boost as a linear equation
Thread Starter
12 Second Club
Joined: Sep 2005
Posts: 354
Likes: 0
From: Springfield, Missouri
boost as a linear equation alright i think this would be the place to post this.
what i was wondering is is/can boost be linear? like a heads/cam car makes 420rwhp and then under 10 psi of boost makes 520rwhp does that mean that 1 psi of boost = 10 rwhp
(i know these numbers are really far off i just am trying to use it as an example)
on dsmtuners.com (old fourm) they said that after 10psi you gain about 10 horse per pound but that would mean that it is linear after 10 according to these guys. i was just wondering.
what i was wondering is is/can boost be linear? like a heads/cam car makes 420rwhp and then under 10 psi of boost makes 520rwhp does that mean that 1 psi of boost = 10 rwhp
(i know these numbers are really far off i just am trying to use it as an example)
on dsmtuners.com (old fourm) they said that after 10psi you gain about 10 horse per pound but that would mean that it is linear after 10 according to these guys. i was just wondering.
TECH Addict
Joined: Aug 2004
Posts: 2,866
Likes: 4
i think there are far too many veriable with boost to say "1 psi equals xbhp". if you cancle out every veriable then yes it might be true, im not sure thouhg. but in really life there is just too much going on to say that. hot air, cold air, the states of you engine, if you got a good or bad batch of fuel in there. and thats not even looking at setups!!! that add a WHOLE new thing to it! lol
i do understand what you mean though, you want to say right i have boosted my motor my this much how much power?? but even if you can figure it out on the dyno and get it bang on every thime, 2 miles down the road its sll gone **** up! lol
hummmm its got me thinking though!
thanks Chris.
i do understand what you mean though, you want to say right i have boosted my motor my this much how much power?? but even if you can figure it out on the dyno and get it bang on every thime, 2 miles down the road its sll gone **** up! lol
hummmm its got me thinking though!
thanks Chris.
Thread Starter
12 Second Club
Joined: Sep 2005
Posts: 354
Likes: 0
From: Springfield, Missouri
yeah thats what i was thinking. when i had my eclipse everyone on that board agreed that after 10psi it was 1 psi = x amount of hp. but then you look at these motors and they are making more on 4 psi then i did at 22 its crazy i know that i just compared apples to oranges but it just blows my mind. thanks for the answer though.
TECH Addict
Joined: Aug 2004
Posts: 2,866
Likes: 4
Originally Posted by waiting2bl33d
yeah thats what i was thinking. when i had my eclipse everyone on that board agreed that after 10psi it was 1 psi = x amount of hp. but then you look at these motors and they are making more on 4 psi then i did at 22 its crazy i know that i just compared apples to oranges but it just blows my mind. thanks for the answer though.
Chris.
How much hp you make with each pound of boost depends on the compressor characteristics and efficiency. It would not be unusual for a race engine to make 200-400 hp more just by adding a couple pounds of boost. All depends on the "sweet spot" of the compressor. To put it another way, if you're already past the "sweet spot" at 10 psi and you're making 500hp, then adding 6# more boost may only make 10hp more.
"boost" is a HORRIBLE way to define the amout of air entering the motor...
...but thats off topic.
the reason they say something like that on that board is because all of their cars are very similar... when you all have the same cars with similar mods, of course they behave the same.. LS1tech does the same thing... for example we go "oh, you have XYZ hp, you should be able to run a x ET... of course you need more data then just the peak hp number.. but since most of us have similar weights, mods, powerband, ect, we know that its around that ballpark.
thoes guys are doing the same thing... a almost stock DSM with x amount of "boost" is around here....
...but thats off topic.
the reason they say something like that on that board is because all of their cars are very similar... when you all have the same cars with similar mods, of course they behave the same.. LS1tech does the same thing... for example we go "oh, you have XYZ hp, you should be able to run a x ET... of course you need more data then just the peak hp number.. but since most of us have similar weights, mods, powerband, ect, we know that its around that ballpark.
thoes guys are doing the same thing... a almost stock DSM with x amount of "boost" is around here....
Staging Lane
Joined: Jan 2006
Posts: 67
Likes: 0
I mean theoretically 14.7psi would double your hp output, if you could keep your intake temp the exact same (100% intercooler efficiency).
At real low 'boost' settings like 4-5psi, you can get real high efficiencies, and at 20-25psi, it's harder to get high efficiencies just because of the compressor efficiency, intercooler efficiency, etc.
So say you have 350hp N/A.
At 4psi you have 85% efficiency, and your hp output would jump to:
(4/14.7)*.85 = 23.1%
350hp*1.244 = 430hp (+80hp)
Now at say 8psi you have 70% efficiency:
(8/14.7)*.75 = 40.8%
350hp*1.408 = 492hp (+142hp)
Now at say 14.7psi you have 55% efficiency:
(14.7/14.7)*.55 = 55.0%
350hp*1.550 = 542hp (+192hp)
The equation is:
(Boost_psi / 14.7)*(1+ (compressor_efficiency*intercooler_efficiency))*NA _hp = final_hp
At real low 'boost' settings like 4-5psi, you can get real high efficiencies, and at 20-25psi, it's harder to get high efficiencies just because of the compressor efficiency, intercooler efficiency, etc.
So say you have 350hp N/A.
At 4psi you have 85% efficiency, and your hp output would jump to:
(4/14.7)*.85 = 23.1%
350hp*1.244 = 430hp (+80hp)
Now at say 8psi you have 70% efficiency:
(8/14.7)*.75 = 40.8%
350hp*1.408 = 492hp (+142hp)
Now at say 14.7psi you have 55% efficiency:
(14.7/14.7)*.55 = 55.0%
350hp*1.550 = 542hp (+192hp)
The equation is:
(Boost_psi / 14.7)*(1+ (compressor_efficiency*intercooler_efficiency))*NA _hp = final_hp
Trending Topics
6600 rpm clutch dump of death Administrator
Joined: Dec 2001
Posts: 4,983
Likes: 13
From: Texas
https://ls1tech.com/forums/forced-induction/511816-holdener-f-i-power-calculation-formula.html
Holdener F/I power calculation formula
--------------------------------------------------------------------------------
Holdener is an contributing author in the Ford Magazines. I saw this formula, and figured some folks might find it useful. You can take the power a motor made in N/A form, and then using this formula come up with a pretty good idea of what it should make in a boosted application. This is good for the folks doing bolt on superchargers.
F/I power = original power * pressure ratio
original power = N/A HP
pressure ratio = (boost pressure/14.7) + 1
Now, from this number, you have to factor in and subtract the hp loss from supercharger. That will vary, and that is a key factor on making this all jive. As an example.
An engine making 365 run at 8.7 psi should make 581 HP
365* ((8.7/14.7) + 1)
365 * 1.59 = 581
When checked on the dyno it actually made 533 (subtract blower loss, in this case 48HP). An average blower can loose you anywhere from 50-75 HP on the average street car. The bigger the blower, etc... the more ploss you may see. This isn't an absolute, but it gives you an idea of where you'll end up at.
F/I power = (original power * pressure ratio) - blower loss
(365 * 1.59) - 48 = 533
Holdener F/I power calculation formula
--------------------------------------------------------------------------------
Holdener is an contributing author in the Ford Magazines. I saw this formula, and figured some folks might find it useful. You can take the power a motor made in N/A form, and then using this formula come up with a pretty good idea of what it should make in a boosted application. This is good for the folks doing bolt on superchargers.
F/I power = original power * pressure ratio
original power = N/A HP
pressure ratio = (boost pressure/14.7) + 1
Now, from this number, you have to factor in and subtract the hp loss from supercharger. That will vary, and that is a key factor on making this all jive. As an example.
An engine making 365 run at 8.7 psi should make 581 HP
365* ((8.7/14.7) + 1)
365 * 1.59 = 581
When checked on the dyno it actually made 533 (subtract blower loss, in this case 48HP). An average blower can loose you anywhere from 50-75 HP on the average street car. The bigger the blower, etc... the more ploss you may see. This isn't an absolute, but it gives you an idea of where you'll end up at.
F/I power = (original power * pressure ratio) - blower loss
(365 * 1.59) - 48 = 533
LS1 Tech Stories
The Best V8 Stories One Small Block at Time
6 Common C5 Corvette Failures and What's Involved In Repairing Them
Pouria Savadkouei
Retro Modern Bandit Pontiac Trans AM Comes With Burt Reynolds' Autograph
Verdad Gallardo
Top 10 Greatest Cadillac V Series Performance Models Ever, Ranked
Pouria Savadkouei
Top 10 Most Powerful Chevy Trucks Ever Made!
Hennessey's New Supercharged Silverado ZR2 Has 700 HP
Verdad Gallardo
Coachbuilt N2A Anteros Is an LS2-Powered C6 Corvette In Italian Clothes
Verdad Gallardo
Awesome K5 Blazer Restomod Comes With C7 Corvette Power
Verdad Gallardo
10 Camaros You Should Never Buy
10 LS Engine Myths That Refuse to Die
Verdad Gallardo
TECH Apprentice
Joined: Oct 2004
Posts: 329
Likes: 0
Theoretically 14.7psi would double your output, but in actuality someting like 17psi do.
So if you had a 34PSI combo on a 300rwhp short block, your talking about 900 rhwp.
If you had 17PSI ontop of somthing that can do 500rwhp by itself you talking 1000rwhp.
So if you had a 34PSI combo on a 300rwhp short block, your talking about 900 rhwp.
If you had 17PSI ontop of somthing that can do 500rwhp by itself you talking 1000rwhp.
Thread Starter
12 Second Club
Joined: Sep 2005
Posts: 354
Likes: 0
From: Springfield, Missouri
alirhgt so. my next question is. (Boost_psi / 14.7)*(1+ (compressor_efficiency*intercooler_efficiency))*NA _hp = final_hp...why 14.7 isn't that 1 BAR? i mean i understand that its an equation and thats how it works i just wanted to know what the 14.7 was i understand everything else and if you could keep everything the same (com. eff, inter. eff) then i guess boost could (on paper) be a linear which is what i was trying to see.
TECH Fanatic
Joined: Nov 2001
Posts: 1,695
Likes: 9
From: Raleigh, NC
Originally Posted by waiting2bl33d
alirhgt so. my next question is. (Boost_psi / 14.7)*(1+ (compressor_efficiency*intercooler_efficiency))*NA _hp = final_hp...why 14.7 isn't that 1 BAR? i mean i understand that its an equation and thats how it works i just wanted to know what the 14.7 was i understand everything else and if you could keep everything the same (com. eff, inter. eff) then i guess boost could (on paper) be a linear which is what i was trying to see.
Teching In
Joined: Jan 2006
Posts: 18
Likes: 0
From: ShenZhen, China
Waiting - I sympathize with your case.
Speaking for myself, I can say for sure that the whole linear equation thing only works when it is _not_ in my favor or is taking the trend line the opposite direction from the intended goals. PITA.
In my line of work, I deal with a lot of problems involving _something_ squared and it is normally a power or thermal issue where it is all working against us and the end consumer. Doing all the calculations and working through a project dead on the money, prototypes, adjustments, windage and a grain of salt thrown in…. In the end, the losses and real world issues such as assembly tolerance and material variance and the voodoo that is Mother Nature always necessitate the need for a kludge factor.
In a motor or FI system or whatever mechanical, the user is going to always deal with a lot of real world issues that really make life tough. Friction, heat of compression, nonlinear flow, the monkey in the wrench, all have a hand in making the real world version of the linear equation here a tough nut to crack.
My opinion is that most of the guys here and in the know really have done an outstanding job of beating the curve in a lot of ways. What I mean is that I look at a lot of the motors (with hard data and dyno data and real world numbers to back it all up) are doing about all that the theory and the numbers added up will do. That is pretty impressive to me.
I was really hoping that your post would get one of the really smart guys to give up the goods on how they do the big numbers and have the big results to match.
Good post- thanks.
Speaking for myself, I can say for sure that the whole linear equation thing only works when it is _not_ in my favor or is taking the trend line the opposite direction from the intended goals. PITA.
In my line of work, I deal with a lot of problems involving _something_ squared and it is normally a power or thermal issue where it is all working against us and the end consumer. Doing all the calculations and working through a project dead on the money, prototypes, adjustments, windage and a grain of salt thrown in…. In the end, the losses and real world issues such as assembly tolerance and material variance and the voodoo that is Mother Nature always necessitate the need for a kludge factor.
In a motor or FI system or whatever mechanical, the user is going to always deal with a lot of real world issues that really make life tough. Friction, heat of compression, nonlinear flow, the monkey in the wrench, all have a hand in making the real world version of the linear equation here a tough nut to crack.
My opinion is that most of the guys here and in the know really have done an outstanding job of beating the curve in a lot of ways. What I mean is that I look at a lot of the motors (with hard data and dyno data and real world numbers to back it all up) are doing about all that the theory and the numbers added up will do. That is pretty impressive to me.
I was really hoping that your post would get one of the really smart guys to give up the goods on how they do the big numbers and have the big results to match.
Good post- thanks.