3. FLYWHEEL ENERGY ISSUES

In discussing the pros and cons of various absorbers I keep mentioning problems with high inertia. To illustrate just how much power flywheel energy can mysteriously "absorbed" let's "build" a crude, dirt-cheap dynamometer with no brake at all! This will be an "inertia dynamometer" because the engine's power output will go into "winding up" a heavy flywheel.

This example uses a flywheel that is large, in relationship to the engine, so accelerating the combination from idle to peak rpm takes several seconds. A fast data acquisition system logs the time periods and rpm changes. From that information we calculate the torque and horsepower the engine supplied to accelerate that known flywheel mass. The formula for determining the torque is:

Torque = JM _* rpm per second / 9.551

where JM represents the Polar Moment of Inertia of our inertia dyno's flywheel.

If we don't know the Polar moment of Inertia for the flywheel (and our flywheel has a constant thickness cross-section) we can calculate it with the formula:

JM = (W _* r ^{^2}) / 32.16 / 2

where W represents the flywheel weight in pounds and r is its radius in feet.

Once you have the torque, it is easy to calculate the horsepower with the standard formula:

Hp = Torque _* rpm / 5252

Keep in mind that the rpm in the last formula must be the average rpm during the sampling period.

Say our example uses a 10-pound flywheel, 8" in diameter (thus it would have a Polar Moment of Inertia of .017 foot-pounds-second²). If the engine was able to accelerate this flywheel from say 4,800 rpm to 5,200 rpm in 2/10 of a second (a rate of 2,000 rpm per second) that would represent a torque of 3.6 pound feet. Since our above example had an average rpm of 5,000, it produced 3.4 Hp during the test. That's all here is to it. Unfortunately, inertial dynamometers alone are useless for doing the steady state testing needed for methodical development of porting, pipes, etc. You can not adjust the load to hold the engine at a given rpm point, it must always be accelerating. Still, inertial testing is handy for working out acceleration and drivability problems.

The real reason for the above math exercise is to illustrate how much power it took to accelerate that small flywheel. If you buy an absorber with a polar moment of inertia in the same rage as our flywheel example above, don't expect to perform sweep acceleration testing. Even accelerating at just 200 rpm per second would consume 10-% of our sample engine’s power! Fortunately, high end computerized data acquisition systems provide composition algorithms to back out the effects of absorber (and crank-train) inertia from acceleration data. On a high inertia dynamometer, compensation is required even for fairly low rate sweep testing.

Angular velocity (w) is simply how quickly something turns, like RPM, or Revolutions per Minute. To keep the units consistent, you want to use radians per second (1 revolution = 6.28 radians) before you plug it into any calculation, but it’s easier to think of RPM. If you go from say, 20,000 rpm to 30,000 rpm in one second, the Angular Acceleration is 10,000 rpm per second. Notice the units are very similar to straight line acceleration. Change in angle per time, per time.

Example:

To increase 1000 rpm in 0.12 seconds is 8,330 rpm/sec. So, a= 872.7 radians/sec/sec. (remember I told you we’d have to whip the units into shape before calculating, and that there are 6.28 radians per revolution and 60 seconds in a minute).

from before,

T = I * a

I = .00121614 kg m² and a = 827.7 rad/sec/sec

so, T = 1.06 kg m²/sec/sec, or 1.06 N m (because 1 Newton = 1 kg m/s². F=ma, remember?!).

Or, if you prefer, that’s 150 oz in. (since 1 newton = .2248 lb, 1pound = 16 ounces, and 1 inch = .0254 meters)

To accelerate that wheel at that rate, the engine had to produce an average of 150 oz in from 20,000 to 21,000 rpm.

How about power?

P = T * w

Power = Torque * engine speed (ref 6, pg 46). For engine speed, we use the midpoint between the two measured points: 20,500 RPM. (In reality, the dyno uses much smaller rpm steps, about 10 or 20 rpm, which makes the approximation valid).

P = 150 * 20,500

= 3079000 oz in * rev / min

Those are strange units, so work them into something familiar:

= 1680 ft lb / sec (since 1 ft = 12 in, 1 lb = 16 oz, and 1 rev = 6.28 radians)

= 3.1 HP (since 1HP = 550 ft lb/sec)

The engine had to develop 3.1 HP to accelerate that inertia wheel, at that rate, at 20,500 rpm.

Now, just record RPM at little time steps during the whole run, say every 500^th of a second, and do this same calculation between each step. Since I can’t write that many numbers that fast, I have the computer record rpm vs. time at 500 Hz for the whole run and store the numbers in an array. Then, to get rid of the high frequency noise that is inherent in a finite bit digitized signal, we pass the data through a low-pass simulated Butterworth filter set at 10 Hz, which leaves just the data we’re interested in. Then the software steps through the data, and calculates Angular Acceleration at each point, and plugs that into the Torque equation. That gives a Torque at each rpm.

3. SAMPLING RATE

A suitable computerized data acquisition system should have a fast sampling rate, especially for testing 4-stroke, single cylinder engines. By fast I mean at least 100 samples, of all sensor channels, per second (100Hz). A 200Hz logging rate is a bit better still. Why? Understand that, between sparkplug firings there is a measurable drop in the instantaneous crankshaft torque and rpm. The crankshaft gets accelerated in the moments after combustion, then begins to slow until almost two revolutions later the plug fires again. You can't feel these rapid highs and lows when driving around the track (with all that vehicle inertia), but the dynamometer will!

You need to time the drum extremely very super accuirately! I cannot stress this enough! A computer cannot do this because its time base "wobbles" enough for the real data to be lost in the noise. You also need a VERY accurate sensor near the edge of the drum of the dual hall effect type. The gap cannot effect the 360 degree pulse. The movement in the bearings is enough to screw up the data sufficiently enough for the signal (the difference in time from one revolution to the next) to be lost. This part is very important to understand. Read it about 50 times. If you get this wrong your curves will be jaged spikes as the data will be lost in the sensor or timing inaccuracy. You are not measuring speed! Well you are as well but thats incidental. You are measuring the difference in time taken for the revolution measured compared to the previous one and using the data to plot a point.

While this is going on your high frequency timing device is counting ignition pulses and averaging them while the drum is being timed. Every 360 degrees the data packet is sent (rpm and drum time) with a handshake so that it never misses a single revolution. You cant miss one.  Now you can plot a power figure for every revolution of the drum because you know its dimensions, and the density of the steel used to make the drum.. And you can plot a torque curve too for the DRUM because you already calculated its rotational inertia. And you know its rpm, as well as its surface (road) speed.  So without rpm data you can plot power v MPH or KPH.  And torque of the drum x mph/kph. To know how to add that same torque curve to the graph you simply need the engine rpm. That tells you how high the torque curve should be. (both cross at 5252rpm on any BHP X LB/FT graph.)

Newton's Second Law says that Force = mass times acceleration,
F = m * a (A linear anology that I couldn't leave out.)

The relationship for rotational motion is; Torque (t) = Rotational Inertia (I) times Angular acceleration (a)
t = I * a

How do you measure Rotational Inertia (I)?

There are a few formulae for working out the moment of inertia of differently constructed drums / rollers. I'll only deal with the one here. The formula for a solid cylindrical drum / roller made of uniform material is;
I = m/2 * r²

You can calculate the mass of the drum or other components by calculating the volume of the piece and using the density figure i.e. 7900kg / m³ for steel.

Using the dimensions of my drum, I'll calculate it's MOI here.

The drum is 360mm wide, has a diameter of 460mm and has 2 stub axles 120mm X 80mm.

Volume for a cylinder = Pr²* h

V = P * .230m² * .360m
V = 0.0598284 m³ * 7900 (density kg/m³) = 472.6450kg

Now for the Stub axles;
V = P * .040m² * .120m
V = 0.0598284 m³ * 7900 (density kg/m³) = 1.5168kg

Now for the MOI
I = m/2 * r²

The drum;
I = 472.6450/2 * .230²
I = 12.5014kg/m²

The stub axles;
I = 1.5168/2 * .120²
I = 0.02184kg/m² now times 2 there are two of them!

Grand Total of;
MOI = 12.5451kg/m²

(I had my drum weighed and it came up at 476kg, the calculated weight is 475.6786kg. Good enough for me.)

(By the way, notice how much the radius of the cylinder affects its rotational inertia. First of all, the mass depends on the radius squared. A cylinder twice the diameter of another has four times the mass. Then, the rotational inertia is that times radius squared AGAIN. This comes from the particle distribution. Two cylinders with the same mass, but one with twice the radius of the other (i.e. not so wide, but with a bigger diameter), has four times the rotational inertia. This is why I settled on the largest diameter I could get hold of.)

Acceleration

a = Dw / Dt &nbsp &nbsp Angular acceleration (a) equals change in angular velocity (w) per change in time.

For the formula to work, the acceleration has to be in radians/second/second. (rad/sec²). There are 2Prads in one revolution.

Angular velocity (w) is simply how quickly something turns, like RPM, or Revolutions per Minute. To keep the units consistent, you want to use radians per second (1 revolution = 6.28 radians (2P) before you plug it into any calculation, but it's easier to think of RPM. If you go from say, 5,000 rpm to 10,000 rpm in five seconds, the Angular Acceleration is 1,000 rpm per second. My front-end program only works in rads/sec, then I convert these to rpm's where needed.

Putting it all together

Here I'll plug in some figures that come from my data acquisition module and step through the manipulation of these figures to get a torque and power value for this particular sample. These two values actually came from a run I did with my TL1000S. When you have a couple of hundred of these samples and plot then, this results as a pretty reasonable graph.

The module transmits hexadecimal values (hex speeds up the module processing time) so the first step is to change these to decimal.
Sample 1, H173D = 5949
Sample 2, H1731 = 5937

What are these numbers? Each one is how long the drum took to do half a revolution in 2 microsecond (ms) increments. So now we have,
Sample 1, 11898ms
Sample 2, 11874ms

I'll move straight onto power now, I'll finish the acceleration calculations off one day.

Power

When we know how much torque is being applied to the drum, and we know the drum speed in revolutions per minute (rpm) or Rads per second (Rad/s) we can calulate how much work is being done or how much power is being produced.

P = t * w (Power = torque by angular velocity).

I'll make up some values and plug them into the formula.
t = 358
w = 209 Rad/s
(2000rpm / 60 = 33.33rps * 2PRad = 209 Rad/s)

P = 358 * 209 = 74,822 watts
74,822 / 1000 = 74.822 Kilowatts (Kw).

(1 watt = 1 Joule / sec, like I said, some of the information supplied will have to be taken for granted as being correct so I don't have to break every bit down into how it is obtained or worked out.)

Converting kilowatts to horsepower or metric horsepower (pferdestarke = ps) is just a simple matter of multiplying the answer by a constant.

Kw to Hp multiply by 1.34
Kw to Metric HP (ps) multiply by 1.35962

So 74.822Kw * 1.34 = 100.3HP

Note, these figures are at the drum, so the bike made 100hp, but the 2000rpm was the drum RPM, the torque was the value seen by the drum. The bike RPM may have been 8000rpm which would make the bike engine torque 89.5

Simplified formula

Have you seen the simplified formulae for Power and Torque figures? For you imperial guys the torque and power curves will always cross at the 5252rpm mark. That is if your engine is making 60hp @ 5252rpm then it is making 60ft/lbs of torque at that point too.

It's a bit harder for the metric guys, the power and torque curves cross at the 9549rpm mark. That is if you are making 80KW @ 9549rpm then the engine is making 80Nm of torque at that point too.

Horsepower = (Torque * RPM)/5252

Why?
It's important to understand that nobody on the planet ever actually measures horsepower from a running engine. What we actually measure (on a dynomometer) is torque, expressed in foot pounds (in the U.S.), and then we *calculate* actual horsepower by converting the twisting force of torque into the work units of horsepower.

Visualize a one pound weight, one foot from the fulcrum on a weightless bar. If we rotate that weight for one full revolution against a one pound resistance, we have moved it a total of 6.2832 feet (Pi * a two foot circle), and, incidently, we have done 6.2832 foot pounds of work.

Remember Watt? He said that 33,000 foot pounds of work per minute was equivalent to one horsepower. If we divide the 6.2832 foot pounds of work we've done per revolution of that weight into 33,000 foot pounds, we come up with the fact that one foot pound of torque at 5252 rpm is equal to 33,000 foot pounds per minute of work, and is the equivalent of one horsepower. If we only move that weight at the rate of 2626 rpm, it's the equivalent of 1/2 horsepower (16,500 foot pounds per minute), and so on. Therefore, the formula applies for calculating horsepower from a torque measurement.

Kilowatts = (Torque * RPM) / 9549

It's much simpler to work out how 9549 constant came to being for the metric one. We need to convert RPM to Rad/s, and the Power is Kilowatts not just Watts, so there is the difference of 1000 times as well. So 1 rpm / 60 = rps = 0.01666 * 2PRad = 0.1047 Now 1000/0.1047 = 9549.29

Putting theory into practice