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
Moderator: Team
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.
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.Stan Weiss wrote: ↑Fri Mar 10, 2023 1:39 pmI 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
andSchmidtMotorWorks wrote: ↑Fri Mar 10, 2023 10:23 amhp = torque x rpmdigger wrote: ↑Fri Mar 10, 2023 4:28 amThere is power consumed by angularly accelerating a body, the power is given by = I x alpha x omegaSchmidtMotorWorks wrote: ↑Thu Mar 09, 2023 11:02 am 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.
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
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.
Maybe you should state a dimensionality that includes acceleration.digger wrote: ↑Fri Mar 10, 2023 6:42 pmandSchmidtMotorWorks wrote: ↑Fri Mar 10, 2023 10:23 amhp = torque x rpmdigger wrote: ↑Fri Mar 10, 2023 4:28 am
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
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
alpha = angular acceleration (rad/s/s) = linear acceleration / radius (m/s/s/r)SchmidtMotorWorks wrote: ↑Mon Mar 13, 2023 12:44 amMaybe you should state a dimensionality that includes acceleration.
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.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.
No, I am stating that neither torque or HP are the correct dimensionalities to describe.
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 bodySchmidtMotorWorks wrote: ↑Mon Mar 13, 2023 11:58 amNo, I am stating that neither torque or HP are the correct dimensionalities to describe.
Energy would be a better choice.
Since we are talking about accelerating a mass, Energy seems like the right way to define it.digger wrote: ↑Mon Mar 13, 2023 4:35 pm
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
if youre talling about acceleration there is time factor involved hence energy and time are where the power comes fromSchmidtMotorWorks wrote: ↑Mon Mar 13, 2023 11:25 pmSince we are talking about accelerating a mass, Energy seems like the right way to define it.digger wrote: ↑Mon Mar 13, 2023 4:35 pm
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
If someone says, "for a given energy, a lesser mass will accelerate faster than a greater mass" the appropriate reply would be "duh".
Energy is sufficient to describe the consequence of reduced mass to acceleration.digger wrote: ↑Tue Mar 14, 2023 4:47 amif youre talling about acceleration there is time factor involved hence energy and time are where the power comes fromSchmidtMotorWorks wrote: ↑Mon Mar 13, 2023 11:25 pmSince we are talking about accelerating a mass, Energy seems like the right way to define it.digger wrote: ↑Mon Mar 13, 2023 4:35 pm
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
If someone says, "for a given energy, a lesser mass will accelerate faster than a greater mass" the appropriate reply would be "duh".
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