Rotating weight.....crankshaft....bobweight

[Rick360]

Wouldn't Superflow be able to provide the inertia factor for their dynos?
Using that info, and the two different dyno pulls at different acceleration rates, couldn't you calculate the inertia factor for the engine? Then take bobweight reductions, and the stroke length (distance from center of rotation) and the weight removal needed to rebalance, and calculate the increase in available power.

They might be able to provide that info and yes, if they had that and you had two very accurate pulls 300rpm/sec and 600rpm/sec you could calculate the total MOI then subtract the dyno components and you'd have the engine MOI, but in reality would be inaccurate due to accuarcy of the dyno.

If you have two engines of equal power, when dynoed on a steady state pull (not accelerated) the only difference is one has a heavier rotating assembly. Then you pull both engines on an accelerated pull, the lighter assembly would now show more power than the heavy one. (correct?) Although both would now show less power than before, because some power is now consumed accelerating the mass of the rotating assembly.

That sounds correct to me.

Question; We have an engine that we have dyno info on. We now take the engine apart replace the 100 gram wrist pins with 200 gram wrist pins, and we drill the rod throws just enough that when we put our new heavier bobweight on and spin it on the balancer, no weight needs removed from or added to the counterweights.(it remained in balance.) In other words, all we did is move weight from rotating to reciprocating.
Assuming no additional crank flex, cylinder loading etc,
Would this engine show a power gain on a steady state pull?
Would this engine show a power gain on an accelerated pull?

Definitely No and probably not.

If what you're getting at is how does reciprocating weight affect the power required to rotate the engine components the answer is it acts like rotating weight. I am not sure that the value would be exactly half like the bobweight would be, but lighter reciprocating components would show power gain on accellerating dyno pulls only and show no gain on a steady state pull.

Thats how I see it anyway.

Rick

[PackardV8]

A physicist once explanined it to me this way:

1. Adding rotating or reciprocating weight is akin to adding weight to the car. It takes more time/and/or horsepower to accelerate it to a given RPM/rate.

2. Once accelerated, the constant horsepower output is not affected by either reciprocating or rotating mass.

3. The crankshaft, connecting rods and wrist pins definitely know the difference between rotating and reciprocating weight. Rotating weight in balance adds very little stress. The constant inertial loads of accelerating and decelerating the reciprocating mass breaks lots of too-heavy parts.

thnx, jv.

[randy331]

Thanks; Rick and packard;

With the power needed to accelerate a given weight,a given rpm. is a fixed amount.
Is there a formula to calculate the tq. or hp. required to accelerate 100 grams 600rpm in one second at 1.74" from the center of rotation?

Or is the formula one that I would need a calculator with the formula allready in it, so I just enter weight,distance from center,and acceleration rate and it gives me the answer?

Thanks; Randy

Excusse the spelling as my spell checker is at school.

[Greenlight]

The formula is:

T = I x a

T = Torque
I = Moment of inertia ( I = m r^2)
m = mass of the object
r = the radius from the center of rotation to the center of mass of the object
a (a.k.a. alpha) = the angular acceleration

Make sure that you use consistent units so that your answer is correct.

Insert your results into this formula to find the HP:

HP = T x RPM / 5252

[panic]

r = the radius from the center of rotation to the center of mass of the object

...and how do you determine that with a crankshaft, let alone a crankshaft and reciprocating components? A point 2" up from the journal center on the rod beam is still mostly rotating weight, but follows a semi-elliptical path with changing radius based on the rod ratio:
near TDC: crank offset + 2"
near BDC crank offset - 2"
other distances ~ rod angle

[randy331]

Greenlight; What unit of measure for mass?
Angular acceleration, Is that inches per sec., feet per sec., if I use inches from center of rotation, I would use inches with angular acceleration?
or is that rpm.?

With a bobweight reduction I wouldn't need to know the MOI. of the engine, as any bobweight reduction would be centered on the rod throw, and and would affect the acceleration the same regardless of the original MOI.? correct???


Thanks; Randy

[John Wallace]

Try this calculator for inertia of piston.
Piston Inertia Calc

I'll have to see about adding the HP to run that weight to the calc.

[TRN]

I converted everything to metric because I can never remember whether its pounds or slugs! The rotational velocity and acceleration is in radians/sec and radians/sec^2. Anyway, I get 0.771 kW, about 1.1hp, for the question asked by randy331. The calculations are covered in rotational kinematics, which is part of a freshman physics class or probably sophomore statics and dynamics. You'll need some calculus, but nothing past basic integrals and derivatives. This stuff was 25 years ago for me, so it's all a bit foggy. And I studied 24/7, and NEVER partied until 4 AM!

[panic]

as any bobweight reduction would be centered on the rod throw

How can this be?

[Rick360]

Anyway, I get 0.771 kW, about 1.1hp, for the question asked by randy331.

At what RPM is this answer for? You should get a torque required to accellerate a given rpm/sec then you calculate the HP from TQ and RPM using the other formula given at the bottom of greenlight's post.

I haven't had a chance to calculate this myself yet, but I'll give it a try tonight if I have time.

Rick

[s/c 266]

ref Wallace calc. rotating inertia of your engine
what's good and what's bad?

I ran my BBC race parts and got some scary numbers.

[TRN]

Rick360
I ran the problem as stated, and you'll have to forgive my oversight. I was so excited that the units ended up correct, I overlooked the obvious problem with starting at 0 RPM, which seems impractical.

I get a rotational inertia of 0.00019533 kg * m^2.
Multiply by angular acceleration of 62.83 rads/sec^2 for torque.
I get 0.012273 kg * m^2/s^2.
At this point, thrilled to death ITS WORKING (the brain) I multiplied by the final angular velocity (62.83 rads/s) and out pops an answer.

So now I need someone to kick the fog out. When you calculate power, do you use the velocity differential, the average velocity, the final velocity, what? I don't recall. I'll think about it on the drive home, and hopefully it clears up.
Since power is at a point in time, not integrated over time (work) my guess is the number I gave is for 600RPM. 6000RPM will be 10 times as much power.

I remind every kid I see to take advantage of education, just like people told me when I was young, probably with the same effect (not much).

[Strange Magic]

Do remember this, since the intended project is ultimately always proven by a contest of speed and accelaration, it is important to always consider component weight prior to your build. The buck doesn't stop at the dyno. There are some serious advantages that come from weight reduction. In many of cases the racer will opt for what he believes is the stronger component based on weight and ultimately will pay the price due to the increased pounds per square inch that the engine block and crankshaft sees. Do also remember that most of your low market priced products are the ones that are the heaviest and greatly assist in the damage caused. Spend a little extra money, hook up with the right engine builder and your mechanical reliabilty and performance will double.

[Greenlight]

Greenlight; What unit of measure for mass?
Angular acceleration, Is that inches per sec., feet per sec., if I use inches from center of rotation, I would use inches with angular acceleration?
or is that rpm.?

With a bobweight reduction I wouldn't need to know the MOI. of the engine, as any bobweight reduction would be centered on the rod throw, and and would affect the acceleration the same regardless of the original MOI.? correct???


Thanks; Randy

As I stated, watch your units. I would suggest using the metric system throughout the entire problem and then convert to the IPS (inch, pound, seconds) system after you are done.

I have an Excel spreadsheet that I use to perform all of the calculations and I use the data that I download from my on board computer to determine the engine acceleration in each gear. In first gear I accelerate at ~5000 rpm/sec (I know this is hard to believe, but I have the data to prove it) and in high gear I accelerate at ~630 rpm/sec.

In general, for an engine operating in the 7k to 9k range, when you remove 100 grams of bobweight from each of the four rod journals and you accelerate the engine at 600 rpm/sec you gain the equivalent of 8 HP.

[Rick360]

Thanks; Rick and packard;

With the power needed to accelerate a given weight,a given rpm. is a fixed amount.
Is there a formula to calculate the tq. or hp. required to accelerate 100 grams 600rpm in one second at 1.74" from the center of rotation?

Or is the formula one that I would need a calculator with the formula allready in it, so I just enter weight,distance from center,and acceleration rate and it gives me the answer?

Thanks; Randy

Excusse the spelling as my spell checker is at school.

I finally had time to work this out.

First convert units to metric
100 gram bobweight change x 4 rod journals = 400gms = 0.4 kg
Crank radius to center of bobweight = 1.74” = 0.044196meters
Accel rate of 600rpm/s^2 = 62.83185 radians/sec^2

MOI = mass x radius^2 = 0.4 x .044196^2 = 0.000781 kg.m^2

Torque = MOI x Accel = 0.000781 kg.m^2 x 62.83185 rad/sec = 0.049092 Nm

Convert back to American/British units 0.049092 Nm = 0.036208 Lb/ft of Torque

HP at 5000rpm = (5000 x 0.036208)/5252 = 0.03447 HP to accelerate 400grams total bobweight at 600rpm/sec at 5000rpm

HP at 7000rpm = (7000 x 0.036208)/5252 = 0.048259HP HP to accelerate 400grams total bobweight at 600rpm/sec at 7000rpm

Not nearly as much as I expected. I went back and recalculated about 5 times thinking there was an error. I can’t find one, please anyone, let me know if I’ve made a mistake. I don’t think you could measure this on any kind of dyno.

The calculated total rotating dyno (engine and dyno) MOI is approx 0.303 kg*m^2 using the HP difference (6hp @ 4500) from your 300rpm/s and 600rpm/s.

Rick