If known: HP & MPH, how to find ET?

Hopefully one of the maths guys will do it...

(If no takers by tomorrow I'll try do it for you (for a couple of sardines!?) mind ..once I post it, they'll all come on to correct it )

Del

This may be simplifyable but it does what your after

((HP/((MPH/234)^3*HP))^0.33)*5.825 = ET

I think!!!

Al

Al,

I poked that into Excel, ooops, doesn't matter if HP is 1 or 5,000, ET stays the same.

Put it in Excel.

A guide would be, with a 3850 lb weight: 400 HP=111 MPH=12 Seconds, 500 HP=119 MPH=11.5 Seconds, Etc are close numbers so can see if formula works.

Ken

I'm not sure what ET is?

John?

Drag Racing,

ET = Elapsed Time, in this case, in the 1/4 mile, from a dead stop.

Sounds like a drag racing enthusiast.

Solve the first equation for "MPH"

Insert that value in last equation to find "ET"

It is just algebraic manipulation, assuming all the equations are correct and the "dimensions" are all correct you will get the correct numerical solution sought. QED

Yeah, the "last" equation is one I want that works. Already knowing MPH and HP what equation can I poke into Excel to give me the ET of the car in the 1/4 mile?

Hi Ken,

Seems to me like you're ignoring traction. Obviously this will have a huge effect on ET. Are you assuming perfect traction for purposes of your calculatiions?

-John

Hi John,

Here are some I used since about 1994 for that:

{DriveWhlTorq|Ft#tq*FstGear*RearRat*.85=DrWhlTqe}

{Wheel Thrust|DrWhlTqe/(Tire~Dia/24)=WhlTrst}

{g force|WhlTrst/Weight=gForce}

They work as written in an HP 200LX in their Solver.

Ken

Seems like you should be able to do it, but after playing with the dimensional analysis for a while, you need the weight in there to account for the energy component of HP. I flipped through John Lawlor's "Auto Math" handbook and did not find any other equations on the subject.

Yeah, I have several auto math books too... all to no avail so far.

Without knowing anything about this esoteric subject, algebra says:

ET = 1363 / MPH

Exactly. HP/Weight cancels out. That's why Mr Chevy gets same answer for different inputs to Excel.

Assuming constant acceleration, velocity = 2*distance/time. Putting in distance = 1/4 mile and converting, I get MPH = 1800/ET, maybe 1363 is found in practice?

Using 1363 and ET = 7 sec say, MPH = 195 and acceleration = 41 ft/sec², which sound like the right order, I believe these things have "sticky" tyres and can accelerate faster than g.

Cheers....Codey

As you say, the formulas given are empiric and their constants cover a multitude of approximations and non-linearities. I think that real-life data can easily be obtained and used for checking derived fornulas such as MPH = 1363 / ET.

Ah, come on dovy. We were getting some "wild" answers! Users on CR4 don't care if the answer is that simple, many try to make it lots more complicated. I wanted to see how complicated it could get (in an answer, not in real life)

If a person knows 1/4 ET, then we can use:

{ET@660=1/4mileET*.8130}

Ken

Hi MrChevy,

As I said, I know nothing about drag (or any other) racing. Your set of equations enables calculating various numbers as functions of other numbers according to the rules of algebra. As these equations are obviously not pure physics (such as v=at) your question as stated was an exercise in algebra.

I did check if my silly little result made any sense and it did.

The wide exchange of views we all enjoyed should help us realize that there is some danger in correct manipulation of empirically fitted equations which (at best) apply only roughly and to a restricted range of variables.

On the other hand, pure physics rarely applies to real world situation. Anyone who wants good practical estimates should judiciouly mix and match.

dovy

Ok here is my try,

As weight is in all the above equations, it seems to me that each of your parameters are dependent on weight somehow. This means to me that it is simply not possible to equate to ET without the weight in there.

Don't forget that each time you replace an unknown parameter, the weight has to be known before you can fill in the empty space.

That is as close as I can get to your question but I have to add that I am not entirely sure I got what you wanted in the first place. (see my signature)

Sorry

Maybe I'm being a bit thick this morning, but doesn't the elapsed time need a bit more in the equation besides speed (mph) and horse power (HP) ????

Maybe distance travelled would be useful?

John.

You are not thick and fully by your senses. It is impossible and not realy a maths question. I think these equations are emperically derived.

Please elaborate MrChevy. We want to know it all now.

These equations are all related to the purest motor sport of all, drag racing. (I'm sure that will ruffle a few feathers. Sorry, NASCAR.) Drag racing is two cars starting from dead stop through one quarter mile. E.T. is the elapsed time for the vehicle to travel the 1/4 mile. MPH is speed at the finish line. Vehicle is assumed to accelerate through the entire length of the track.

Yes, these equations are empirically derived, so things like distance traveled are contained within the constants. The problem we have with Mr. Chevy's question, is he wants to relate HP & MPH directly to ET. You can directly relate MPH to ET, but without any weight in the equation, HP becomes meaningless. There has to be some mass for the power to work against. There are only two unique equations given, with two forms of each. Trying to algebraically combine will cancel HP out.

See, I thought so. Pleased to see that my maths brain is still supple enough to overcome this trick question.

Naughty MrChevy

I do like drag racing, is this equation the same for bikes?

I suppose that it is now down to the skill of the rider to make it up to the theoretical maximum of their bikes or cars. Otherwise it would be just "comparing notes" at the start line. That would be fun.

I have some friends who do it redularly and they say that they easily outperform much better bikes just on start skills.

Interesting thing this equation.

Thanks CSM Engineer. (Central Sugar Manufacturer?)

If it works for Chevy's, it should work for bikes. And you're correct, reaction time can make a world of difference at the finish line. Drivers and crew chiefs (the guys responsible for setting the car up for track conditions) play a far larger role in winning races than horsepower does.

Yes, same for bikes.

ET races: Doesn't matter how fast or slow the cars are that race each other (or bike/car etc). Example: you get to write what "time/ET" your car runs the 1/4mile in on the windshield. Any number you want. Your choice.

1. You go faster than that time you chose and you lose.

2. You go too slow and you lose.

How do you get that time? Make some practice runs. Let's say you run 13.01/13.02/13.00 and then time trials are shut off and you have to decide what you are going to write on your window (for them to punch into the clocks). You decide on 13.00 and write (or have them write, if you forgot to bring your bottle of white shoe polish with you) 13.00 on your windshield and passenger window.

The guy next to you pulls up and s/he has 17:00 wrote on their window. Easy win... ooops, NOT in ET drags. The guy at the start line pokes in both of your times and then you stage and then the timer starts counting the time down. WOW, the person next to you takes off and your 1st yellow light hasn't even came on!!

They are WAY down the track before your side of the tree starts counting down.

Yep, your green light comes on for you to go when the other car has been gone EXACTLY 4 seconds! 17:00-13:00=4:00 seconds.

You both then have EXACTLY the SAME amount of time to reach the finish line.

You run a 12.99 and beat them by a foot across the finish line... they ran a 17:02. Oops, you went faster than 13:00... YOU LOSE.

So, yes, a LOT depends on how close you "cut" the (green) light. Leave too quick and you RED lighted... doesn't matter how fast the guy you are racing goes then, fast, slow, don't matter, they just have to make it through the lights at the end of the 1/4 to win that round since you lost the instant you red lighted.

Ken

Isn't there a lot of 1/8 mile drag racing out there?

Good reply CSM! And then we have handicapping and don't forget the 9.90 cars, that was really something, first time I seen them. WHAM, FULL THROTTLE... then the motor died (that's what it sounded like)... and then took off again. And ran right on 9.90. Those guys/gals run 9.89/9.90/9/91 time after time after time. But that is what electronics does for you!

Ken

He said, 1/4 of a mile. F=Ma where F is the force M is mass and a is acceleration.

BUT, whether this is a linear or non-linear acceleration, mph and hp. are not good enough. He also needs to have the mass. As he is trying to calculate acceleration (he also said, "from dead stop") at full HP he will keep on accelerating until coming to a constant speed with zero acceleration. it will most probably be be a nonlinear curve, unless power is being reduced proportionally as speed builds up. ET will be the time it took to pass the 1/4 mile point. and will be a particular point on the acceleration curve.

This how I see it.

Wangito.

Hi John,

It is all about drag racing, and the 1/4mile. dovy gave the answer that all of us drag racers have used for years and years. We have such a large mix of users on here and "guest" throws out these "weird" questions without enough information to make an honest try at an answer...

I don't know where the 1363 number came from, had to be from years of info boiled down into that answer, and no matter how fast or slow the car, it the gear ratio is correct and you obtain good traction, then the answer is going to be really close.

My Nova, (just a street car, drive it back and forth to work 10 years ago, no motor in it since June 1999) would run 12.00 but at 111MPH. Poking that into the equation (don't need HP or Weight or anything else, everything is in the 1363 number) we get: 1363/12=113.58MPH telling us it was lying down on the top end (and it was, from me running too small of a carb, was only running a 750CFM Holley) but with a bigger carb I would have went faster too, and that was a no-no. Anything 11.99 and quicker required a rollcage.

With everything "right" (within reason) a 111MPH=12.28ET ... .28 of a second is a LONG time! So, using that formula, a person can tell if his car is running pretty close or not.

Ken

Thanks jhammond I haven't messed with that in soooo long.

Another variable is street or strip. Traction on the strip was much better.

Brad

Thanks for the link. All of the equations I use have been around since way before the internet (as we know it now, say since the early 1990's is when it got going like it is now). Some of the formulas I was using back in the 70's if I remember correctly, with a slide rule (non-electronic calculator!) Nice looking car you have.

Ken

MrChevy I think the problem is your acceleration is not constant because Traction and Aerodynamics are not constant. Also is your 500 hp at the rear wheels or shaft?

Aerodynamics take a major shift a approximately 70 miles per hr, so are not linear.

Next is RPM constant or dynamic? If my memory serves me right Torque not HP is the major factor, thus RPM is another dynamic.

I don't know whether this helps or over complicates, The math was to many years ago.

Brad

You don't.

See my answer to electroman.

I maybe wrong though.

Wangito.

HP isn't particularly good when determining how fast a car will travel down the strip. Really what you should be concentrating on is torque, and where its peak occurs.

Consider this: HP=torque*RPM/5252. If you had say 100 ft-lbs of torque at 1000 RPM, mathematically you would have the same HP as 50 ft-lbs at 2000 RPM. Saying you have 500 HP doesnt mean much if that peak power is at a high RPM for a number of reasons:

1) It means your torque is pretty low

2) You have to wait a long time to shift gears because you want to take advantage of your power band

3) Unless you're racing in nascar or formula 1 and have an engine designed to rev up to 10k RPM, you'll probably blow the engine's valvetrain above 7000 rpm.

Its torque that moves the car. Thats why drag cars run such tall rear end gears, because they add more torque multiplication. Most of your weekend racers are probably running 4.10 or higher rear end ratios because of the boost to torque.

There is software which can give you a more accurate idea of how fast your car will go. COMP cams puts it out (although I think its the same software that was put out years ago under the name "Desktop Dyno", and that company must've got bought out by COMP) and you can order it through summitracing.com. Theres a dyno simulator, a drag race simulator, and an "all in one" package. The "all in one" package runs about $200, but if all you're interested in is the drag racing, that software goes for about $75. And there are more basic, watered down versions available for about $40. I have the desktop dyno program where I virtually designed a small block chevy V-8 and could virtually test different cam/head/etc perameters, with good results. Ended up with a 500 HP/500 ft-lbs (peaks) chevy 406 small block that I built for relatively cheap because I could pick the right parts the first time. Slapped that in a 72 El Camino and now I blast the doors off the honda kids with thier giant wings and farty exhausts...

I've been around drag racing for a long time and as I recall every time someone came up with a formula the racers made a shambles out of it. -- JHF

Sorry I haven't been on here guys. I played too hard when I was younger, now the body says I get to pay for it. The last 4-5 days have been really bad, body in so much pain the water running out of my eyes makes it hard for me to see the monitor.

I take one of these and sometimes the pain is still so much it makes my eyes water so much I can't see clearly. Gonna have to ask the Dr for something stronger, taking 120 of these a month is not doing the job and I had to sign a Federal Narcotics contract with the Dr to get them. After a year or so, I am getting tired of pain.

That's why I haven't been on here for a few days.

Lot's of good response in this thread, thank you all!

Ken

It's too much of a drag to go through all the replies in detail - but if the formulae are correct, we can rewrite them as (MPH/234)^3 = HP/Weight = (5.28/ET)^3.

Which means: ET=1235.5/MPH

Of course, the average speed is exactly related to the ET, but that would give an equation for the 1/4 mile of:
ET=3600/4/MPH=900/MPH
So my guess is that this is the velocity at the end of the drag run - in which case the formulae must rely on a load of assumptions.

In so far as the formulae are accurate, ET and MPH would be different ways of measuring the same variable. (Clearly, this MPH is not the average velocity which is directly related to ET, or this reduction would read

So if you want to calculate HP, you always need to know weight, but you can use either MPH or ET as the other input.

Regarding one of the replies, you will always be limited by traction when starting, but you will be limited by power at higher speeds, so even in principle the formula cannot be exact. I suspect it is also unreasonable to assume that tyre and aerodynamic losses are related to weight in a totally consistent way...

...try this empirically derived equation: ET = 1.05 + 1102.79/(MPH - 10)

...from my 11/73 "Drag Strip Dyno" article in HOT ROD Magazine about F = M*A:

How to determine the MINIMUM rear wheel horsepower (HP) required to accelerate weight (WT) through ¼-mile distance (S) at time (T):

1) Force (F) equals Mass (M) times Acceleration (A): F = M*A, where M = WT(lbs)/32.17 and A = 2*S/T² ...

2) Work (W) equals Force (F) times Distance (S): W = F*S, where W = (WT/32.17)*(2*S/T²)*(S) = (WT/32.17)*(2*S²/T²) ...

3) Power (P) equals Work (W) divided by Time (T): P = W/T, where: P = (WT/32.17)*(2*S²/T²)/T = (WT/32.17)*(2*S²/T³) ...

4) 1 HP equals 550 ft-lbs/second and S equals 1320 ft (¼-mile): HP = (WT/32.17)*(2*S²/T³)/(550 ft-lb/sec) = (WT*2*S²)/(32.17*T³ * 550 ft-lb/sec)...where K = (2*1320²)/(32.18*550) = 196.9536836 ≈ 197 ...so: HP = (WT*K)/T³ ...

5) Thus: HP = (WT*197)/T³ ...

6) Conversely: T = (K*WT/HP)^(1/3) = (197*WT/HP)^(1/3)

7) Thus: T = 5.818*(WT/HP)^(1/3) ...where: Kt = 5.818

8) From ET*MPH equals Kt-times-Kv: Kv = (ET*MPH)/Kt = (1350/5.818) = 232.039 ~ 232 ...

9) Thus: Kt ≈ 5.82 and Kv ≈ 232

10) Finally: ET ~ 5.82*(WT/HP)^(1/3) and MPH ~ 232*(HP/WT)^(1/3)

I'm realize I'm very late to this forum and this might never be seen but I figured I'd throw my two cents worth in. Personally I believe a formula could predict the minimum elapsed quarter-mile time, but it would be very unreliable taking a actual quarter-mile time and trying to convert it to horsepower. Obviously traction has a lot to do with it.

In a old direct connection manual I had they had a chart utilizing trap speed and equating it to horsepower. I've run my 70 road runner on street tires, and on slicks. The elapsed times were over a second difference between sticky street tires and slicks. The trap speed barely varied between the two sets of tires.

The site that jhammond suggests it is great. I tried many of the varying formulas based on trap speed . The direct connection chart estimated 560 hp, averaging all the formulas that seems to have been on the conservative side however most were reasonably close.

Attempting to base horsepower elapsed times seem to diverge considerably from the figures obtained based on miles per hour.

Read this paper:

[URL]http://www.gregraven.org/hotwater/calculators/ET-and-MPH-Equations.php[/URL]