Landracing Forum
Introductions => Formulas => Topic started by: Stan Back on April 06, 2010, 06:56:28 PM
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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
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I found this one
http://robrobinette.com/top_speed.htm
170.025mph on 600hp need 715 hp to get 180mph
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Here's Sumner's calculation Excel sheets, including the one you're looking for.
http://purplesagetradingpost.com/sumner/bvillecar/bville-spreadsheet-index.html#Horse%20Power%20Needed (http://purplesagetradingpost.com/sumner/bvillecar/bville-spreadsheet-index.html#Horse%20Power%20Needed)
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The Robinette one gave some unbelievable figures -- and as usual I couldn't access Sumner's spreadsheets. I guess I'll have to depend on the Shearers for the best answer.
Stan
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Stan,
Just mash the gas pedal and hang on.....I am sure you have all the power you need.
Charles
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I think it goes... all the power you can afford dose not go as fast as you want.
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Stan,
Send me the numbers and I will crunch them for you.
Tom G.
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I think it goes... all the power you can afford dose not go as fast as you want.
Here's the formula - it's corollary is based on boat ownership -
($ x APR) + 5W = S x .99, where
$=$
APR = variable
5W = lodging
S = speed
.99 = "I'll be back next year" coefficient
:cry:
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Payback is hill.
Stan
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Did you want that formula to work forwards or backwards? :evil:
Geo
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Yes
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Stan,
Send me the numbers and I will crunch them for you.
Tom G.
Stan,
Are you sure those are the right numbers you sent me? I come up with a minus 7. Sorry.
Tom G.
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If you are only going 160 on 500 hp you need a different car- 'less you are driving a freightliner or something! :cheers:
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If you are only going 160 on 500 hp you need a different car- 'less you are driving a freightliner or something! :cheers:
it's called a rolling brick :-D
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The power is proportional to the speed. To find the power take the ratio of the two speeds and cube that number. 180/160 = 1.125, 1.125 cubed =1.438 so 500 hp x 1.438 = 712 hp, rounded up
So how big is this car? It has one heck of a CdA!
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check your calculator --or your typing willie buchta
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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
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146,600 = 439 HP
JL222 :cheers:
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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
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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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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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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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http://craig.backfire.ca/pages/autos/drag (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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DRAG goes up as Square of velocity. HORSEPOWER goes up as CUBE.
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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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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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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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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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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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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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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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Not to high-jack the thread, but this is a great topic and I LOVE the formulas. IF they are accurate, I need far less power to meet my goals than I previously thought.
And power is certainly not the problem I have...it's money to make the truck safe to drive that's the problem.
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Is there a formula that involves weight, or does that not matter when you already have hp and top speed? Is there a reasonably accurate formula for motorcycle?
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Dr Mayfield,
Mayf, for all the Listers, has a site with many specifics including the Cd, frontal area etc... of many production cars and trucks,
along with simple explanations...
http://www.mayfco.com/dragcoef.htm
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theoretical vehicles set theoretical records every theoretical time.....
Kent
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Kent, you should add that you're speaking theoretically, I theorize.
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But it's so true! :-D :-D :-D
Pete
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Dr Mayfield,
Mayf, for all the Listers, has a site with many specifics including the Cd, f :evil:rontal area etc... of many production cars and trucks,
along with simple explanations...
http://www.mayfco.com/dragcoef.htm
Mike,
He forgot Roadsters!!!!!!!!!!!! :evil: :-o :-D
Tom G.
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And he runs one ------------ just a little newer than the rules allow!!! :? :? :-D :-D
Pete
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Interestingly enough Tom, I stole an old method from my high school Photography instructor to calculate the frontal area for #300. He was a Air Force Photo Intelligence Interpreter.
Shoot a pix of the of the intended vehicle with a known value object in the frame. From there you can calculate the frontal area of the vehicle. In Vietnam they knew what the bore of rifle the VC was carrying, therefore he could tell you what size shoe he (the VC) was wearing.
I have since lost the excel file I was working with to approximate the HP necessary to put that brick over the record, and we really never knew how much HP we were producing because the LSR tires would never hod traction on a chassis dyno. but alas, we know what happened to that engine...I still have some (very broken) parts of it in the "offering to the gods of speed" section of my LSR chapel.
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DRAG goes up as Square of velocity. HORSEPOWER goes up as CUBE.
Really?
Maybe someone can explain how this works...
If most of the drag on a car at speed (like >100 MPH) is aero, and aero drag goes up as a second power function of the increased velocity, how come HP needed would go up as a third power function?
In other words, if air resistance is the largest single factor in drag, and if it goes up 4X when I double the speed... where's the other "horsepower vampire" hiding here?
Just tryin' to understand...
-Bill
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In other words, if air resistance is the largest single factor in drag, and if it goes up 4X when I double the speed... where's the other "horsepower vampire" hiding here?
Just tryin' to understand...
The speed itself is the 3rd factor.
If you could drive in a vacuum the power required to go faster would go up directly in proportion to the speed. With air resistance you also increase drag at the square of the speed. speed squared x speed = speed cubed
Larry
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In Vietnam they knew what the bore of rifle the VC was carrying, therefore he could tell you what size shoe he (the VC) was wearing.
Plus or minus 100%...
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Bill- what you were missing is that power is force times velocity.
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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
Aerodynamics is not the only thing that plays a role here. There are other frictional things coming into play. But based simply on aerodynamics -- which is probably the major biggie --, the horsepower required increases as the cube of the speed. Of course, available power is not the only important thing. Appropriate gearing is also critical to success.
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I said that 14 years ago? Yikes!!! I musta stole that from someone else.