Un-sprung wheel weight calculator

[bluedevil]

I have read about 100 different ways to calculate reductions of weight of the wheels. Most actual hard numbers I was able to find are car related. Im trying to find out the actual hard numbers on lets say a 1.8lb to 3.9lb reduction of weight on the wheels and how it equates to power gain or change in actual spinning weight? Anyone have a calculator or a simple equation in relation to motorcycles for only a few lbs of change?

Not looking to debate... just a simple way to put a numerical value on shaving some weight off of wheels/ sprockets/rotors etc.....

Thanks for any input......

[rybo]

as reltated to bicycles

http://en.wikipedia.org/wiki/Bicycle_performance

Look down, while it doesn't address "unspurung" weight, it does address "rotational" weight, which is a pretty big deal on motocycles.

As the article states, the only time KE (kinetic energy) comes into play is when velocity changes. This is the WHOLE game on the motorcycle on the racetrack. We are always seeking to accelerate, brake or turn, so in that case velocity is almost ALWAYS changing.

s

[Sol Performance]

your moment of inertia exn would be

I=(1/2)(m)(r^2)

where
m=weight of wheel/g (IE 3 pounds/32.174)
g=32.174

r=radius of wheel

This gives a way for basic comparison for a simple "short cylinder..

Edit: or you can just look at the link rybo just put..haha

[bluedevil]

ok so lets say your OEM Rear wheel weighs 15.4lbs
your aftermarket Rear wheel weighs 13.4lbs

total loss of 2lbs of non spinning weight....... How many lbs of spinning weight does that equate to losing? Is it 1 for 1 or? 10:1 ?

What if any in hard numbers benefit are you gaining by the drop of the 2lbs?



going further: Lets say the front and rear wheels together:
OEM F+R = 25.2lbs

AM F+R = 22.2lbs
Loss of combined weight of 3 lbs of non-spinning weight.... How will that equate to spinning weight? (Kinda the same question just re-worded with some actual data inputs...)
:wink:

[rybo]

your moment of inertia exn would be

I=(1/2)(m)(r^2)

where
m=weight of wheel/g (IE 3 pounds/32.174)
g=32.174

r=radius of wheel

This gives a way for basic comparison for a simple "short cylinder..

Edit: or you can just look at the link rybo just put..haha

Oscar

That's a start, but doesn't account for speed of rotation, right? To determine KE you would have to add speed into the equation. For example the 3 lb drop that Dion quotes is static weight has a much larger effect at 100MPH (or how ever many RPM) than it does at 10 MPH...

This is where you engineer types come in....

s

[T Baggins]

All's I know is:

Lighter wheels are for cheaters!

And they crash very expensively... lol :lol:

[Sol Performance]

Reading your question again the easy answer would be just to have your power/weight ratio (HP/pounds) for before and after and that gives you the most direct answer. But it doesn't really tell you what it might do for the the track and the dynamics of how it helps.

I think what you are looking for instead would be in the terms of the bastard child of dynamics..torque is basically a measurement of how much energy is required to turn some thing

The energy of a wheel turning at a velocity of w (rad/sec) is
E = (1/2) I w^2

Using the inertia exn we had previously talked about
I=(1/2)(m)(r^2)

thus E= (1/4) (m) (R*w)^2

R*w is just the velocity of the rolling wheel with radius R
(This is basically what your stock speedo goes off of)

so R*w= v (mph) which we will have to convert to ft/sec later.

so now we have,
E = (1/4) (m) (v^2)

For a wheel and tire combo with a weight of 35 lbs (Remember m=w/g, g=gravity 32.174 ft/sc^2) and 60 mph (88 ft/sec)

E = (1/4) (35/32.174) (88^2) = 2106 lbfft^2/sec^2

This is where you could stop if you really wanted to but if you want to torture yourself further you would go on..

it's actually easier doing it in metric
35lbs =15.875733 kg
60mph=26.8 m/sec
gravity is now 9.8 m/s^2

which makes the above
E= (1/4)(15.875733/9.8 )(26.8^2)= 290.88 Joules

This is how much torque is needed to turn one wheel weighing 35 lbs.

To get HP you need to have some sort of time value. So if you wanted to see what you would need to spin the wheel and tires up to 60 in a matter of 2 seconds you would need:

P = (2 x 290.88 Joules)/2 sec = 290.88 watts

1 hp = 745.699872 watts
so 290.88 watts * (1hp/745.7 watts) = .39 HP

or 3 seconds would be

P= (2 x 290.88 )/ 3 sec = 193.92 watts = .26 HP

Note: This for just the wheel and/or tire. It does not take the rest of the bike or transmission lag into consideration at all. So there is the basis for a comparison for the weight in a perfect and yet unperfect world.

Additionally, if you wanted to find out how much this effects you in a turn we would do so by using one of the oldest exns known to man:

F=ma

well, we would have to modify it just a bit to fit your example though..

F=m((v^2)/(r))

where:
m= mass of bike (remember also that m=w/g, where w=weight of bike (pounds) and g=gravity (32.174 )

v=velocity of bike (need the value in ft/sec!!)

r=radius of the turn you are in (Taking an example turn, say turn 1 at hpr. If you stood at the apex and followed the curve all the way around the turn to make a complete circle.. this would be the distance from the apex to the center of the circle, probably be by the inside wall or a little further in the paddock or stands)

So then plug all the numbers is an you have a basis for comparison from one weight to another (additionally, one turn to another too).

Example:
If you had a 350 lb bike, going 75 mph (110 ft/sec) , on a turn with a radius of 100ft.

F=(350/32.174)((75^2)/100)) == (10.878)((110^2)/100) == 10.878(121) ==1316.238 lbs. of force

If you dropped the weight down to 345 lbs in the same turn the end result would be:
1297.476 lbs force.

What does that number actually mean? That means that if there was a piece of wire or rope where one side was attached to your bike and the other was somehow "attached" to a post at the center of the that turn, that would be the force you would put on that wire. Basically how much force you would put on it trying to break free from it.

Obviously the force doesn't just make you just fly off the track because of you leaning into the turn and also why some tires are better than others to handle this "centrifugal" force. (yes, that was a shameless plug, don't hang me!)

Hope this helps a little bit!

Disclaimer: Me thinking before I get coffee, can and might affect some reasoning..

[Bartman]

The other thing to remember is where is the weight, on the rim or the hub because that will make a huge difference as to rotating weight.
I can tell you from personal experience that a 10ish lbs drop will feel like a 10 HP gain as well as the handling benefits.
I remember that Roadracing world did a article with all the answers to your questions a couple of years back, you can prolly find it.

[bluedevil]

Reading your question again the easy answer would be just to have your power/weight ratio (HP/pounds) for before and after and that gives you the most direct answer. But it doesn't really tell you what it might do for the the track and the dynamics of how it helps.

I think what you are looking for instead would be in the terms of the bastard child of dynamics..torque is basically a measurement of how much energy is required to turn some thing

The energy of a wheel turning at a velocity of w (rad/sec) is
E = (1/2) I w^2

Using the inertia exn we had previously talked about
I=(1/2)(m)(r^2)

thus E= (1/4) (m) (R*w)^2

R*w is just the velocity of the rolling wheel with radius R
(This is basically what your stock speedo goes off of)

so R*w= v (mph) which we will have to convert to ft/sec later.

so now we have,
E = (1/4) (m) (v^2)

For a wheel and tire combo with a weight of 35 lbs (Remember m=w/g, g=gravity 32.174 ft/sc^2) and 60 mph (88 ft/sec)

E = (1/4) (35/32.174) (88^2) = 2106 lbfft^2/sec^2

This is where you could stop if you really wanted to but if you want to torture yourself further you would go on..

it's actually easier doing it in metric
35lbs =15.875733 kg
60mph=26.8 m/sec
gravity is now 9.8 m/s^2

which makes the above
E= (1/4)(15.875733/9.8 )(26.8^2)= 290.88 Joules

This is how much torque is needed to turn one wheel weighing 35 lbs.

To get HP you need to have some sort of time value. So if you wanted to see what you would need to spin the wheel and tires up to 60 in a matter of 2 seconds you would need:

P = (2 x 290.88 Joules)/2 sec = 290.88 watts

1 hp = 745.699872 watts
so 290.88 watts * (1hp/745.7 watts) = .39 HP

or 3 seconds would be

P= (2 x 290.88 )/ 3 sec = 193.92 watts = .26 HP

Note: This for just the wheel and/or tire. It does not take the rest of the bike or transmission lag into consideration at all. So there is the basis for a comparison for the weight in a perfect and yet unperfect world.

Additionally, if you wanted to find out how much this effects you in a turn we would do so by using one of the oldest exns known to man:

F=ma

well, we would have to modify it just a bit to fit your example though..

F=m((v^2)/(r))

where:
m= mass of bike (remember also that m=w/g, where w=weight of bike (pounds) and g=gravity (32.174 )

v=velocity of bike (need the value in ft/sec!!)

r=radius of the turn you are in (Taking an example turn, say turn 1 at hpr. If you stood at the apex and followed the curve all the way around the turn to make a complete circle.. this would be the distance from the apex to the center of the circle, probably be by the inside wall or a little further in the paddock or stands)

So then plug all the numbers is an you have a basis for comparison from one weight to another (additionally, one turn to another too).

Example:
If you had a 350 lb bike, going 75 mph (110 ft/sec) , on a turn with a radius of 100ft.

F=(350/32.174)((75^2)/100)) == (10.878)((110^2)/100) == 10.878(121) ==1316.238 lbs. of force

If you dropped the weight down to 345 lbs in the same turn the end result would be:
1297.476 lbs force.

What does that number actually mean? That means that if there was a piece of wire or rope where one side was attached to your bike and the other was somehow "attached" to a post at the center of the that turn, that would be the force you would put on that wire. Basically how much force you would put on it trying to break free from it.

Obviously the force doesn't just make you just fly off the track because of you leaning into the turn and also why some tires are better than others to handle this "centrifugal" force. (yes, that was a shameless plug, don't hang me!)

Hope this helps a little bit!

Disclaimer: Me thinking before I get coffee, can and might affect some reasoning..

well that clears it up .... thanks. LOL :roll:

The other thing to remember is where is the weight, on the rim or the hub because that will make a huge difference as to rotating weight.
I can tell you from personal experience that a 10ish lbs drop will feel like a 10 HP gain as well as the handling benefits.
I remember that Roadracing world did a article with all the answers to your questions a couple of years back, you can prolly find it.

I found some info that states: 1oz reduction on the wheels, is equal to 20 lbs of moving weight @ 100 mph. So a reduction of 1 lb really equals 320 lbs @ 100 mph??

[Jim 'smooth' Brewer]

Using the inertia exn we had previously talked about
I=(1/2)(m)(r^2)
...

That makes the assumption that all mass (m) is at radius r. It seems we need mass distribution as a function of r, then integrate over that distribution .. sort of what Bartman is suggesting. The gnarly equation (from Wikipedia) is ..



where r is the radius vector of a point within the body, ρ(r) is the mass density at point r, and d(r) is the distance from point r to the axis of rotation. The integration goes over the volume V of the body.

At minimum it would be tough to figure ρ(r) for our wheels, so probably just summing up the individual components mass @ r and using Oscar's formulas for each would be pretty close (like disk, disk screws, tire, rim, etc.)

Simple statements like 1oz equals 20 lbs @ 100 mph doesn't really make sense unless you know where the 1oz is. It wouldn't help shaving 1oz off of your wheel bearings.

Here's a decent article with some motorcycle wheel comparisons ..
http://www.sportrider.com/sportbike_...son/index.html

But, yeah, they're for cheaters anyway. [-X

[rybo]

Dion,

Unfortunately there just isn't an easy answer for this one. This is the problem with weight in motion, there are a lot of variables and that requires a lot of math to sort out.

S

[polar x]

Dion,

Unfortunately there just isn't an easy answer for this one. This is the problem with weight in motion, there are a lot of variables and that requires a lot of math to sort out.

S

DION!!!!! Just take a big poo and thats instant HP... 8) :wink: 8)

[rybo]

Dion,

Unfortunately there just isn't an easy answer for this one. This is the problem with weight in motion, there are a lot of variables and that requires a lot of math to sort out.

S

DION!!!!! Just take a big poo and thats instant HP... 8) :wink: 8)

Speaking of "weight in motion"!

[Jim 'smooth' Brewer]

This is the problem with weight in motion, there are a lot of variables and that requires a lot of math to sort out.

So, let's do it. Dion, what are you trying to do?
Figure out if you should buy titanium sprocket screws?
Drill out your disk rotors?
Buy some BST carbon wheels off ebay?

All will help some - some will help more, but they all cost differently.

Cooner, you know launching an anaconda doesn't help unsprung weight. Sheesh ...