Static < Rotational Weight=?
01-11-2009, 10:50 AM
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Static < Rotational Weight=? I'm trying to find out the exact formula on Static weight conversion to Rotational weight. To put into a better example or basically the question that is bothering me how much of a difference is from a wheel previously weighted in static form at 25 pounds to a lighter wheels of 15 pounds however when both are in motion?
Should we take into account the diameter of the rim too??
like how much is the difference in weight of the two of a Static weighted 15 ich 25 pounds rims to a 20 inch 25 (static) pound rim in Rotational weight?
would these two above weight the same because both have the same Static weight or the 20" rim weight more because of having more mass away from the center line(in diameter) when in motion?
Oh I almost forgot all of this happening say at 20 mph.And to make it fun lets calculate all of the above examples at 100 mph too!!.
Please lets keep this educational....
Should we take into account the diameter of the rim too??
like how much is the difference in weight of the two of a Static weighted 15 ich 25 pounds rims to a 20 inch 25 (static) pound rim in Rotational weight?
would these two above weight the same because both have the same Static weight or the 20" rim weight more because of having more mass away from the center line(in diameter) when in motion?
Oh I almost forgot all of this happening say at 20 mph.And to make it fun lets calculate all of the above examples at 100 mph too!!.
Please lets keep this educational....
01-11-2009, 11:27 AM
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It's actually very difficult to get an exact answer for something like a wheel, especially with a tire on it. A formula for calculating rotational mass is a=T(2/md+d/2I) where a is acceleration, T is the torque the car applies to the wheel, mass is the mass of the wheel, d is the diameter, and I believe I is the polar integral of the mass at any point times the distance from the center squared. So that means that weight at the outside of the wheel matters a lot more than weight towards the center. So a 20" wheel would have a much higher rotational mass than on a 15" wheel.
The difference in weight between static and rotational is 2X on an object with evenly distributed weight, so it's probably 3 or 4 times as much for something like a wheel.
The difference in weight between static and rotational is 2X on an object with evenly distributed weight, so it's probably 3 or 4 times as much for something like a wheel.
01-11-2009, 05:17 PM
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First off, inertia only comes into play during changes in speed and or cornering.
This is Newtons First Law of motion (known as the law of inertia)- an object in motion or at rest will remain in motion or at rest unless it is acted upon by an outside force.
When looking at the effects of wheel intertia during a vehicle acceleration event you would need to define the rate of acceleration (at tires contact surface) and know the polar moment of inertia of the wheels and outside diameter of the tires. Once those are defined, the calculation is simple.
The result will tell you how much torque is required to accelerate the wheel at that rate. Then you can compare what effect it has.
The hard part will be defining the polar moment of inertia of the wheel because the location of the mass is just as critical as the mass itself.. (i.e. you can have 2 different 20" wheels that weight 30 lbs and come up with different inertia values)
This is Newtons First Law of motion (known as the law of inertia)- an object in motion or at rest will remain in motion or at rest unless it is acted upon by an outside force.
When looking at the effects of wheel intertia during a vehicle acceleration event you would need to define the rate of acceleration (at tires contact surface) and know the polar moment of inertia of the wheels and outside diameter of the tires. Once those are defined, the calculation is simple.
The result will tell you how much torque is required to accelerate the wheel at that rate. Then you can compare what effect it has.
The hard part will be defining the polar moment of inertia of the wheel because the location of the mass is just as critical as the mass itself.. (i.e. you can have 2 different 20" wheels that weight 30 lbs and come up with different inertia values)
01-11-2009, 06:37 PM
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First off, inertia only comes into play during changes in speed and or cornering.
This is Newtons First Law of motion (known as the law of inertia)- an object in motion or at rest will remain in motion or at rest unless it is acted upon by an outside force.
When looking at the effects of wheel intertia during a vehicle acceleration event you would need to define the rate of acceleration (at tires contact surface) and know the polar moment of inertia of the wheels and outside diameter of the tires. Once those are defined, the calculation is simple.
The result will tell you how much torque is required to accelerate the wheel at that rate. Then you can compare what effect it has.
The hard part will be defining the polar moment of inertia of the wheel because the location of the mass is just as critical as the mass itself.. (i.e. you can have 2 different 20" wheels that weight 30 lbs and come up with different inertia values)
This is Newtons First Law of motion (known as the law of inertia)- an object in motion or at rest will remain in motion or at rest unless it is acted upon by an outside force.
When looking at the effects of wheel intertia during a vehicle acceleration event you would need to define the rate of acceleration (at tires contact surface) and know the polar moment of inertia of the wheels and outside diameter of the tires. Once those are defined, the calculation is simple.
The result will tell you how much torque is required to accelerate the wheel at that rate. Then you can compare what effect it has.
The hard part will be defining the polar moment of inertia of the wheel because the location of the mass is just as critical as the mass itself.. (i.e. you can have 2 different 20" wheels that weight 30 lbs and come up with different inertia values)
Yeah I looked up on Newtons laws but ofcourse the answer wasn't there to eat it. It is a little more elaborate than that.
01-11-2009, 06:43 PM
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[QUOTE]It's actually very difficult to get an exact answer for something like a wheel, especially with a tire on it. A formula for calculating rotational mass is a=T(2/md+d/2I) where a is acceleration, T is the torque the car applies to the wheel, mass is the mass of the wheel, d is the diameter, and I believe I is the polar integral of the mass at any point times the distance from the center squared. So that means that weight at the outside of the wheel matters a lot more than weight towards the center. So a 20" wheel would have a much higher rotational mass than on a 15" wheel.
The difference in weight between static and rotational is 2X on an object with evenly distributed weight, so it's probably 3 or 4 times as much for something like a wheel.
QUOTE]
Indeed the same weight with a longer ratio should be more weight when measuring as rotational weight. I wounder if anyone has made an deep research experiment with mass in a rotational status . Thank you very much for your insight that formula above shed some light into was I was looking for.
The difference in weight between static and rotational is 2X on an object with evenly distributed weight, so it's probably 3 or 4 times as much for something like a wheel.
QUOTE]
Indeed the same weight with a longer ratio should be more weight when measuring as rotational weight. I wounder if anyone has made an deep research experiment with mass in a rotational status . Thank you very much for your insight that formula above shed some light into was I was looking for.
