gear rear end. I didnt have to downshift to a lower gear on the highway to pass a car, and I could just leave it in second gear at a light and take off like that.

Wheel hop is not spinning tires!! If you could spin tires at will with your new rear then why are you complaining that everytime you try to do a burnout or just get on it the car wheel hops? Get the supporting suspension mods to fix the issue or better tires! Its that simple! You have to pay to play. My 02 T/A wheel hops while doing burnouts so I dont do burnouts in it! I already broke 1 rear end and it was probably due to me wheel hopping at the track!

[Slow-V6]

Originally Posted by V

Myth #1: Dynamometers only actually measure torque. Power is an abstract quantity that can't be measured, and must be calculated from the torque.
This belief is false and it will not be hard to show that. However, it involves a discussion of how dynamometers actually work, and before we can go there, we have to begin with a brief review of some fundamental physics. I'll try to avoid equations as much as possible, and instead I'll try to instill an intuitive understanding of the basic concepts, which is something that classroom lectures and textbooks often fail to accomplish for many people.
which agrees exactly with the result that I got when I compared the power!............................................

That is alot to read and I hate reading!

[WildBillyT]

Since it's late and I would probably just drool on my keyboard instead of type a legible response:

http://www.blueink.com/CLASS/physcom1/gear.htm

Cliffs notes:

Simply put, torque at work (such as at a wheel) is your motor's torque times your gear ratio.
Motor Torque x gear ratio = torque at the wheel

Lets say we have a 10rmps motor that is capable of 5 oz Torque (we know this from our motor spec.)

Lets say we have 2 gears. Our input gear (attached to our motor) has 10 teeth Our output gear has 50 teeth

Our Gear ratio is 5:1

Motor Torque x gear ratio = torque at the wheel

5oz x 5:1 = 25 oz

What if our gear ratio were 1:3 ?

5oz x 1:3 = 1.6oz

OR

http://www.carcraft.com/techarticles...rks/index.html

In addition to changing the direction of power flow by 90 degrees (from the driveshaft to the axles), the purpose of the rearend gears is to multiply the torque delivered by the engine and transmission. Gears can be thought of as complex levers. In other words, they provide a mechanical advantage that multiplies work--in this case, torque--to help the engine's power move the vehicle. Lower gears are like a longer lever: They provide more mechanical advantage. Higher gears are like a shorter lever: They provide less mechanical advantage. It's similar to when you use a long breaker bar instead of a short ratchet handle to remove tight lug nuts. Just like a long bar puts more torque on a lug nut, lower axle gears provide more torque to the wheels.

Mark,

Allow me to be blunt for a minute.

Cut the **** and upgrade your suspension parts from the cost cutting stamped steel factory garbage.

[Slow-V6]

Originally Posted by WildBillyT

Since it's late and I would probably just drool on my keyboard instead of type a legible response:

http://www.blueink.com/CLASS/physcom1/gear.htm

Cliffs notes:



OR

http://www.carcraft.com/techarticles...rks/index.html

So your rear wheel hp is not only based on how much power you have it is based on what gear ratio you are running? So with a lower gear you are not gaining more power it is just freeing up more power by being more eff?

[WildBillyT]

Originally Posted by Slow-V6

So your rear wheel hp is not only based on how much power you have it is based on what gear ratio you are running? So with a lower gear you are not gaining more power it is just freeing up more power by being more eff?

Gear ratio will drop out when calculating horsepower. RPM adjusts to make up for the extra torque in the equation.

Think of it this way. An engine on a stand will make an set amount of torque at a given RPM.

If you put that in a car with a manual transmission and put it in 6th gear at a stop you won't go anywhere. Not enough torque. Put it in first and it's smoke city. No change was made to the engine. It still makes the same amount of torque at a given RPM regardless of what's behind it. The torque is multiplied by the gears. But when you are calculating horsepower, RPM is factored in. With more torque there is less RPM so HP stays the same.

I'm sure you get the idea even though I'm sure I'm missing something.

Here's a proof of it done by someone more skilled than I:
http://craig.backfire.ca/img/gear-formulas.png

[Slow-V6]

Originally Posted by WildBillyT

Gear ratio will drop out when calculating horsepower. RPM adjusts to make up for the extra torque in the equation.

My bad. It is late. I ment to say TQ instead of hp.

[V]

i figured my last post would fit in well with all the other gibberish.

[Mark42]

Originally Posted by Slow-V6

So do you gain power from gears like you stated earlier or not?

The engine is making the same hp and tq now as it did before the gear change. When I said it was because of more hp, I was referring to the fact that the rpm is higher with the lower gears, therefore the engine is into a higher hp rpm range than before. Not that installing gears magically creates more hp output.

Originally Posted by Slow-V6

Wheel hop is not spinning tires!! If you could spin tires at will with your new rear then why are you complaining that everytime you try to do a burnout or just get on it the car wheel hops?

Well, yes you can have spinning tires and wheel hop at the same time. Unfortunately for me, that is exactly what I experience, and the pavement proves it by showing a foot of solid rubber mark, then a 3" gap, then 6" of rubber, 3" gap, etc.

I'm not trying to do smokey burnouts. Just trying to launch the car as hard as I can to get the best 0-60 time. In my other cars, the best acceleration usually came with a small amount of wheel spin.

I don't want to throw money at a lot of heavy duty suspension upgrades, like LCA relocation brackets, new tubular LCA's, new springs, heim ends, etc, when a stock V8 camaro/firebird making more hp and tq than my car, can launch hard and have no wheel hop. The factory setup works for thousands of V8 Fbodies, and it should work for my V6 making considerably less hp and tq.

So experience tells me that there is probably something wrong, worn or broken that is preventing the rear suspension from performing properly. So before spending hundreds on race grade parts, I want to be sure that the parts I have are good, because they should be doing the job.

Regarding the discussion of torque and the affect of gears, think of this: A bolt that requires 90 ft/lbs of torque to unscrew it has a 1 foot long box end wrench on it with a 10 lb weight attached to the end of the wrench. That is 10 ft/lbs of torque and will not unscrew the bolt. Hang 100 lbs on the box wrench, and the bolt will unscrew. But if you put a 10 foot extension on the box wrench, and hang the 10 lb weight on it, it will also unscrew the bolt.

Why? Its still just the same 10 lbs pulling on the bolt. So why did 10 lbs break the bolt free on a 10 foot long wrench, but not a 1 foot long wrench?

Because of leverage. The extension multiplies the torque x 10, creating 100 ft/lbs of torque from the same 10 lb weight.

Gears do the same they multiply the available torque. Same as the trans does, etc.

But even with the approx. 20% increase in gear ratio, I think the stock parts should handle the job.

Thanks for all your responses. You have made some great points that have made me rethink a few things. And that's what these forums are all about, no?

[sweetbmxrider]

fbodies with stock suspension are notorious for having wheel hop issues. don't know where you sourced that info from.

[Slow-V6]

Originally Posted by Mark42

The engine is making the same hp and tq now as it did before the gear change. When I said it was because of more hp, I was referring to the fact that the rpm is higher with the lower gears, therefore the engine is into a higher hp rpm range than before. Not that installing gears magically creates more hp output.




Well, yes you can have spinning tires and wheel hop at the same time. Unfortunately for me, that is exactly what I experience, and the pavement proves it by showing a foot of solid rubber mark, then a 3" gap, then 6" of rubber, 3" gap, etc.

I'm not trying to do smokey burnouts. Just trying to launch the car as hard as I can to get the best 0-60 time. In my other cars, the best acceleration usually came with a small amount of wheel spin.

I don't want to throw money at a lot of heavy duty suspension upgrades, like LCA relocation brackets, new tubular LCA's, new springs, heim ends, etc, when a stock V8 camaro/firebird making more hp and tq than my car, can launch hard and have no wheel hop. The factory setup works for thousands of V8 Fbodies, and it should work for my V6 making considerably less hp and tq.

So experience tells me that there is probably something wrong, worn or broken that is preventing the rear suspension from performing properly. So before spending hundreds on race grade parts, I want to be sure that the parts I have are good, because they should be doing the job.

Regarding the discussion of torque and the affect of gears, think of this: A bolt that requires 90 ft/lbs of torque to unscrew it has a 1 foot long box end wrench on it with a 10 lb weight attached to the end of the wrench. That is 10 ft/lbs of torque and will not unscrew the bolt. Hang 100 lbs on the box wrench, and the bolt will unscrew. But if you put a 10 foot extension on the box wrench, and hang the 10 lb weight on it, it will also unscrew the bolt.

Why? Its still just the same 10 lbs pulling on the bolt. So why did 10 lbs break the bolt free on a 10 foot long wrench, but not a 1 foot long wrench?

Because of leverage. The extension multiplies the torque x 10, creating 100 ft/lbs of torque from the same 10 lb weight.

Gears do the same they multiply the available torque. Same as the trans does, etc.

But even with the approx. 20% increase in gear ratio, I think the stock parts should handle the job.

Thanks for all your responses. You have made some great points that have made me rethink a few things. And that's what these forums are all about, no?

Have you owned other 4th gen f-bodies? Everyone I have owned wheel hopped until I started to do suspension mods! I have owned 2 V6 models and 2 V8 models and they all do pretty much the same for me with the stock suspension and a 3.73 or lower gear! The 3rd gens stock suspension was alot better then the 4th gens but they also made less power. I didnt get any wheel hop from my 4 3rd gens I have owned.


You said you didnt have this problem before the new rear. Maybe its the New rear end? Maybe its not setup correctly, or the Eaton posi works different then a stock F-body LSD? Who knows! All that everyone knows is just like how the stock 10 bolts are crap and should be changed out for a stronger rear end that the stock suspension componants are crap that if you want to run your car you need to change them out also!!

[LTb1ow]

Mark42... you are one crazy guy.

My V8 would wheel hop/spin and do everything BUT go fast.

LCA relo brackets, LCAs, and panhard bar solved that. Simple. The wheel does not need to be reinvented.

[Slow-V6]

Originally Posted by LTs1ow

Mark42... you are one crazy guy.

. The wheel does not need to be reinvented.

Nore does the Oil filtration system!!

[Mark42]

Originally Posted by sweetbmxrider

fbodies with stock suspension are notorious for having wheel hop issues. don't know where you sourced that info from.

From my own Camaro and my friends Camaros and Firebirds. They didn't have wheel hop and all but my Camaro were 4 gen V8's. But for the most part, those cars were stock, with no upgrades other than wheels and exhaust. Don't know if that makes a difference.

[LTb1ow]

So your friends stock V8s don't have wheel hop, but your V6 does.

Amazing.

[Mark42]

Originally Posted by Slow-V6

Nore does the Oil filtration system!!

I do believe there is another thread for beating on me about oil systems! This one is for beating on me about suspension systems!

Originally Posted by LTs1ow

Mark42... you are one crazy guy.

My V8 would wheel hop/spin and do everything BUT go fast.

LCA relo brackets, LCAs, and panhard bar solved that. Simple. The wheel does not need to be reinvented.

Ok, ok, ok!!!!!

So, just hypothetically speaking, lets assume that somewhere at some time, there was a Camaro or Firebird that had wheel hop due to an inadequate factory suspension design. Remember, this is just conjecture, and is not making a case that any F-bodies have an issue in this area.

So if I were theoretically need to do brackets, and I'm not racing, and not pushing monster power, what brackets would you recommend? Aren't longer control arms needed.

[sweetbmxrider]

any of the major suspension companies will do, spohn, bmr, umi, etc. you can get them in bolt in or weld in versions.

they retain stock length control arms.

[NastyEllEssWon]

josh from js performance is a umi dealer. check with him to see what kind of deals he can get you.