SO I have been told a few times over the years that for every 10 Lbs you shave off your bike you gain 1 horsepower. Is this true? How can I find the formula for weight to horsepower ratio on the Busa?
Weight to HP ratio?
- Thread starter CaBusaGirl
- Start date
Well I've been trying to google it and I can't find any solid answers. Where did you get 7/1?
hellbent
Registered
I'm sure you realize you don't actually gain hp by dropping weight. The overall performance of any machine is directly affected by the power to weight ratio. If your bike were to weigh 500lbs and had 100hp then you would have a 1:5 power to weight ratio or 1hp for every 5lbs. of weight. The better you can get this ratio, by adding hp or dropping weight, the better your bike will perform. Think of the difference between a 300hp truck and a 300hp bike.
its a very relativistic thing... your obviously not "gaining" any engine horsepower... but there is a relative improvment in performance which people approximated to the effect of adding 1hp....
that said the rule of thumb which used to be 1hp = 10lbs applies only in a very very narrow set of circumstances.....
in other words, 10lbs of weight lost on a car is not going to reflect the same gains as 10lbs of weight lost on a motorcycle...
in a very basic way it comes down to Force = Mass*Acceleration....
or more directly F/M = A
so, as an example....and im going to ignore units for the sake of simplicity
a decent sports car can put out 400 HP... but weights say 3200 lbs...
a motorcycle on the other hand is outputting say 160hp but only weights 500 lbs
so a loss of 10 lbs from the motorcycle will obviously have a more profound effect on the acceleration of the bike vs the car...
in essence your horsepower to weight ratio is actually describing F/M = A
though as mentioned, i ignored some details for the sake of simplicity..
that said the rule of thumb which used to be 1hp = 10lbs applies only in a very very narrow set of circumstances.....
in other words, 10lbs of weight lost on a car is not going to reflect the same gains as 10lbs of weight lost on a motorcycle...
in a very basic way it comes down to Force = Mass*Acceleration....
or more directly F/M = A
so, as an example....and im going to ignore units for the sake of simplicity
a decent sports car can put out 400 HP... but weights say 3200 lbs...
a motorcycle on the other hand is outputting say 160hp but only weights 500 lbs
so a loss of 10 lbs from the motorcycle will obviously have a more profound effect on the acceleration of the bike vs the car...
in essence your horsepower to weight ratio is actually describing F/M = A
though as mentioned, i ignored some details for the sake of simplicity..
Yeah, I know the hp is there regardless. Was having a hard time putting it into words but I think you know what I meant. For every ____ weight you shave off a bike then the quality of your hp get's better by ____ ???
...
I know what I am trying to say but am having a hard time here...
hellbent
Registered
I know what you're getting at. You trying to find out how much weight to shave off to gain the performance equivelent of a 1 hp increase. My exwife caught on to this concept real quick and mentioned that instead of spending all the money on a new exhaust, my fat butt could just lose 20lbs. I was not amused, ate 2 cheeseburgers, and added nitrous. 
Less weight means each "pony" you have does less work. Imagine a shopping cart full of bricks, and your trying to push it. Nothing you can do to yourself at that moment will make you stronger, but less bricks in the cart, it will be as if you were stronger, easier to push.
I shaved 45lbs of unnecessary weight and it does make a difference, trust me. I do not know the ratio, but I do know the handling and responsiveness of the bike.
Bubba
jdsbusa
Registered
I know what you're getting at. You trying to find out how much weight to shave off to gain the performance equivelent of a 1 hp increase. My exwife caught on to this concept real quick and mentioned that instead of spending all the money on a new exhaust, my fat butt could just lose 20lbs. I was not amused, ate 2 cheeseburgers, and added nitrous.
A diet IS the cheapest way to add HP. You save money by not modding your bike and save money on the grocery bill at the same time.
busaville
Registered
I've heard the same thing.........For every 7 lbs. lost equals to 1 hp. gain and vise versa.... so to speak.
Amode
Registered
Horsepower Design Equations Formulas Calculator - Power Weight Ratio 
Yeah, I know the hp is there regardless. Was having a hard time putting it into words but I think you know what I meant. For every ____ weight you shave off a bike then the quality of your hp get's better by ____ ???
...
I know what I am trying to say but am having a hard time here...
Amode
Registered
Don't forget when you do your calcs, Gen1 are diff than Gen2.
As soon as I read the title I knew someone would drag me into this.
Let's use Newton's F=ma to answer the question.
First we need to ensure we all understand what Force (F) and mass (m) are.
Horsepower is not force. Horsepower is a rate at which force can be applied. Torque is a measure of rotational force. Torque is the work an enging/gearing system can do, horsepower is how fast the system can deliver that work.
A pound is a measure of force, not mass. A pound is calculated by multiplying the mass of an object by the local force of gravity, which on Earth is 32.2 feet per second squared. What this means to us is that if we have a hypothetical 700 lb. motorcycle/rider combination, the mass is actually a number far smaller...after all, if we put our motorcycle/rider combination into space it wouldn't weigh 700 lbs, but it would still have the same mass.
Dividing 700 lbs. by 32.2 ft/second squared equals 21.7. To keep things simple with unit conversions I'm going to skip over all that, but you'll have to take my word that our 21.7 value is actually 21.7 slugs, which is the British unit of mass.
The Gen 2 torque at the crankshaft is claimed to be 115 lb*ft. Gearing multiplies torque, however, so we have to account for both the gearbox ratio and the final drive ratio.
Second gear ratio, or torque multiplier is 1.937. Final drive ratio (stock) I believe (correct me if I'm wrong) is 2.35...based on a 17-tooth sprocket and a 40 tooth sprocket. There's a reson I've selected second gear which I'll explain later.
So our "F" for the equation is 115 lb*ft*1.937*2.35. For simplicity's sake I'm only using second gear.
F=523.5 lb*ft
m=21.7 slugs
Now for the fun part...
F=ma
523.5 lb*ft = 21.7 slugs(a)
523.5 lb*ft/21.7 slugs = a
24.1 = a
Again for simplicity's sake I've ignored unit conversion. Again I ask your understanding as you read that I'm just going to add a feet/second squared to our answer.
a= 24.1 ft/s*s.
How does our result stack up against real world numbers?
From wikipeda, Cycle World is quoted as having run a 9.75 @142 m.p.h. quarter mile. We don't know all the details behind that run (did their bike/rider combo weigh 700 lbs.? How much torque did they lose due to friction deficiencies?) We also know the torque multiplication of the gearbox isn't a constant rate during a quarter mile run. Our formula also doesn't account for parasite drag because we're keeping things simple here.
Taking these things into consideration, let's see what speed our math says the Hayabusa should be able to accelerate to in 9.75 seconds.
The final velocity of an object can be found by taking the initial velocity and simply adding the rate of acceleration multiplied by the time available for acceleration. In formula terms it looks like this:
V(f) = V(i)+at
so
V(f) = 0+24.1 ft/s*s*9.75s
V(f) = 235 ft/s
To convert that into miles per hour, multiply by 0.681818182.
V(f) = 160.22 miles per hour
160 m.p.h. is what our math says the Hayabusa should be able to achieve in 9.75 seconds...assuming zero wheelspin, zero aerodynamic drag, a 700 lb total weight, and that we never had to shift out of second gear. The last one is tricky because obviously that's not what is going to happen in the real world. I selected the second gear torque multiplication ratio in an attempt to provide a reasonable average torque multiplication value for the gearbox. With more time a better number could be calculated.
I think given the limitations we've placed on our calculation for time's sake that 160 m.p.h. in 9.75 seconds is close enough to 147 m.p.h. in the same time period to say our math is correct given the limitations we're working with.
So...now we have all the tools we need to calculate how much more "a" we can expect given whatever "m" reduction we want to use.
And that's really what we wanted to know...isn't it?
Last edited:
megawatt
Registered
F=MA...well done...and you must have way too much time on your keyboard..So, please continue with the math....so if we lose 10 lbs...what does that equate to in accleration or horsepower? I am a medical scientist and never went past calculus or first year physics....but I can tell you have2hip
2hip
As soon as I read the title I knew someone would drag me into this.
Let's use Newton's F=ma to answer the question.
First we need to ensure we all understand what Force (F) and mass (m) are.
Horsepower is not force. Horsepower is a rate at which force can be applied. Torque is a measure of rotational force. Torque is the work an enging/gearing system can do, horsepower is how fast the system can deliver that work.
A pound is a measure of force, not mass. A pound is calculated by multiplying the mass of an object by the local force of gravity, which on Earth is 32.2 feet per second squared. What this means to us is that if we have a hypothetical 700 lb. motorcycle/rider combination, the mass is actually a number far smaller...after all, if we put our motorcycle/rider combination into space it wouldn't weigh 700 lbs, but it would still have the same mass.
Dividing 700 lbs. by 32.2 ft/second squared equals 21.7. To keep things simple with unit conversions I'm going to skip over all that, but you'll have to take my word that our 21.7 value is actually 21.7 slugs, which is the British unit of mass.
The Gen 2 torque at the crankshaft is claimed to be 115 lb*ft. Gearing multiplies torque, however, so we have to account for both the gearbox ratio and the final drive ratio.
Second gear ratio, or torque multiplier is 1.937. Final drive ratio (stock) I believe (correct me if I'm wrong) is 2.35...based on a 17-tooth sprocket and a 40 tooth sprocket. There's a reson I've selected second gear which I'll explain later.
So our "F" for the equation is 115 lb*ft*1.937*2.35. For simplicity's sake I'm only using second gear.
F=523.5 lb*ft
m=21.7 slugs
Now for the fun part...
F=ma
523.5 lb*ft = 21.7 slugs(a)
523.5 lb*ft/21.7 slugs = a
24.1 = a
Again for simplicity's sake I've ignored unit conversion. Again I ask your understanding as you read that I'm just going to add a feet/second squared to our answer.
a= 24.1 ft/s*s.
How does our result stack up against real world numbers?
From wikipeda, Cycle World is quoted as having run a 9.75 @142 m.p.h. quarter mile. We don't know all the details behind that run (did their bike/rider combo weigh 700 lbs.? How much torque did they lose due to friction deficiencies?) We also know the torque multiplication of the gearbox isn't a constant rate during a quarter mile run. Our formula also doesn't account for parasite drag because we're keeping things simple here.
Taking these things into consideration, let's see what speed our math says the Hayabusa should be able to accelerate to in 9.75 seconds.
The final velocity of an object can be found by taking the initial velocity and simply adding the rate of acceleration multiplied by the time available for acceleration. In formula terms it looks like this:
V(f) = V(i)+at
so
V(f) = 0+24.1 ft/s*s*9.75s
V(f) = 235 ft/s
To convert that into miles per hour, multiply by 0.681818182.
V(f) = 160.22 miles per hour
160 m.p.h. is what our math says the Hayabusa should be able to achieve in 9.75 seconds...assuming zero wheelspin, zero aerodynamic drag, a 700 lb total weight, and that we never had to shift out of second gear. The last one is tricky because obviously that's not what is going to happen in the real world. I selected the second gear torque multiplication ratio in an attempt to provide a reasonable average torque multiplication value for the gearbox. With more time a better number could be calculated.
I think given the limitations we've placed on our calculation for time's sake that 160 m.p.h. in 9.75 seconds is close enough to 147 m.p.h. in the same time period to say our math is correct given the limitations we're working with.
So...now we have all the tools we need to calculate how much more "a" we can expect given whatever "m" reduction we want to use.
And that's really what we wanted to know...isn't it?
WTF did you just say?
Similar threads
