How much more power would i need?

AstroGuardian

Senior member
May 8, 2006
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Hi everyone,

I know that there are many really smart ppl here so i will ask.

Is there any way i could know how much more engine power i would need so i can increase speed from 200 to 250 km/h? Is there a formula i can use to calculate this?

Thx all
 

Demo24

Diamond Member
Aug 5, 2004
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Thats not really possible to tell you with what you gave. Have to factor in wind resistance of the vehicle and gearing of the transmission, and a few other things as well.
 

AstroGuardian

Senior member
May 8, 2006
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I am talking about taking as much factors as possible. But i think wind resistance would be the most important factor.

Also i am talking about 200-250 km/h in final gear.
 

punjabiplaya

Diamond Member
Nov 12, 2006
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if you give us some information about the car and what surroundings you're at, we can at least give an educated guess
 

punjabiplaya

Diamond Member
Nov 12, 2006
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New horsepower = old horsepower * (new top speed / old top speed)^cubed

edit: I have no idea how accurate this is
 
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bommy261

Golden Member
Dec 17, 2005
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too many variables. although the difference between 120 mph and 155 mph is probably 50-100 hp, neglecting gearing, wind resistance, and assuming a weight around 3k lbs. just a guess. the difference between 155-180 is much more severe. 180-200 even more so.. the faster you go, the more wind resistance, i am not familiar with that fluid equation for calculating resistance, but I assume its a linear curve, maybe a slight exponential.
 

exdeath

Lifer
Jan 29, 2004
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I'll add that it typically takes double the power for each 1 second faster in the quarter mile, all else held constant.

eg: 500 HP gets you ~10s-11s, 1000 HP gets you 8s-9s, 2000 HP gets you 6-7s, all the way down to 8000 HP to run 4s-5s.

It climbs quickly due to fluid drag (eg: air) being a cube of velocity.
 
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punjabiplaya

Diamond Member
Nov 12, 2006
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too many variables. although the difference between 120 mph and 155 mph is probably 50-100 hp, neglecting gearing, wind resistance, and assuming a weight around 3k lbs. just a guess. the difference between 155-180 is much more severe. 180-200 even more so.. the faster you go, the more wind resistance, i am not familiar with that fluid equation for calculating resistance, but I assume its a linear curve, maybe a slight exponential.

fairly linear, until you get past ~90mph
 
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JCH13

Diamond Member
Sep 14, 2010
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New horsepower = old horsepower * (new top speed / old top speed)^cubed

edit: I have no idea how accurate this is

If all you account for is aero, this is basically correct. FWIW aero is probably the most significant factor, though there are many things that will change as you go faster.

You should also make sure that in your highest gear you can actually spin your engine fast enough to get to 250kph, i.e. that you're not gear-limited to below 250kph. For example, to exactly make 250kph at a red-line of say 6,000 rpm, your engine would have to be at 4800rpm at 200kph.

This is an interesting problem because your car's power output better increase with RPM at an appropriate rate. Say you needed 200hp to hit 250kph, your engine will need to output 200hp at our made-up 6,000 rpm, not at some lower RPM.
 

JCH13

Diamond Member
Sep 14, 2010
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too many variables. although the difference between 120 mph and 155 mph is probably 50-100 hp, neglecting gearing, wind resistance, and assuming a weight around 3k lbs. just a guess. the difference between 155-180 is much more severe. 180-200 even more so.. the faster you go, the more wind resistance, i am not familiar with that fluid equation for calculating resistance, but I assume its a linear curve, maybe a slight exponential.

D:

If you neglect gearing and wind resistance then you can get to any speed with any power level.

Weight doesn't make a difference in top speed, just how fast you get there.

Power loss through friction is not linear, if it was linear the power required to get from 150-175kph would be the same increase in power as 175-200kph. It is also not exponential. It is cubic for high Reynolds number situations like a car driving fast in air.
 
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bommy261

Golden Member
Dec 17, 2005
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D:

If you neglect gearing and wind resistance then you can get to any speed with any power level.

Weight doesn't make a difference in top speed, just how fast you get there.

Power loss through friction is not linear, if it was linear the power required to get from 150-175kph would be the same increase in power as 175-200kph. It is also not exponential. It is cubic for high Reynolds number situations like a car driving fast in air.

what i should have said, was assuming your car could reach that speed at a reasonable RPM based off its current gearing. But a cubic is an exponential last time I checked. You are right about that acceleration as well.
 

JCH13

Diamond Member
Sep 14, 2010
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what i should have said, was assuming your car could reach that speed at a reasonable RPM based off its current gearing. But a cubic is an exponential last time I checked. You are right about that acceleration as well.

Maybe you're checking the book of stupid.

The 3 in the function f(x)=x^3 is the exponent of a cubic function.

The function f(x)=4^x is exponential, where x is the exponent.

See this.

a function in which an independent variable appears as an exponent
 

bommy261

Golden Member
Dec 17, 2005
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Either way the growth is still exponential. I could care less which is the exponent and which is the exponential.
 

JCH13

Diamond Member
Sep 14, 2010
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Either way the growth is still exponential. I could care less which is the exponent and which is the exponential.

D:

You just don't get it...

Wiki: Exponential growth.

The very first image is a graph comparing, guess what, linear growth, cubic growth, and exponential growth. They are very different functions with very different implications.
 

punjabiplaya

Diamond Member
Nov 12, 2006
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also, cubic != exponential

math ftmfw

edit: edit: I'm stupid, not doing series/sequences
 
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bommy261

Golden Member
Dec 17, 2005
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You want a cookie? From now on for all matters dealing with growth i shall defer to you. In all my years of calculus ive always referred to any growth outside of linear to be expoential. Thanks for the review in 7th grade math tho.
 

JCH13

Diamond Member
Sep 14, 2010
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You want a cookie? From now on for all matters dealing with growth i shall defer to you. In all my years of calculus ive always referred to any growth outside of linear to be expoential. Thanks for the review in 7th grade math tho.

A cookie would be great, thanks. I just finished lunch and a dessert would be nice :thumbsup:

You're welcome on the review of "7th grade math." It was clear you needed a refresher since you've been getting it wrong for all your "years of calculus."