Hoping for a sanity check on inertial dyno roller/flywheel design (herein referred to as "roller"). If you feel like reading/thinking through all of this, please feel free to let me know if I have anything wrong here.
Otherwise, feel free to use this for your own dyno, if you trust my math I can share a spreadsheet if anyone is interested.
I am hoping to build a (hopefully portable) dyno at some point, mostly for 50cc engines -- but I am quickly realizing that engine size/power might not be the most important design input parameter for an "ideal" 2T dyno; rather the mass of the bike and rider may be the only important parameter. More specifically, for a chassis dyno with a thin-wall roller, the roller mass should be equal the mass of the bike+rider. Things can get a little more complicated for a engine dyno that operates from the transmission output shaft, with corrections needed for sprocket ratio and roller vs. tire diameter, but same basic concept.
The core assumption comes from what I remember reading here: that for 2T development purposes, it seems there is some consensus that the "best" way to use a dyno is to replicate on-track usage; in other words, steady-state readings are not as useful as inertial readings while replicating on-track acceleration, in (large) part due to heat rates (e.g. exhaust temp). Please correct me if I am wrong, but if this is correct, then the dyno roller's inertia should replicate the mass of the bike and rider.
The math:
The equivalent mass (i.e. converting to linear kinematics) of a dyno roller is given by the following equation:
mequivalent = Iroller * (1 / rroller)2
Assuming the roller's circumferential speed matches the rear tire's circumferential speed, as on a chassis dyno, we just need to get the roller's equivalent mass to match the on-track mass.
With the equivalent-mass equation above, and the equations for roller inertia:
I ~= mthin roller * rthin roller2, or
I = 1/2 * msolid roller * rsolid roller2,
the roller radius cancels itself out - so assuming the roller's circumferential speed matches the tire's circumferential speed:
mthin roller ~= mbike+rider, or
msolid roller = 2 * mbike+rider
Correction for engine dyno:
If, however, the circumferential speeds differ (as can be the case with an output-shaft driven roller), the equivalent mass does change, by a factor of the speed ratio (roller circumferential speed / tire circumferential speed, or roller radius / tire radius) squared. So if the roller has the same mass as bike+rider but half the diameter of the tire and is driven at the same angular velocity as the tire, it acts roughly like 1/4 of the mass of bike+rider - of course only applicable for an output shaft-driven roller (not a chassis dyno).
So two key takeaways here: thin-wall roller is most mass-efficient (of course); and to best simulate on-track usage (for a chassis dyno with said thin-wall roller), the roller mass should simply be roughly equal to the bike + rider mass.
Therefore if I want portability, like being able to put the dyno in the back of a truck by myself, I need to overdrive the roller to be able to reduce its mass sufficiently (so a chassis dyno will not work due to the required roller mass).
Personal Example:
I am working on a 50cc moped with a single-speed centrifugal clutch transmission. Let's say that currently the specifications are:
- Primary Ratio: 16:57 = 1:3.5625
- Chain Drive: 10:60 = 1:6
- Total Gear Ratio: 21.375
- Tire Radius, rolling: 0.281m
- Mass of bike and rider: 127kg
- Maximum Engine Speed: 15,000rpm
- This gives a top speed of 46.2mph (74.3km/h).
Let's also assume my roller specifications, for an engine dyno driven by the transmission output shaft, are:
- Mass: 20kg
- Material: A36 steel, 7800kg/m3, 250MPa Yield Strength
- Radius: 15cm
- Thickness: 1.5cm
- Length: 19.09cm
- Inertia: 0.407kg/m2
If I spin this roller at the same angular velocity as my tire, i.e. if I use 10:60 sprockets for the roller drive, then the equivalent mass of the roller is 5.16kg. Not nearly enough! This is derived from:
mequivalent = Iroller * (omegaroller / vtire)2
...where the previous variables relate to engine speed by:
omegaroller = omegaengine * gear ratioroller, total, and
vtire = omegaengine * rtire / gear ratiotire, total
Thus the equivalent roller mass equation is:
mequivalent, roller = Iroller * (Zrear, tire * Zfront,roller / (Zfront, tire * Zrear, roller))2 * (1 / rtire)2
...where the "Z" values are the tooth counts for each sprocket.
To get the equivalent mass to match up as closely as possible, I need to swap the roller shaft sprocket to a 12-tooth to overdrive it 5x as fast. Doing so gives an equivalent mass of 129kg, with a roller angular velocity of 3509rpm at 15000rpm on the crankshaft. A hoop stress calculation shows ~23.7MPa:
stress = mroller * rroller2 * omegaroller2
If I gear the moped for top speed rather than acceleration, say close to 80mph, I might use a sprocket ratio more like 12:42 for the bike, in which case my equivalent mass with a 12:9 sprocket ratio for the rollers is about 112kg (this is probably the practical limit to sprocket sizes). Hoop stress here is about 60MPa, so seems like it would still be safe.
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