Drivetrain When does 15 + 2 = 17.3
When does 15 + 2 = 17.3
If the stock SC is turning 10,000 rpm and 15% smaller SC pulley is added,
the SC will turn 11,500 rpm (10000 x 1.15).
If a 2% larger crank pulley is added, the SC will turn 11,730 rpm
(11500 x 1.02).
15 + 2 = 17.3
15 + 3 = 18.45
15 + 4 = 19.6
the SC will turn 11,500 rpm (10000 x 1.15).
If a 2% larger crank pulley is added, the SC will turn 11,730 rpm
(11500 x 1.02).
15 + 2 = 17.3
15 + 3 = 18.45
15 + 4 = 19.6
Your one percent off... The oversize of the crank pulley does not directly effect the overdrive of the supercharger. Our cars have a single belt driving all of the pulleys, with each pulley having a different size and weight. Knowing this, the ratios of each pulley's effect on each other are different. Meaning a 15% overdrive pulley matched with a 2% bigger crank pulley create about a 16.3% overdrive on the supercharger. Hope this helps
!
-Brian
-Brian
Because you are adding percentages (or ratios) which are derived from different denominators.
The stock crank size is 5.4" and the stock SC pulley is 2.58"
102% of 5.4 and 85% of 2.58 ... do the math...
MINI Cooper SC Crank/Pulley Options
__________ Crank________ Supercharger__ Crank 2 SC______ Redline
__________ Pulley________Pulley________ Ratio____________RPM
Stock______5.4__________ 2.58__________ 2.09____________ 6250
__________ Over________ Reduction____ New____________Eff. %
__________ Crank________Pulley________Ratio__________Reductio n______ SC RPM @ redline
Stock______0%__________ 0%____________2.09____________0%____________ 13081
JCW_______0%__________ 14.7%_________ 2.45__________ 14.7%__________ 15336
__________ 0%______ ____15%__________ 2.46__________ 15.0%__________ 15390
__________ 2%__________ 15%__________ 2.51__________ 16.7%__________ 15698
__________ 4%__________ 15%__________ 2.56__________ 18.3%__________ 16005
__________ 0%__________ 17%__________ 2.52__________ 17.0%__________ 15761
__________ 2%__________ 17%__________ 2.57__________ 18.6%__________ 16076
__________ 4%__________ 17%__________ 2.62__________ 20.2%__________ 16391
__________ 0%__________ 19%__________ 2.58__________ 19.0%__________ 16150
__________ 2%__________ 19%__________ 2.64__________ 20.6%__________ 16473
__________ 4%__________ 19%__________ 2.69__________ 22.1%__________ 16796
The stock crank size is 5.4" and the stock SC pulley is 2.58"
102% of 5.4 and 85% of 2.58 ... do the math...
MINI Cooper SC Crank/Pulley Options
__________ Crank________ Supercharger__ Crank 2 SC______ Redline
__________ Pulley________Pulley________ Ratio____________RPM
Stock______5.4__________ 2.58__________ 2.09____________ 6250
__________ Over________ Reduction____ New____________Eff. %
__________ Crank________Pulley________Ratio__________Reductio n______ SC RPM @ redline
Stock______0%__________ 0%____________2.09____________0%____________ 13081
JCW_______0%__________ 14.7%_________ 2.45__________ 14.7%__________ 15336
__________ 0%______ ____15%__________ 2.46__________ 15.0%__________ 15390
__________ 2%__________ 15%__________ 2.51__________ 16.7%__________ 15698
__________ 4%__________ 15%__________ 2.56__________ 18.3%__________ 16005
__________ 0%__________ 17%__________ 2.52__________ 17.0%__________ 15761
__________ 2%__________ 17%__________ 2.57__________ 18.6%__________ 16076
__________ 4%__________ 17%__________ 2.62__________ 20.2%__________ 16391
__________ 0%__________ 19%__________ 2.58__________ 19.0%__________ 16150
__________ 2%__________ 19%__________ 2.64__________ 20.6%__________ 16473
__________ 4%__________ 19%__________ 2.69__________ 22.1%__________ 16796
Originally Posted by Brian@WebbMotorsports
Your one percent off... The oversize of the crank pulley does not directly effect the overdrive of the supercharger. Our cars have a single belt driving all of the pulleys, with each pulley having a different size and weight. Knowing this, the ratios of each pulley's effect on each other are different. Meaning a 15% overdrive pulley matched with a 2% bigger crank pulley create about a 16.3% overdrive on the supercharger. Hope this helps
!
-Brian
-Brian
. Ratios = say the stock S/C pulley was 2.58"diam for the sc pulley and 5.4" diam for the crank that would equat to a ratio of 2.0930. If you reduce the size of the S/C pulley you get 2.193" and if you increase the crank by 2% you get 5.508" Diam. Thus giving you a ratio of 2.5116.
(2.5116-2.0930)/2.5116 = 16.67% difference.
Confused how you got 16.3%

I however doubt these are the actual sizes and are probably something easier to manufcture. So it's probably pretty close to a 17% actual.
P.S. Come on Bust out the calipers lets see what it is truely
Shouldn't be (2.5116-2.0930)/2.0930 = 20% difference.
The SC is turning 120% faster then stock.
The SC is turning 120% faster then stock.
Originally Posted by FlynHawaiian
Wouldn't it be higher than this. Knowing that a stock crank pulley rotates at 5.5-lbs versus the lightweight one which rotates at 0.9 lbs. Doesn't that mean there is less work required to turn the pulley? Thus creating more hp? I'm interested how you get your 16.3 calcs though
.
Ratios = say the stock S/C pulley was 2.58"diam for the sc pulley and 5.4" diam for the crank that would equat to a ratio of 2.0930. If you reduce the size of the S/C pulley you get 2.193" and if you increase the crank by 2% you get 5.508" Diam. Thus giving you a ratio of 2.5116.
(2.5116-2.0930)/2.5116 = 16.67% difference.
Confused how you got 16.3%
I however doubt these are the actual sizes and are probably something easier to manufcture. So it's probably pretty close to a 17% actual.
P.S. Come on Bust out the calipers lets see what it is truely
. Ratios = say the stock S/C pulley was 2.58"diam for the sc pulley and 5.4" diam for the crank that would equat to a ratio of 2.0930. If you reduce the size of the S/C pulley you get 2.193" and if you increase the crank by 2% you get 5.508" Diam. Thus giving you a ratio of 2.5116.
(2.5116-2.0930)/2.5116 = 16.67% difference.
Confused how you got 16.3%

I however doubt these are the actual sizes and are probably something easier to manufcture. So it's probably pretty close to a 17% actual.
P.S. Come on Bust out the calipers lets see what it is truely

Originally Posted by FlynHawaiian
Doesn't that mean there is less work required to turn the pulley? Thus creating more hp?
As to creating more HP? That is a fallacy, It should be worded as "freeing" up horsepower your engine already creates because its doing less work. By wording it as creating more HP, people are going to think they get X more HP by adding a $140 part and thats not true. The HP is already there.
Originally Posted by ted leist
Shouldn't be (2.5116-2.0930)/2.0930 = 20% difference.
The SC is turning 120% faster then stock.
The SC is turning 120% faster then stock.
It's an old math issue...
Like 120 is 20% more than 100, but 100 is 16.666% less than 120
Percentage change is
(new - old ) / old
if you take (new - old) / new then ratio percentage would match pulley
the pulley difference
2.58" --> 2.193" 15% difference
(2.193 - 2.58) / 2.58 = -15% (15% reduction in size)
ratio difference... the ratio is acutal increasing
2.09 --> 2.46 17.3% difference
(2.46 - 2.09) / 2.09 = 17.3% ( increase in ratio )
(2.46 - 2.09) / 2.46 = 15% (but this isn't accurate)
Trending Topics
your correct in that. Hmm, I always get those backwards O_o. Thats the last time I ask a person for a fomula :P Funny cause I had it at 20% in the begining.
Okay the only value I have from that website is that the 15% pulley is 2.19" what is the 2% over one? Lets play with formulas
.
And your right, it's "freeing up" the hoursepower that is already there if your using a stock size one. If your using the 2% your freeing up hp as well as increasing it slightly because your giving more boost.
Same can be said about exhaust mods too, your not increasing the horsepower your only freeing it up so the engine can be more efficient. I think the efficency of a gas engine is what less than 50%? I mean Ideally your putting in way more energy (ie fuel) to make the engine work when it's not super efficent. Kinda makes you wounder if the engine was 80% efficent how much power would it have?
Okay the only value I have from that website is that the 15% pulley is 2.19" what is the 2% over one? Lets play with formulas
.And your right, it's "freeing up" the hoursepower that is already there if your using a stock size one. If your using the 2% your freeing up hp as well as increasing it slightly because your giving more boost.
Same can be said about exhaust mods too, your not increasing the horsepower your only freeing it up so the engine can be more efficient. I think the efficency of a gas engine is what less than 50%? I mean Ideally your putting in way more energy (ie fuel) to make the engine work when it's not super efficent. Kinda makes you wounder if the engine was 80% efficent how much power would it have?
Math is very cool and fun. Whats even funnier is that this is low level math. And I always have problems with, give me calc or diff eqs I can get it
. Regradless it's still better to go with the 15% and the 2% it so seems. Not only are you "feeing" up more power, by creating less work. Your inducing more boost into the system. Ever so slightly. How much are we talking about with the 15 and the 2 %? like an extra 5-psi boost or less? 
So whats the correlation between boost and hp. how many psi of boost = xxxhp? Is this an exponetial function or purly linear?
. Regradless it's still better to go with the 15% and the 2% it so seems. Not only are you "feeing" up more power, by creating less work. Your inducing more boost into the system. Ever so slightly. How much are we talking about with the 15 and the 2 %? like an extra 5-psi boost or less? 
So whats the correlation between boost and hp. how many psi of boost = xxxhp? Is this an exponetial function or purly linear?
Originally Posted by FlynHawaiian
And your right, it's "freeing up" the hoursepower that is already there if your using a stock size one. If your using the 2% your freeing up hp as well as increasing it slightly because your giving more boost....
Same can be said about exhaust mods too, your not increasing the horsepower your only freeing it up so the engine can be more efficient.
Both Catbacks and headers also are freeing up HP and NOT making HP. True.
For the Bugatti Environ, it's about a third....
Originally Posted by FlynHawaiian
I think the efficency of a gas engine is what less than 50%?
matt
The perfect internal combustion engine is only 50% efficient. That includes no friction or heat loss etc. Most engines are in the 30% range. There are some (like the Miller cycle) that are in the 40% range but there's no current production car I know of that uses the Miller cycle. The design of the ICE from the start is basically flawed.
Originally Posted by Dr Obnxs
The motor releases about 3000 HP of chemical energy, 1000 is used for the cooling system, 1000 goes out the exhaust pipe as wasted heat, and about 1000 is available for rocket like acceleration.
matt
matt
. YES BLISS. MAKE IT HAPPEN
Sorry about not posting sooner, but I have been side tracked with reading the Webb exhaust and swaybar threads. I got the estimate of 16.3% from Randy, which is why I stated "about". I will talk to him about your calculations, and see what he thinks.
-Brian
-Brian
Don't forget the importance of belt thickness:
https://www.northamericanmotoring.co...5&postcount=17
Stock:
(5.460+0.126)/(2.58+0.126)=
5.586/2.706 = 2.06
15+2:
(5.569+0.126)/(2.193+0.126)=
5.695/2.319 = 2.46
So, for each revolution of the crank, the stock setup would turn 2.06 times and the 15+2 setup would turn 2.46 times, or 19.4% faster.
https://www.northamericanmotoring.co...5&postcount=17
Stock:
(5.460+0.126)/(2.58+0.126)=
5.586/2.706 = 2.06
15+2:
(5.569+0.126)/(2.193+0.126)=
5.695/2.319 = 2.46
So, for each revolution of the crank, the stock setup would turn 2.06 times and the 15+2 setup would turn 2.46 times, or 19.4% faster.
Stepping into the abyss...
Pully diameters don't matter if you have accurate percentages. It's all about linear belt velocities.
So, if you have a bigger crank pully and a smaller SC, the relative spin rates at a given RPM =
RPM (SC new) = RPM (SC Old)*(1+ Crank Pully Oversize in %)/(1-SC Underdrive pully in %).
So for a 2% crank and a 15% SC, you get 1.02/.85 or 1.20
Flame away!!!
Matt
So, if you have a bigger crank pully and a smaller SC, the relative spin rates at a given RPM =
RPM (SC new) = RPM (SC Old)*(1+ Crank Pully Oversize in %)/(1-SC Underdrive pully in %).
So for a 2% crank and a 15% SC, you get 1.02/.85 or 1.20
Flame away!!!
Matt
Originally Posted by FlynHawaiian
And your right, it's "freeing up" the hoursepower that is already there if your using a stock size one. If your using the 2% your freeing up hp as well as increasing it slightly because your giving more boost.
Originally Posted by Dr Obnxs
Pully diameters don't matter if you have accurate percentages. It's all about linear belt velocities.
So, if you have a bigger crank pully and a smaller SC, the relative spin rates at a given RPM =
RPM (SC new) = RPM (SC Old)*(1+ Crank Pully Oversize in %)/(1-SC Underdrive pully in %).
So for a 2% crank and a 15% SC, you get 1.02/.85 or 1.20
Flame away!!!
Matt
So, if you have a bigger crank pully and a smaller SC, the relative spin rates at a given RPM =
RPM (SC new) = RPM (SC Old)*(1+ Crank Pully Oversize in %)/(1-SC Underdrive pully in %).
So for a 2% crank and a 15% SC, you get 1.02/.85 or 1.20
Flame away!!!
Matt
At 10000 SC RPM, with a 15% reduction SC pulley and a stock crank pulley, would it work like this:
New SC RPM = 10000 * ((1 + 0.0) / (1 - 0.15)
= 10000 * 1 / 0.85
= 10000 * 1.1765
= 11765
I may be applying the equation incorrectly, but that doesn't seem just right to me.= 10000 * 1 / 0.85
= 10000 * 1.1765
= 11765
Or maybe it's a misunderstanding about the numbers. Does a 15% pulley produce ~17.6% more SC speed?
It's the ratio game......
Originally Posted by eMINI
No flames, just trying to understatnd the math:
At 10000 SC RPM, with a 15% reduction SC pulley and a stock crank pulley, would it work like this:
Or maybe it's a misunderstanding about the numbers. Does a 15% pulley produce ~17.6% more SC speed?
At 10000 SC RPM, with a 15% reduction SC pulley and a stock crank pulley, would it work like this:
New SC RPM = 10000 * ((1 + 0.0) / (1 - 0.15)
= 10000 * 1 / 0.85
= 10000 * 1.1765
= 11765
I may be applying the equation incorrectly, but that doesn't seem just right to me.= 10000 * 1 / 0.85
= 10000 * 1.1765
= 11765
Or maybe it's a misunderstanding about the numbers. Does a 15% pulley produce ~17.6% more SC speed?


