Flywheel physics
06-03-2009, 12:44 PM
#1
jersey fresh
Thread Starter
Join Date: Dec 2005
Location: Boston, MA
Posts: 3,688
Likes: 0
Received 0 Likes
on
0 Posts
Flywheel physics
Thought i'd open the discussion up for the physics of a flywheel assembly.
(warning, math ahead)
You start with the follow equation:
Torque = (D(I OMEGA) / DT
Omega is your rotational speed to Torque, and your flywheel will increase the moment of inertia I.
If Torque increased your derivative of Omega will shrink if I is larger.
NONE MATH TERMS: If I (the weight of your flywheel) is increased or decreased, in order to solved the equation either Omega must change or your torque must change. Since the torque of the engine is not going to be changed, the Omega will be smaller. Basically the heavier the flywheel the slower you'll accelerate.
BUT, the flywheel is a device of stored energy, the heavier it is, the more energy it can store. Storage of energy has several equations depending on the shape of the "flywheel."
The lighter the flywheel, the harder it will be to initially go from a stop "clutch wise." Your lightened flywheel will not store as much energy as the stock one, so to balance the equation you'll have to give it "more gas" upon take off from a stand still.
In the car world, you have to balance drivablity, clutch wear, gas mileage, driver ability, etc when designing a flywheel's dimensions and weight. Do you want smooth easy take offs and a lower idle RPM at the cost of acceleration? Or do you want maximum acceleration at the cost of higher idle RPM's and higher RPM launching speeds? Or somewhere in the middle?
(feel free to comment or correct any maths present, it's been awhile since i reviewed rotational energy equations.)
(warning, math ahead)
You start with the follow equation:
Torque = (D(I OMEGA) / DT
Omega is your rotational speed to Torque, and your flywheel will increase the moment of inertia I.
If Torque increased your derivative of Omega will shrink if I is larger.
NONE MATH TERMS: If I (the weight of your flywheel) is increased or decreased, in order to solved the equation either Omega must change or your torque must change. Since the torque of the engine is not going to be changed, the Omega will be smaller. Basically the heavier the flywheel the slower you'll accelerate.
BUT, the flywheel is a device of stored energy, the heavier it is, the more energy it can store. Storage of energy has several equations depending on the shape of the "flywheel."
The lighter the flywheel, the harder it will be to initially go from a stop "clutch wise." Your lightened flywheel will not store as much energy as the stock one, so to balance the equation you'll have to give it "more gas" upon take off from a stand still.
In the car world, you have to balance drivablity, clutch wear, gas mileage, driver ability, etc when designing a flywheel's dimensions and weight. Do you want smooth easy take offs and a lower idle RPM at the cost of acceleration? Or do you want maximum acceleration at the cost of higher idle RPM's and higher RPM launching speeds? Or somewhere in the middle?
(feel free to comment or correct any maths present, it's been awhile since i reviewed rotational energy equations.)
06-03-2009, 12:52 PM
#2
The anti-ricer
iTrader: (1)
Join Date: Oct 2005
Location: Hillsdale, NJ
Posts: 1,888
Likes: 0
Received 0 Likes
on
0 Posts
Thought i'd open the discussion up for the physics of a flywheel assembly.
(warning, math ahead)
You start with the follow equation:
Torque = (D(I OMEGA) / DT
Omega is your rotational speed to Torque, and your flywheel will increase the moment of inertia I.
If Torque increased your derivative of Omega will shrink if I is larger.
NONE MATH TERMS: If I (the weight of your flywheel) is increased or decreased, in order to solved the equation either Omega must change or your torque must change. Since the torque of the engine is not going to be changed, the Omega will be smaller. Basically the heavier the flywheel the slower you'll accelerate.
BUT, the flywheel is a device of stored energy, the heavier it is, the more energy it can store. Storage of energy has several equations depending on the shape of the "flywheel."
The lighter the flywheel, the harder it will be to initially go from a stop "clutch wise." Your lightened flywheel will not store as much energy as the stock one, so to balance the equation you'll have to give it "more gas" upon take off from a stand still.
In the car world, you have to balance drivablity, clutch wear, gas mileage, driver ability, etc when designing a flywheel's dimensions and weight. Do you want smooth easy take offs and a lower idle RPM at the cost of acceleration? Or do you want maximum acceleration at the cost of higher idle RPM's and higher RPM launching speeds? Or somewhere in the middle?
(feel free to comment or correct any maths present, it's been awhile since i reviewed rotational energy equations.)
(warning, math ahead)
You start with the follow equation:
Torque = (D(I OMEGA) / DT
Omega is your rotational speed to Torque, and your flywheel will increase the moment of inertia I.
If Torque increased your derivative of Omega will shrink if I is larger.
NONE MATH TERMS: If I (the weight of your flywheel) is increased or decreased, in order to solved the equation either Omega must change or your torque must change. Since the torque of the engine is not going to be changed, the Omega will be smaller. Basically the heavier the flywheel the slower you'll accelerate.
BUT, the flywheel is a device of stored energy, the heavier it is, the more energy it can store. Storage of energy has several equations depending on the shape of the "flywheel."
The lighter the flywheel, the harder it will be to initially go from a stop "clutch wise." Your lightened flywheel will not store as much energy as the stock one, so to balance the equation you'll have to give it "more gas" upon take off from a stand still.
In the car world, you have to balance drivablity, clutch wear, gas mileage, driver ability, etc when designing a flywheel's dimensions and weight. Do you want smooth easy take offs and a lower idle RPM at the cost of acceleration? Or do you want maximum acceleration at the cost of higher idle RPM's and higher RPM launching speeds? Or somewhere in the middle?
(feel free to comment or correct any maths present, it's been awhile since i reviewed rotational energy equations.)
O and there is definitely a difference in gas required to take off, as well as the increase in acceleration!
(mine is 8.98lbs...ty Ray)
Thread
Thread Starter
Forum
Replies
Last Post
