Reciprocating weight tests ?

[Stan Weiss]

A less powerful car may need a lighter rotating mass for better acceleration.. but if you grab that flywheel it doesnt make torque.


IMG20230310185004.jpg



-juhana

I am sure you will see the difference in torque/HP when you see how much farther you can throw the lightened flywheel verse the regular wight flywheel.

Stan

[juuhanaa]

A less powerful car may need a lighter rotating mass for better acceleration.. but if you grab that flywheel it doesnt make torque.


IMG20230310185004.jpg



-juhana

I am sure you will see the difference in torque/HP when you see how much farther you can throw the lightened flywheel verse the regular wight flywheel.

Stan

I bet my 800kg 60hp opel would accelerate differently, but which tire flies further or goes faster? Hint: I havent had a problem with the flywheel.

Does anyone know if the car would have gone faster with a lighter flywheel? My experience is that when the car falls on its face and is slow slow (torque curves and butt dyno), it doesnt matter at that point anymore how heavy the flywheel is.



-juhana

[digger]

I'm curious, who believes that DVs claim that reducing mass increases rear wheel power?
Hint; The work being done was reduced, the power was not increased.

This can be computed more precisely than dyno or track testing can.
If dyno or track testing contradicts the computation, the testing is flawed.

There is power consumed by angularly accelerating a body, the power is given by = I x alpha x omega

where

I = mass moment of inertia (kgm^2)
alpha = angular acceleration (rad/s^2)
omega = rotational velocity (rad/s)

take a 10kg flysheel 12" in diameter accelerating from 2000-7000 in 7 seconds

the power consumed at 7000rpm is

I = 1/2mr^2 = 0.5*10*0.1524^2 = 0.116
omega = 7000 * 2 * pi / 60 = 733 rad/s
alpha = (733 - 2000 * 2 * p / 60) / 7 = 75 rad/s^2

power = 0.116 * 733 * 75 = 6377 watts = 6.3 kW

hp = torque x rpm

and

torque = I x alpha

which is the rotational equivalent to F=ma

ergo

hp = I x alpha x rpm with appropriate constants/conversion factor for units

[juuhanaa]

The flywheel turns when the starter engages it, not when the crankshaft turns it.



-juhana

[Geoff2]

What turns the f/wheel when the starter disengages......

[juuhanaa]

What turns the f/wheel when the starter disengages......

A good flowing headers, balanced port with a 350fps at the seat and 225 on average? Well it does it for a 1000kg car without the need to slip the clutch. I use less counter weight crank for a lighter piston rod combo and diesel crank for a forged piston h-profile rod combo.

Where is bigchief with signature quote when you need it most..



-juhana

[SchmidtMotorWorks]

There is power consumed by angularly accelerating a body, the power is given by = I x alpha x omega

where

I = mass moment of inertia (kgm^2)
alpha = angular acceleration (rad/s^2)
omega = rotational velocity (rad/s)

take a 10kg flysheel 12" in diameter accelerating from 2000-7000 in 7 seconds

the power consumed at 7000rpm is

I = 1/2mr^2 = 0.5*10*0.1524^2 = 0.116
omega = 7000 * 2 * pi / 60 = 733 rad/s
alpha = (733 - 2000 * 2 * p / 60) / 7 = 75 rad/s^2

power = 0.116 * 733 * 75 = 6377 watts = 6.3 kW

hp = torque x rpm

and

torque = I x alpha

which is the rotational equivalent to F=ma

ergo

hp = I x alpha x rpm with appropriate constants/conversion factor for units

Maybe you should state a dimensionality that includes acceleration.

[digger]

hp = torque x rpm

and

torque = I x alpha

which is the rotational equivalent to F=ma

ergo

hp = I x alpha x rpm with appropriate constants/conversion factor for units

Maybe you should state a dimensionality that includes acceleration.

alpha = angular acceleration (rad/s/s) = linear acceleration / radius (m/s/s/r)

are you really disputing that it takes torque and power to accelerate an object with rotational inertia from rpm A to rpm B within a time period of t?

https://en.wikipedia.org/wiki/Torque

[juuhanaa]

This is a twisted way of thinking about increasing hp. Brakes are for braking, while the engine produces the best hp under braking.. It depends on us if we want the engine to accelerate quickly and like so use it as efficiently as possible. Just carry on, but i would like to remind what the OP was asking..

Any links to any dyno tests of reciprocating weight along with the potential reduction in rotating weight for balance ?

If it's dyno results, it would have to be with the load held for a period of time, otherwise mathematical.

Oh almost forgot, folkrace is mostly about the engine and getting the car hooked! Shaky brakes and rust holes are included in the sale price.

IMG20230313172830.jpg

-juhana

[SchmidtMotorWorks]

are you really disputing that it takes torque and power to accelerate an object with rotational inertia from rpm A to rpm B within a time period of t?

No, I am stating that neither torque or HP are the correct dimensionalities to describe.

Energy would be a better choice.

[digger]

are you really disputing that it takes torque and power to accelerate an object with rotational inertia from rpm A to rpm B within a time period of t?

No, I am stating that neither torque or HP are the correct dimensionalities to describe.

Energy would be a better choice.

It takes a force to linearly accelerate an object from velocity A to velocity B, so going from rpm A to rpm B requires a torque to rotationally accelerate the body

You are doing it within a defined time period so the change in kinetic energy divided by time gives power

[SchmidtMotorWorks]

It takes a force to linearly accelerate an object from velocity A to velocity B, so going from rpm A to rpm B requires a torque to rotationally accelerate the body

You are doing it within a defined time period so the change in kinetic energy divided by time gives power

Since we are talking about accelerating a mass, Energy seems like the right way to define it.

If someone says, "for a given energy, a lesser mass will accelerate faster than a greater mass" the appropriate reply would be "duh".

[digger]

It takes a force to linearly accelerate an object from velocity A to velocity B, so going from rpm A to rpm B requires a torque to rotationally accelerate the body

You are doing it within a defined time period so the change in kinetic energy divided by time gives power

Since we are talking about accelerating a mass, Energy seems like the right way to define it.

If someone says, "for a given energy, a lesser mass will accelerate faster than a greater mass" the appropriate reply would be "duh".

if youre talling about acceleration there is time factor involved hence energy and time are where the power comes from

the reality is if on a transient power run of chassis dyno increasing rotational mass results in less power at the wheels, the engine is making the same power but a portion is being consumed in accelerating mass.

if you have no frictional losses and the gearing is 1:1 the flywheel torque will still NOT equal wheel torque as it is not in static equilibrium. A sum of moments will show you this

[juuhanaa]

This is BS

Screenshot_2023-03-14-08-49-17-39_92460851df6f172a4592fca41cc2d2e6.jpg

-juhana

[SchmidtMotorWorks]

It takes a force to linearly accelerate an object from velocity A to velocity B, so going from rpm A to rpm B requires a torque to rotationally accelerate the body

You are doing it within a defined time period so the change in kinetic energy divided by time gives power

Since we are talking about accelerating a mass, Energy seems like the right way to define it.

If someone says, "for a given energy, a lesser mass will accelerate faster than a greater mass" the appropriate reply would be "duh".

if youre talling about acceleration there is time factor involved hence energy and time are where the power comes from

the reality is if on a transient power run of chassis dyno increasing rotational mass results in less power at the wheels, the engine is making the same power but a portion is being consumed in accelerating mass.

if you have no frictional losses and the gearing is 1:1 the flywheel torque will still NOT equal wheel torque as it is not in static equilibrium. A sum of moments will show you this

Energy is sufficient to describe the consequence of reduced mass to acceleration.