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Author Topic: Hp needed for % of Speed Increase  (Read 17277 times)
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sheribuchta
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« Reply #15 on: April 07, 2010, 11:53:34 PM »

check your calculator --or your typing                                       willie buchta                                                   
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jl222
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« Reply #16 on: April 07, 2010, 11:55:05 PM »

 Salty Dog... I mean Stan evil

  From Carroll Smith's book [Tune to Win]  Drag HP=  CD X Frontal area X ( Velocity in mph) to the 3rd power divided by
                                                                                                 146,600

  His example  Drag HP = (.65) X ( 17 FT) x (180 mph) to the 3rd power OR 180 X180X 180
                                                  --------------------------------
                                                             146,600                                                     =  439 HP


                           JL222 cheers
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jl222
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« Reply #17 on: April 08, 2010, 12:08:08 AM »

 With out putting a lot of time in, Bonneville Pro came up with 166 mph but thats at 4500 ft

                    JL222

  Smith was using a formula 5000, car dropping the weight from 2350 to 1200lbs the speed was 170
« Last Edit: April 08, 2010, 10:14:32 AM by jl222 » Logged
Stan Back
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« Reply #18 on: April 08, 2010, 10:33:07 AM »

We're not talking about any specific car, here.  Nor any Cd.  There's a formula that sez that if you get the car (bike, whatever) to go, say 180 with, say 200 HP, how fast you'll go with 250 HP.  I think that Avanti comes closest, but I think it has to do with the square of the percentage of increase, not the cube.  I don't need anyone to figure it out for me -- I'm good with formulas -- I just don't have that one.

Stan
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« Reply #19 on: April 08, 2010, 11:39:48 AM »

  
 There would have to be a specific car [ cd and frontal area] it takes less HP to make a small slippery car go faster than a big dragey one.


              JL222
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Stan Back
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« Reply #20 on: April 08, 2010, 01:46:22 PM »

No, no, no.

There's a formula that sez that if you get the car (bike, whatever) maxed out to go, say 180 with, say 200 HP, how fast you'll go with 250 HP.

In other words -- but not so different -- "I went 180 MPH best with 150 HP after years of trying, doing my best.  If I had 200 HP, how fast would I go?"

Stan
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bbarn
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« Reply #21 on: April 08, 2010, 02:03:18 PM »

http://craig.backfire.ca/pages/autos/drag

Stan,

Look down the page to the "Acceleration and Top Speed" section, there is a formula below that. I have an Excel spreadsheet someplace that has a simpler formula already in it but I can't find it.
 
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jacksoni
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« Reply #22 on: April 08, 2010, 04:07:01 PM »

DRAG goes up as Square of velocity. HORSEPOWER goes up as CUBE.
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Jack Iliff
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4-barrel Mike
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« Reply #23 on: April 08, 2010, 04:48:07 PM »

Or a Sum summerized in the explanation to his spreadsheet, to go twice as fast, you need eight times the horsepower.

Mike
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« Reply #24 on: April 08, 2010, 05:59:07 PM »

There you go!

Now if I could only read the spreadsheets.

I'm about ready to give up on this.

But I think that the percentage of increased HP cubed is equal to the percentage of increase in velocity squared -- right?

Stan
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« Reply #25 on: April 08, 2010, 06:01:00 PM »

Really -- I'm thankful of all the offers of where to find the formula and how to plug it in here and there -- but I don't have a passport (or machine) to get there.  I just wanted the formula.

Stan
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4-barrel Mike
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« Reply #26 on: April 08, 2010, 06:26:43 PM »

What's the formula(s) for when you can go 160 on 500 HP, how fast you can go with 600 and/or if you need to go 180, how much increased HP you need.

You know -- the square of the increase(?) is half the angle of the dangle or whatever?

Help!!!

Simple Stan

180 mph / 160 mph = 1.125 times faster
1.125 cubed (1.125 x 1.125 x 1.125) = 1.423828125
500hp for 160 mph * 1.423828125 = 711.9140625 hp for 180 mph

Mike

Assuming that I entered and copied everything correctly
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« Reply #27 on: April 08, 2010, 06:32:21 PM »

F=MA, E=MC2

In all seriousness, the Sten Block machine has a realistic shot at the Dirty Two Club, if the Inkrez predictions are correct!

Phone me in Atlanta if ya do!
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« Reply #28 on: April 08, 2010, 08:03:23 PM »

Now that I've got this cubic foot of enriched unobtainium, how do you lean it out? 
Edsel = Motor Cars for 2?

Stan
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« Reply #29 on: April 09, 2010, 01:37:38 AM »

It is proportional to the cube of the speed.  The force due to drag is proportional to the square.  To get power you have to multiply by the rate.  Therefore the power is proportional to the cube of the speed.

If it takes 100 hp to go 100 mph then to go 200 it is the ratio of (200/100) cubed times 100 hp or 800 hp.

The calculator works fine.
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