9 replies to this topic

[Gnielsen]

Pound for pound, reducing rotating weight will result in a greater performance improvement than reducing non-rotating, where performance improvement is measured by the resultant change in vehicle acceleration. This statement is generally true, but there appears to be no shortage of misunderstanding when it comes to quantifying this mass benefit ratio (MBR). What is the magnitude of the benefit ratio? This is a simple question, with a surprisingly complicated answer. For about a dozen years my work as an engineer involved the analysis of motor vehicle dynamics, at which time I had the opportunity to examine this very question: The answer to which is reproduced here.

Quantifying the MBR, unavoidably, requires the application of some math and physics. Consequently, I have structured this article into two parts:

• Part 1 describes the physical phenomenon and presents the equation that quantifies the MBR, all while introducing as few equations as possible. Some interesting findings resulting from this phenomenon are summarized in part 1.

• In Part 2 the equation quantifying the MBR is derived from Newton’s Law’s of motion. Part 2 also includes a discussion of the effect of air drag on the MBR.

Part 1
Figure 1 below depicts a model of a car and wheel. This car has non-rotating mass M1, and the wheel represents rotating mass M2. A torque is generated by the engine and transmitted to the driving wheels and tire via the transmission and differential. The torque at the wheel (T_w) applies a force to the road surface of magnitude F(t/r) = T_w /R . [note the subscript in the term F(t/r) means that the force is applied is from the tire to the road]. From Newtons third law, the road exerts a force on the tire (F(r/t), also shown in figure 1, that is equal in magnitude but opposite in direction to the force of the tire on the road. It is important to note that the only external force acting on the car is the force of the road pushing on the tire F(r/t) - neglecting air drag which is addressed later.

 wheel car 2.jpg   23.15K   0 downloads

Figure 1.

In order to accelerate the car in the direction shown the applied torque has two jobs to do at the same time: it must produce the force F(r/t) that accelerates the car and tire’s linear inertia (i.e. M1 and M2), and a portion of it must also apply a torque to the wheel to overcome its rotational inertia, such that the wheel’s axle accelerates forward at the same rate as the car. From Newton’s second law:

F(r/t) = (M1 +M2) a Equation 1

Equation 1 applies to linear acceleration only. The rotational analogy of Equation 1 takes a similar form. Recall from Newton’s third law: F(r/t) = -F(t/r), i.e. the force applied by the road to the tire is equal in magnitude but opposite in direction to the force of the tire acting on the road F(t/r). Therefore, the torque applied by the tire to the road is equal in magnitude but opposite direction to the torque applied by the road to the tire, i.e. T(r/t) = -T(t/r).:

T(r/t) = Iα Equation 2

• Where T(r/t) is the torque applied from the road to the tire. This torque has a force component equal to F(r/t) and a distance component equal to the radius of the wheel ®.
• The Greek letter alpha (α) is the rotational acceleration of the wheel. This rotational acceleration is directly proportional to linear acceleration according to the equation:

α = a/R Equation 3

• The capitol letter I in Equation 2 represents the rotational inertia, which is the wheel’s resistance to spinning up. For a wheel shaped object the equation for rotational inertia is:

I = (M2)r² Equation 4

Where M2 is the mass of the wheel and r is its radius of gyration.

The radius of gyration represents the distance, measured from the axle toward the tire’s outer circumference, at which the entire mass of the wheel (M2) would be located such that this idealized point-mass model of the wheel has the same rotational inertia as the actual wheel. The value of r must be greater than 0, and less than the outer radius of the tire, i.e. less than R.

Inspection of equations 1 and 2 reveals that the applied torque at the wheel is responsible for producing F(r/t), i.e the force that linearly accelerates the vehicle. If the transmission and differential convey horsepower with 100% efficiency, then the output torque of the transmission/differential is given by the gear ratio. For example, assume a gear ratio (for the combined transmission and differential) of 12:1, i.e. for every 12 rotations of the engine the differential output shaft makes one revolution. In this case the output torque at the differential’s output shaft is 12 times greater than the input torque to the transmission. This torque multiplication is represented by the Greek letter Eta (η), and in this example η =12.

The approach used to quantify the MBR was to develop the applicable equations of motion and then reduce the rotating weight while holding all other relevant parameters constant. The amount of rotating weight removed from the vehicle is given by “delta M2” or ΔM2. The vehicle acceleration (a) was calculated for the case where an arbitrary amount of rotating mass (ΔM2) is removed from the drivetrain; and then the amount of non-rotating weight needed to be removed from car (ΔM1) was calculated to produce the same level of acceleration as that achieved by removing rotating weight. Finally, the MBR is given by the ratio ΔM1: ΔM2 as per Equation 5. The derivation of equation 5 is shown in Part 2. The result is:

 equation 5 mfa.jpg   6.04K   0 downloads

Equation 5

By inspection of equation 5, and given that r is always greater than 0 and less than R, the MBR can never be less than 1.

By selecting the appropriate torque multiplication factor η, this equation can apply to any rotating element within the drivetrain for the purpose of calculating the MBR. When weight is removed from a rotating component in the drive train, the appropriate η is the torque multiplication factor that exists in the drivetrain, between the tire’s contact patch and the location in the drive train where weight was removed. For example, if we remove weight from the wheel, because there is no gearing between it and the contact patch, η = 1.

Alternatively, when weight is removed from the fly wheel, the torque multiplication factor of the transmission and differential must be accounted for. For a differential with a final drive ratio of 3.939:1; and gear ratios of 2.583 for first gear, and 1:1 for fourth gear: η = 10.17 when in first gear, and η equals 3.93 in fourth gear.

Applying the above result to calculate the MBR applicable to first and fourth gears, and assuming r = 0.75R:

 first and fourth gears mfa.jpg   9.24K   0 downloads


A standard mini flywheel is listed as weighing 16.71 lbs, and a steel lightweight flywheel 11.02 lbs, for a difference of 5.69 lbs. Therefore, the equivalent amount of mass that would need to be removed from the mini’s non-rotating mass to produce the same level of acceleration as that achieved by switching to the lightweight flywheel is (6.62)(5.69) = 37.7 lbs when in first gear. This mass benefit ratio gets smaller with each gear change. When in fourth gear, the 5.69 lbs removed from the flywheel is equivalent to removing 18.3 lbs non-rotating weight.

The equation for the mass benefit ratio was calculated based on the assumption that the drivetrain could transmit power with 100% efficiency. Power losses in the drivetrain can be represented by multiplying the torque multiplication factor η by the efficiency of the power transfer in the drive train.


Conclusions to Part 1

• The mass benefit ratio is never less than 1.
• The maximum MBR that can be achieved is strongly dependent upon the torque multiplication factor η, i.e. the gear ratio. The appropriate value of the torque multiplication factor to use when evaluating the MBR changes depending on the location within the drive train where rotating mass is being removed.
  
• If removed from the wheel/tire, then η = 1.
• If mass is removed from the flywheel, then η depends on the differential ratio and the gear ratio.

The largest value of η results from selecting first gear and using a high gear ratio differential. With each shift from first gear to fourth, the value of η decreases and hence the MBR also decreases.

Part 2


The basic model of the physical phenomenon of interest is shown in figure 1. It can be used to find the MBR, but for the wheel only. In order to develop an expression for the MBR that would apply to all rotating elements in the drivetrain – from the engine output shaft to the tire’s contact patch – this model needs to be expanded to include the drivetrain, as shown in figure 2:

 transmission 2.jpg   23.21K   0 downloads


Figure 2

Tracing the torque path from the engine to the drive wheels begins with the torque output of the engine (Te). The torque delivered from the engine to the transmission (the transmission input torque, or T_tx) is reduced by the amount needed to overcome the flywheel’s rotational inertia. The torque delivered to the wheel is that which is equal to the input torque at the transmission multiplied by the “torque multiplication factor”, represented by the Greek letter Eta (η), which is simply the gear ratio of the transmission/differential.

The only external force acting on the vehicle is F(r/t): it is this force that linearly accelerates the vehicle. The magnitude of this force is equal to the applied torque at the wheel (Tw) divided by the radius of the wheel. The greater this amount of torque Te, the greater will be the applied force F(t/r), and hence the greater will be the vehicle acceleration – e.g. more horsepower equals more torque equals more acceleration.

Another strategy for improving vehicle acceleration is to reduce the weight of the vehicle. A reduction in non-rotating weight will result in a proportional increase in the vehicle acceleration. Reducing rotating weight in the drivetrain will result in the drivetrain having less rotational inertia, which means there is less resistance to rotational acceleration of the drive train, which results in more torque being delivered from the engine to the tire contact patch.

In order to isolate the effect of reducing rotating or non-rotating mass on vehicle acceleration, it was necessary to hold all other variables constant. A key assumption has to do with the phenomenon of slip. Slip refers to the speed of the tire’s contact patch relative to the road. For the car that skids to a stop with the wheels locked (i.e. not rotating), the tire’s contact patch travels at the same speed as the car, relative to the road. This condition is referred to as 100% slip. A car with a free rolling tire has 0% slip. In this case the speed of the tire’s contact patch, relative to the road, is zero. Slip can also occur when accelerating forward. A vehicle achieves the maximum level of acceleration, for a given level of torque, when operating at between 0% and 20% slip. The following derivation is based on the vehicle accelerating with 0% slip. The case where acceleration takes place with non-zero slip is commented upon later.

The torque delivered to the transmission is equal to the engine’s output torque minus that required to rotationally accelerate (i.e. spin up) the flywheel. Or,

T_{tx =} (T_e - Iα) Equation 6


The transmission’s output torque is equal to the transmission input torque multiplied by the torque multiplication factor η of the transmission/differential (i.e. the gear ratio). The torque delivered to the wheel is equal to the transmission output torque. Thus:

T_w = η(T_e - Iα)

Recall the force F(r/t) is given by the equation T = FR, or:

F_rt® = η(T_e - Iα)

And solving for F(r/t):

F_rt = (η/R)(T_e - Iα) Equation 7

Equation 7 expresses the magnitude of the force that is responsible for linearly accelerating the vehicle, F(r/t), as a function of the engine’s output torque (Te), where R is the radius of the wheel; Iα is the rotational inertia of rotating elements comprising the drive train; and η is the torque multiplication factor of the transmission and differential (the gear ratio).

By selecting the appropriate torque multiplication factor, this equation can apply to any rotating element within the drivetrain for the purpose of calculating the MBR. When weight is removed from a rotating component in the drive train, the appropriate η is the torque multiplication factor that exists in the drivetrain, between the tire’s contact patch and the location in the drive train where weight was removed. For example, if we remove weight from the wheel, because there is no gearing between it and the tire contact patch, η = 1. Alternatively, when weight is removed from the fly wheel, the torque multiplication factor of the transmission and differential must be accounted for, and in this case η = 10.17 (for a gear ratio of (2.583)(3.939) = 10.17).

The expression (η/R)(T_e - Iα) in equation 7 is the force applied by the road to the tire. When this force is applied to the vehicle the mini begins to accelerate at a rate given by equation 1. I.e.:

(η/R)(T_e - Iα) = (M1 +M2)a Equation 8

And, substituting the expressions (M2)r² for I, and α = a/R, into equation 8 leads to:

 equation 8 9 10 11mfa.jpg   15.42K   0 downloads

Equation 10 is used to calculate the linear acceleration of a vehicle, expressed as a function of the torque generated by the engine T_e. By inspection of Equation 10, the force that causes the vehicle to linearly accelerate, i.e. the force of the road acting on the tire (Fr/t), is given by:

F_rt = (ηT_e)/R Equation 11

and the effective mass of the vehicle accelerated by F_rt, accounting for the rotational inertia of the drivetrain, is:

M_effective = M1 +M2(1 + ηr²/R²) Equation 12

Regardless of the amount of torque produced by the engine, the maximum amount of torque that can be applied by the wheel to the road, and hence the maximum level of vehicle acceleration, is limited by the maximum level of friction that can be generated at the tire’s contact patch, i.e. the coefficient of friction between the tire and road. The coefficient of friction is denoted by the Greek letter Mu (μ), and the maximum friction force is giiven by:

F_friction = μk(M1 +M2)g Equation 13

Where g is the acceleration of gravity, and k is the percentage of total vehicle weight that acts on the drive wheel. If a mini had a locked differential, and for a 60/40 weight distribution, k = 0.6. If there is only one drive wheel, then k is equal to about 0.3.

Substituting the maximum force F_friction from Equation 13 into Equation10 will produce an expression for the maximum rate of acceleration that can be achieved:

 amax for mfa.jpg   6.03K   1 downloads Equation 14

An explicit expression for the MBR can be derived by re-writing Equation 10. Subtracting arbitrary masses ΔM1 from M1, and ΔM2 from M2, and then setting the vehicle accelerations resulting from these changes in mass, as calculated by equation 10, to be equal to one another, i.e:

 equations 15 mfa.jpg   15.11K   0 downloads Equation 15


Expanding the above expression into its individual terms, produces the following expression for the mass benefit ratio:

MBR = ΔM1/ΔM2 = (1 + ηr²/R²) Equation 16

There is only one level of applied torque that will produce the condition where the force of the road pushing on the vehicle exactly matches the available force of friction. When a vehicle achieves this level of applied torque it will accelerate at the maximum rate possible – sometimes referred to as “threshold acceleration”. Any torque level less than this amount will mean that the maximum level of acceleration has not been reached; and any applied torque that exceeds this critical value will cause the tires to slip. The problem with slip is that as slip increases, the coefficient of friction begins to reduce.

For the following analysis threshold acceleration is assumed to occur at 0% slip. Clearly it would be very difficult for even the best driver to apply just the right amount of torque so that the maximum level of acceleration is maintained for any meaningful distance. In modern cars, the task of accelerating at or near 20% slip is performed by launch control, or ABS when braking.
Threshold acceleration actually occurs at about 20% slip because the coefficient of friction effectively increases from 0% slip to about 20% slip, and then it decreases rapidly from 20 to 100% slip. The case where the vehicle accelerates at 20% slip can be modelled by assuming threshold acceleration occurs at 0% slip and increasing the value of the coefficient of friction used in the analysis.

Braking is simply accelerating in the direction opposite to velocity. Because the braking force is applied at the wheel, η = 1 when braking at 0% slip, and η = 0 at 100% slip. The value of the percentage of the total vehicle weight acting on the road, k, is 1 (when all four brakes are working). Reducing rotating weight at the wheel will reduce it’s rotational inertia. This means that, for a given rate of deceleration, less work is done to rotationally decelerate the wheel. Less work equals less heat generated by the brakes, implying lightweight wheels have the potential to postpone the onset of brake fade.

Air Drag

The magnitude of the aerodynamic drag force increases in proportion to the square of the vehicle’s velocity, and it acts to retard a vehicle’s rate of forward acceleration. At very low speeds the air drag force is negligible and the MBR is given by equation 16. At a high speed, the air drag force will be equal in magnitude to the force of the road acting on the tire, F(r/t). In this case the vehicle acceleration is zero, and the car has reached its terminal velocity.

For a mini weighing 1850 lbs (840 kg), accounting for air drag, the mini’s terminal velocity is calculated as follows:

The force of air drag, F_ad, as a function of velocity (V), is given by the equation:

F_ad = ρSC_DV² Equation 17

Where: ρ = air density = 1.225 kg/m³
S = frontal area of mini = 1.9 m²
C_D = drag coefficient = 0.50

At terminal velocity the force of friction is equal in magnitude to the drag force, i.e. the force as per equation 13, is to equal the air drag force of equation 17. For a coefficient of friction of 0.75, one drive wheel, and a 60/40 fore/aft weight distribution, the terminal velocity is calculated to be 126 mph (203 km/h). It is important to note that this numerical result, while helpful for illustrating the effect of air drag on MBR, should not be viewed as being accurate owing to uncertainties inherent in modelling aerodynamic drag.

As a vehicle accelerates from a stop to its terminal velocity, the portion of F(r/t) that needs to be allocated to offset the effect of air drag increases from 0% to 100%. At 0%, all of force F(r/t) is available to accelerate the vehicle and the MBR is given by equation 16. At 100% F(r/t) = F_ad and the vehicle is no longer accelerating, consequently the MBR = 1. Evidently, the MBR is speed dependent. Its maximum value is given by Equation 16 when accelerating at (and near) zero velocity. Because the drag force increases with the square of velocity, the MBR will decrease in proportion to the square of velocity, reaching the minimum value of MBR = 1 at terminal velocity.

By equation 17 the air drag force is proportional to air density. Air density decreases with an increase in elevation. For example, the air density at 5000 ft elevation (e.g. Denver Colorado) will be about 13 % less than that at sea level. This translates to a higher terminal velocity, which implies that the rate at which the MBR decreases, as a function of increasing velocity, at elevation, will be less than its rate of decrease at sea level.

The MBR is proportional to the elevation at which the vehicle is operated: it increases with an increase in elevation. This statement, in isolation, is not incorrect. However, the loss of torque resulting from decreased volumetric combustion efficiency, due to less dense ambient air at elevation, would have a much greater detrimental effect on vehicle acceleration than that resulting from any increase in MBR with elevation.  equation 8 9 10 11mfa.jpg   15.42K   0 downloads  amax for mfa.jpg   6.03K   1 downloads

Attached Files

•  mbr mfa 2.jpg   5.83K   0 downloads

Edited by Gnielsen, 14 March 2024 - 02:18 PM.

[Spider]

Thanks Gnielsen, many thanks for taking the time to put this highly detailed post together.

I'm looking forward to reading through it !

 

[Gaz66]

I wish I knew what all that meant 🤔

[Java_Green]

Thank you Gnielsen for the clarification. I do find it reasonable and sound. Was a bit surprised of the stated Cd-value, it is the same value as for a VW bus from the 50:is... One thing I noted you neglected was the weight shifting due to yaw. In an comparison of MBR it should be neglected but it is relevant to include when talking about threshold acceleration.
Again, I find it interesting and even though I know the theory, I haven't concluded it like this. Unfortunately, I guess this is a bit too abstract for the majority of the audience within this forum and thereby hard to grasp.

[Gaz66]

Or....for the hard of understanding like myself.....lighten as many rotating parts as possible as well as lightening as much fixed stuff as possible.
Your Mini won't get much faster than that! 😁😁

[stuart bowes]

heavy is bad

light is good

 

no but more seriously, admirable though it is to have that level of understanding about physics and so on (and fair play to you).. is it not more of a practical reality that your choices are basically limited to what tryes fit on what rims, to fit over your brakes and provide the amount of poke you can get away with.  which in itself seems complicated enough without adding in any further variables which in all probability you can't do anything to change anyway

 

not intending to detract from the quality of work you put in or how interesting the subject is to those who understand it 

Edited by stuart bowes, 14 March 2024 - 09:44 PM.

[Gnielsen]

Thank you Gnielsen for the clarification. I do find it reasonable and sound. Was a bit surprised of the stated Cd-value, it is the same value as for a VW bus from the 50:is... One thing I noted you neglected was the weight shifting due to yaw. In an comparison of MBR it should be neglected but it is relevant to include when talking about threshold acceleration.
Again, I find it interesting and even though I know the theory, I haven't concluded it like this. Unfortunately, I guess this is a bit too abstract for the majority of the audience within this forum and thereby hard to grasp.

Hi Java Green. The Cd value was one I found on the internet for a mini - I had to pick a number so I chose that one. As for weight shift and its effect on acceleration, you are correct that the article is silent on this point. I did consider it and the effect is cryptically reported in the uncertainty of the result by modelling the k value as “about” 0.3.

[Gnielsen]

I wish I knew what all that meant

what does it all mean? I think Stewart Bowes comment sums it up perfectly “ heavy is bad, light is good”.

[Spider]

 

Thank you Gnielsen for the clarification. I do find it reasonable and sound. Was a bit surprised of the stated Cd-value, it is the same value as for a VW bus from the 50:is... One thing I noted you neglected was the weight shifting due to yaw. In an comparison of MBR it should be neglected but it is relevant to include when talking about threshold acceleration.
Again, I find it interesting and even though I know the theory, I haven't concluded it like this. Unfortunately, I guess this is a bit too abstract for the majority of the audience within this forum and thereby hard to grasp.

Hi Java Green. The Cd value was one I found on the internet for a mini - I had to pick a number so I chose that one. As for weight shift and its effect on acceleration, you are correct that the article is silent on this point. I did consider it and the effect is cryptically reported in the uncertainty of the result by modelling the k value as “about” 0.3.

 

 

I haven't yet read through to that part, however, for what ever it's worth, I had a Cd figure for a stock 1967 model 850 mini of 0.48 that came from a wind tunnel facility (I think in the US ?).

[Gnielsen]

heavy is bad
light is good

Hi Stewart. In a nutshell you’re right. Remove as much weight as you can from as many places as you can provided other factors aren’t compromised . Eg you’re better off with performance tires that may weigh more than a smaller tire depending on traction, etc. On the other hand, if you’re trying to decide where to best spend your money, eg choosing between a carbon fibre hood v. A lightweight flywheel, a sense of the mass benefit ratio might come in handy.

Cheers

 
no but more seriously, admirable though it is to have that level of understanding about physics and so on (and fair play to you).. is it not more of a practical reality that your choices are basically limited to what tryes fit on what rims, to fit over your brakes and provide the amount of poke you can get away with.  which in itself seems complicated enough without adding in any further variables which in all probability you can't do anything to change anyway
 
not intending to detract from the quality of work you put in or how interesting the subject is to those who understand it