ESE's works engine tuner

[Flettner]

Thats an ugly picture
At least its under cover now, hoping to modify it so it can be an inertia dyno as well as load dyno, depending on what you want at the time.

[Flettner]

Pissed around and made a press tool for making tight 90 degree water pipe bends from 1.6mm alloy sheet. Just tig along the edges. 19.6 mm outside dia, 20mm radius.

[SwePatrick]

If someone ever needs a guide how to easily install ignitech ignitionsystems:

[SwePatrick]

And a little progress video.
My Kartengine is coming along nicely.
It has dynoed so far, very rough tuned 42.3rwhp.(howgrinded meteringrod in my 42mm lectron)
And!
Keep in mind, i have just thrown together a pipe from 'feeling', no calculations at all, just for fun.
I´m about to build a pipe designed for this engine, coming up in the future.

https://youtube.com/shorts/OKxjDYwWk44?feature=share

[wobbly]

I am in the process of getting Wiseco to do a proper single ring race piston for the KR180 drag race engine.
It has a 59mm bore , and this is unusual except for several models of 125 engines that were converted with a big bore to become 150cc.
It looks as though the MOQ is going to be 12 units, most of which I dont want or need.
The only people I can think of who would want this piston ( apart from Thai drag racers ) are the several people here in NZ that converted their RS250
engines into RS300 racebikes by putting on NSR150 cylinders that had the same 59mm bore.
I cant remember who these guys were , one of them used to post on here, anyone know who - please PM me, or post it here.

[F5 Dave]

That'll be Sketchy and Richard of i-tools . Both those bikes are probably parked. Sketchy had been fooling fairly extensively with 1/2 an FZR250 and turbo. Had his partner in crime over for a beer Friday and was supposed to go see his turbo at kiatoki on Monday but the 300 Beta had to be ridden.

Amusing aside for electric start 2 strokes.
I washed it and changed dash button battery. Start it to check.
But it wouldn't catch. Making some weird noises. Then hissing after an attempt. WTF? What have I done?!?

Oh. Bung in muffler still from wash. Doh. The hisses explained. Pipe pressure. Not ideL for crank seals but started right up after my embarrassing discovery.

[ApolloMotoMoto]

So, now that I have become the math guru for more than one dyno operator, I need some help from higher level gurus than myself.


How to calculate the moment of inertia of a flywheel...


I have found a few formulas floating around for the mythical "drop test".

Lets see if I can describe what we did, and the math I have found for some "peer review"....


We have a flywheel pinned on a shaft.

The shaft is supported between two pillowblock bearings on a frame.

The shaft also has a sprocket pinned on it.

A complete loop of chain is placed around the sprocket.
(the same chain and sprocket that will be used to drive the flywheel when it comes time to connect an engine)

The chain "drapes" down below the sprocket about 2 feet.

All of this is supported on the frame high enough that the draping chain doesnt touch the ground.

A 3ld weight is hung from a link of "draping" section of chain (near the top).

This 3ld weight is support by a persons hand.

The stopwatch is started the moment the weight is released.

The stopwatch is stopped the moment the weight touches the ground.
(during the full drop, the chain is in a perfectly straight line)

The weight was 3lds (as already noted)
The height of the drop was 1.4375 feet
The time of the drop was 7.3 seconds
The radius of the sprocket was 0.375 feet


I stumbled onto this formula, but I do not know if it correct???

"The formula to calculate the moment of inertia (inertia) of a flywheel using a hanging weight is typically expressed as: I = (m * g * h) / (ω²); where "m" is the mass of the hanging weight, "g" is the acceleration due to gravity, "h" is the height from which the weight falls, and "ω" is the angular velocity of the flywheel."

A a few images are attached to provide some more detail.


My questions:

1. Is the above formula actually applicable to the test I have performaned, and is the formula an accepted method of calculating moment of inertia?

2. Does the drop test I performed provide sufficient data to perform a moment of inertia calculation?

3. We have currently convinced ourselves that using the sprocket and chain during the drop test will ALSO be measuring the resulting moment of inertia of those components as the sum of everything that is being "spun" by the dropping weight. Is this a reasonable assesment?

4. We have further convinced ourselves that this is a good idea, as the engine will be spinning the inertia of these componenets during a pull. Are we missing something?

[husaberg]

Use an acceleration time that is equivalent to what you are going to have when you ride it. Ideal case: the acceleration time in each gear is the same on the dyno and on the road. For that to be the case the roller has do be twice the weight of the bike including rider.

As long as you are unsure about jetting and ignition timing, fuel supply etc, use a shorter acceleration time as it is less likely to destroy an engine with lighter load. Once you are more comfortable with the setting you can switch to longer pulls, using a higher gear. Extra long / longer acceleration times would be the equivalent to driving uphill and vice versa.

You can do what 2 stroke stuffing did and use a ABB electric motor and let Abb do the calcs for you.

[wobbly]

I used this test as well as doing the manual math inertia calculations supplied by SportDevices.
But the drop test is much more accurate if you expand the drop height , and thus the time base.
I wound string several times around the inertia weight up to a pulley in the workshop roof ( 6M height ) with a 1KG weight.
Holding the dyno wheel by hand , as I let it go I hit the dyno run record button , and hit end when the weight hit the floor , this gave the wheel speed and time info needed.
Doing this several times and averaging the result gave accurate info that also includes the roll friction.
The answer gave almost perfect correlation to the manual calc's of the inertia wheel, starter flexplate and brake disc and axle that all form the total run up inertia.

[Noxin]

I just calculated all parts individual inertia and added them together.
Then I spun the flywheel and measured the negative friction and wind drag power when it slowed down.
Then I just add this power curve when testing a motorcycle or moped.

https://youtube.com/shorts/pbGIH3S1t14

[ApolloMotoMoto]

I used this test as well as doing the manual math inertia calculations supplied by SportDevices.
But the drop test is much more accurate if you expand the drop height , and thus the time base.
I wound string several times around the inertia weight up to a pulley in the workshop roof ( 6M height ) with a 1KG weight.
Holding the dyno wheel by hand , as I let it go I hit the dyno run record button , and hit end when the weight hit the floor , this gave the wheel speed and time info needed.
Doing this several times and averaging the result gave accurate info that also includes the roll friction.
The answer gave almost perfect correlation to the manual calc's of the inertia wheel, starter flexplate and brake disc and axle that all form the total run up inertia.

Okay, we did the "manual calc" for all the sub assemblies of the flywheel (hub - spokes - rim) and got 266 pounds per quare foot.

This was done for JUST the flywheel so far, not the sprocket and the chain which are also attached in the drop test.

The math from the drop test produces a resulting 280 pounds per square foot, or 5% higher than the flywheel all by itself.

That seems to be a fairly good correlation; the extra 5% should be the measured friction loss of the pillow block bearings plus the additional inertia of the chain and sprocket.

[ApolloMotoMoto]

Alright, next question...


The measured torque to HP math;

I ran into something I thought I understood, but now I am not so confident about my understanding;

Torque = Total Dyno Inertia multiplied by the measured Angular Acceleration (α) of the flywheel for the given set of data-points.

T = Iα

Now that you have the Torque value for a given set of data-points;

Horsepower = Torque multiplied by (Revolutions Per Minute divided by 5252)

HP = T * (RPM / 5252)


Okay, here is my question;

For the above HP equation, when using an inertia dyno;

RPM of the ENGINE...???

or

RPM of the FLYWHEEL...???


I am having an internal argument in my head and I cant come to a resolution about which one it has to be >.<

Can someone with more empirical knowledge of the math behind an inertia dyno measurement point me on the right track please?

[Vannik]

Alright, next question...

Can someone with more empirical knowledge of the math behind an inertia dyno measurement point me on the right track please?

On the same component where you calculated the torque.

[ApolloMotoMoto]

On the same component where you calculated the torque.

Thank you Neels!

[ApolloMotoMoto]

The student returns with another question

Can someone please check my math?



We did calculations for Flywheel Moment of Inertia.

I = (k)mr^2

Sub-divided each part, used the mass for that part, and the k constant that made sense for that part; k=1 for the rim, k=0.5 for the hub, etc.

Using pounds for mass and feet for radius, this is how we got to 266 ld ft^2 for the flywheel by itself.
Using the same units for the drop test, we ended up with 280 ld ft^2.

Okay, now I have a value for Inertia in the torque equation T=Ia

All bright eyed and bushy tailed, I calculated the angular acceleration of the flywheel for a test pull that was just conducted with a 212cc 4 stroke engine that should make 8.1ft ld ft of torque peak, and about 6.5 hp peak. (Honda Generator Engine Clone)

I dont have the recorded pull data to see the actual curve shape, but I know the start and end engine RPM's of the pull, I know the time of the pull in seconds, and I know the gear ratio between engine and flywheel shaft.

RPM range = 1000 Engine RPM to 5000 Engine RPM
Pull Time = 15 seconds
Gear Ratio = 6:1 (10t on engine crankshaft, 60t on flywheel, no transmission, no clutch)

Thus, averaged across the pull, the engine accelerated at a rate of 266.66 RPM's per second.
Using the 6:1 gear reduction, this means the flywheel accelerated at an average rate of 44.5 RPM's per second.

The software has a sampling frequency of 5 samples/second.

So, I decided to look at 0.2 second "steps" of time for the rate of angular acceleration (a). Using the average acceleration rate of the Flywheel I whitled down to:

44.5 RPM's per second = 0.74 RPS's per second = 0.148 RPS's per data point (per 0.2 seconds) = 0.93 angular velocity (w) per data point

So, if this is averaged across the whole run, each data point will have the same (w), and thus the same delta(w).
Same thing for time, every data point is 0.2 seconds "wide", so delta(t) will always be 0.2.

Finaly;

a = delta(w) / delta(t)
a = 0.93 / 0.2
a = 4.65


Now, armed with all of the mathematical powers needed to divine the measured engine power.... everything falls appart.

T=Ia
T=280 * 4.65
T=1302

1302 ft lds of Torque measured at the flywheel shaft?!?!?!
-okay, thats a little big, but thats flywheel shaft torque, so lets divide it by the 6:1 gear ratio to find out what the engine did....

1302 ft lds / 6 = 217 ft lds of "average" Engine Torque across the 15 second pull.

Thats a fever dream, not a measurement....


After considering my life decisions for a few hours, I think I stumbled into my problem:


I = (k)mr^2 only works with SI units.

The output of that formula when we use pounds and feet is absolute bunk for executing the T=Ia formula.

So, I redid the Moment of Inertia Calcuation using SI units.

I end up with 11.8 kg m^2 after converting the weight of the flywheel in lds to kgs (110ld flywheel), and converting the radius to meters (2ft radius flywheel).
-lets try this one more time:

T=Ia
T=11.8 * 4.65
T=54.87

54.87 Nm of Torque measured at the Flywheel shaft, that sounds more like reality.
54.87 Nm lds / 6 = 9.15 Nm of "average" Engine Torque across the 15 second pull.
9.15 Nm converts to 6.75 ft lds.

6.75 "average" ft lds across a pull from 1000 - 5000, using a 4 stroke engine that makes ~8.1 ft lds of peak torque at 2,500 RPM?
-considering this 6.75 ft lds is "uncorrected power", and the ~8.1 ft lds number is the manufacturer advertised, SAE corrected output.

That sounds like a real measurement.

Now, lets check HP using flywheel RPM like Neels just advised.

To do this, I took the 54.87 Nm of flywheel shaft torque and converted it to ft lds;
54.87Nm converts to 40.47 ft lds

Using 40.47 for torque and flywheel RPM for RPM, I calculated HP at 1000 Engine RPM (166 flywheel rpm) and 5000 Engine RPM (833 flywheel rpm).

HP = T * (RPM / 5252)

If the Torque curve is averaged to the "flat line" of 6.75 ft lds of torque at every RPM step like I have done;

HP at 1000 RPM = 1.3 HP
HP at 5000 RPM = 6.4 HP

That is "in the ballpark" as they say.

Final sanity check;

The pull did not go to 5252 RPM, but if I keep extending the averaged 6.75 ft lds of torque out to 5252;

HP at 5252 RPM = 6.75
Torque at 5252 RPM = 6.75


HP and Torque "cross over" at 5252 RPM's!

Woohoo!


Can someone please check my math and give me a positive confirmation that I = (k)mr^2 only works with SI units for the purpose of using in the formula T=Ia?