Author Topic: Brand new at this please help !  (Read 83991 times)

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Offline taper41

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Re: Brand new at this please help !
« Reply #180 on: June 04, 2012, 07:40:15 PM »
is there a calculator or a program that I can use to find the absolute top speed for this same scenario thats exactly what I need

Offline Jack Gifford

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Re: Brand new at this please help !
« Reply #181 on: June 05, 2012, 01:27:44 AM »
No.

Oops... I intended to remain silent...
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Offline fastman614

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Re: Brand new at this please help !
« Reply #182 on: June 05, 2012, 02:13:11 AM »
No.

Oops... I intended to remain silent...

Jack, Jack, Jack.....

But, if there IS, I want a copy of, or a link to it, as well!
No s*** sticks to the man wearing a teflon suit.

Offline Jon

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Re: Brand new at this please help !
« Reply #183 on: June 05, 2012, 03:27:58 AM »
You can do it in excel if you ignore the fact that distance is limited and traction can also be limited.

Inputs are drag, torque curve data points, gear ratios and tyre rolling radius.
By using the torque curve, gear ratios and tyre radius you get torque available at the tyre contact patch.
Overlay that with the drag line.
The gap between the drag line and the wheel contact torque is what is available for acceleration, if the lines cross no more acceleration.

I've never put in weight or coefficient of friction but that would show where traction limited fairly easily.

Cheers
jon
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Offline tortoise

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Re: Brand new at this please help !
« Reply #184 on: June 05, 2012, 11:29:33 AM »
By using the torque curve, gear ratios and tyre radius you get torque available at the tyre contact patch.


Nit-pick: That would be a linear force, not torque.

« Last Edit: June 05, 2012, 12:56:40 PM by tortoise »

Offline taper41

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Re: Brand new at this please help !
« Reply #185 on: June 05, 2012, 01:29:20 PM »
there is a calculator i found but i dont know if the math is right it seems different, I am not sure if this calculator can tell me what i wanna know. It's on this website .

www.apexgarage.com/tech/horsepower_calc.shtml?cd=.5&a=10&w=2000&v=400&hp=1222.8

Offline taper41

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Re: Brand new at this please help !
« Reply #186 on: June 05, 2012, 03:56:15 PM »
As many of you have asked for me to do my own research I have, although very basic I it involves the factors that I have been asking about.

If any of you would like to respond, letting me know if either of these test are valid or if they do not make sense at all and why, that would be great.

I started off with very basic factors

bhp= 2000                                  
drag= .4
front area= 20
weight= 4000

top speed result = 325.76 mph
increased bhp only, by 25% = 2500 = top speed of 350.88 mph


same starting specs

top speed = 325.76
decreased drag only, by 25% = .3 = top speed of 354.98 mph


To double check my work I did the opposite of the first test, by decreasing the bhp by 25% and increasing drag by 25%, while the other factors once again remained constant.

bhp = 2000
drag = .4
front area = 20
weight = 4000

top speed result = 325.76 mph
decrease bhp only, by 25% = 1500 = 295.945 mph

top speed result = 325.76
increase drag only, by 25% = .5 = 304.225 mph


If these are viable, they both show that drag plays a more significant part when it comes to maximum speed, than horsepower does. 
Do these test have any truth to them at all or are these the wrong test to use all together?
« Last Edit: June 05, 2012, 05:32:16 PM by taper41 »

Offline hotrod

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Re: Brand new at this please help !
« Reply #187 on: June 05, 2012, 05:28:43 PM »
It is close -- as is usually the case in this sort of problem you have to be very careful of definitions.

You never specified exactly what you meant by reducing drag by 25% I interpret that as reducing the "drag force", and how much power it would take to reach the same speed with the newer drag number.

That said the difference in top speed is only 1.1% different.

Working it for the same top speed to determine the amount of power it would take to get to the same speed you get:

CD=0.3
FA = 20
weight=4000
speed = 325.76
power required at new lower drag = 1544.8 = equals 77.24% of the original power.

That is  2.9% more than 75% of the original power. If reducing drag by 25% was more important than power you would require less than or = 75% of the power to go the same speed with the new lower drag number.

What does this tell you?

Most likely what is telling you is that you are within the rounding errors of the calculations that they are using for the spread sheet and for all practical purposes there is no difference between changing power by 25% or drag force or CD or frontal area by 25%, the resulting speed will be essentially identical regardless of which you do.

You'd have to work those same numbers in the actual airdrag formula with high precision (ie air density, CD and FA to 4 decimal places etc.) so that the precision of the calculation is significantly larger than the difference you are trying to verify.

Larry
« Last Edit: June 05, 2012, 05:30:36 PM by hotrod »

Offline racer

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Re: Brand new at this please help !
« Reply #188 on: June 05, 2012, 05:29:35 PM »
Well stated....

Retired Millwright Local 102, Oakland, CA.

See ya on the salt.....

Ray

 
Fastman, here is my post that taper thought contradicted you. I don't think you'll argue with it.
Quote from: taper
if  you decrease the aerodynamic drag by 25% or add power by 25% would the result differ or be the same, if it was on the same vehicle?
Set unit values arbitrarily so power/drag ratio is 1.
Decrease drag by 25%, power/drag is 1/.75=1.33.
Increase power by 25%, power/drag is 1.25/1=1.25
Ergo, result differs, decreasing drag 25% gives superior result.
I think we've done all we can.

When dealing with engineers, there comes a time when engineers are best left to argue amongst themselves.... and we Maintenance/repair/installation technicians (who by trade designation, call ourselves millwrights) quietly go ahead and install or repair the item being argued about.... we then let the engineer(s), by process of elimination, figure out by what (or whose) method it was done... while we just stand back with a knowing smile .... knowing that the process will repeat and that the outcome(s) will be accomplished my much the same method the next time...

So, in Toronto, Ontario (Canada) the millwrights union local was incorporated in April of 1958. In 2008, they had stickers made for their hardhats that said something like - Millwrights Local 1454 - Incorporated April 15, 1958.... Making Engineers Look Good For Over 50 Years!

I don't profess a whole lot of real engineering expertise... I understand the math.... I can do A LOT of practical "on the job" engineering but with aerodynamics and how to make cars go fast, I know about 3 or 4 things....

1- Make sure that your car is built to "known" specifications (looking at other "fast" dragster type lakesters, I copied them)

Enhance the aerodynamics by what you see on any other fast cars. (With what I had to work with, I copied, to the best of my ability, the rear tail extension and top "fin" that Al Teague had on his streamliner - Al did not have a rear tail fin anywhere near as pronounced as most other fast cars)

3- "Buy" or otherwise procure as much horsepower as you can - even going in to debt to do so! (I bought a  "new" Dodge R5P7 engine that was massaged beyond NASCAR legality)

4- I chose a transmission with custom gear ratios that help keep the engine running within 7% of peak horsepower from the top of second gear on up...

It has resulted in record setting performance.... beyond this is guesswork on my part... and the farther away from these above that the guesswork takes place, the "less educated" of guesswork it is.

Like I inferred and perhaps said earlier, Taper41 - I wish you well in your schooling... I also hope you take a lesson or two from this exercise... If, for instance you compiled the more educated replies, made calculations and answers etc based on those replies and then asked for further input in the vent that you were not understanding the outcomes, you could have "ramped" the intellectual level of this thread up a bit.

I did also say that some of the information has already been filed away for my future reference. I will be working over what I have saved.... for this I am thankful as some of the replies and postings, while not fully understood by me, had information that I never knew and I will be attempting to make some practical application of it, if for no other reason than to make some meaningful "fudge" numbers in my desktop dyno program.

I believe that this discussion may never be "over" .... if Taper41 comes back on here and posts some interesting facts obtained via his studies, I for one will be interested in what he will have to say....

Wishing you all "happy posting"....
Dave

Ray C Wheeler

Offline taper41

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Re: Brand new at this please help !
« Reply #189 on: June 05, 2012, 05:37:49 PM »
It is close -- as is usually the case in this sort of problem you have to be very careful of definitions.

You never specified exactly what you meant by reducing drag by 25% I interpret that as reducing the "drag force", and how much power it would take to reach the same speed with the newer drag number.

That said the difference in top speed is only 1.1% different.

Working it for the same top speed to determine the amount of power it would take to get to the same speed you get:

CD=0.3
FA = 20
weight=4000
speed = 325.76
power required at new lower drag = 1544.8 = equals 77.24% of the original power.

That is  2.9% more than 75% of the original power. If reducing drag by 25% was more important than power you would require less than or = 75% of the power to go the same speed with the new lower drag number.

What does this tell you?

Most likely what is telling you is that you are within the rounding errors of the calculations that they are using for the spread sheet and for all practical purposes there is no difference between changing power by 25% or drag force or CD or frontal area by 25%, the resulting speed will be essentially identical regardless of which you do.

You'd have to work those same numbers in the actual airdrag formula with high precision (ie air density, CD and FA to 4 decimal places etc.) so that the precision of the calculation is significantly larger than the difference you are trying to verify.

Larry


So by your calculations, and the test you used the outcome is showing that power is more significant for top speed by apx. 3% ?

Offline hotrod

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Re: Brand new at this please help !
« Reply #190 on: June 05, 2012, 05:57:00 PM »
No it is showing that there is no meaningful difference between airdrag and power when it comes to top speed!

When the difference in calculated performance is that small, it is not statistically significant, because in this case you don't know anything about how precise the calculations are that they used for the spread sheet.

For example the error in measuring the engines horsepower on the dyno will be larger than the differences in those calculations.
A 1% error in the 20 square foot frontal area is 28.8 square inches, an area about the size of 1/4 of a sheet of copier paper (or a small side view mirror).

If there was a real meaningful difference between power and drag the numbers would vary by 10% or more.

One of the things that they don't teach engineers any more is judging the significance of their calculations. They are conditioned by working with a computer that all those digits behind the decimal point mean something. They seldom do. The real world is not as neat as people would like it to be, and it takes a great deal of care to get calculations for things like air drag to give you really precise answers as your answer cannot be any more accurate than the lowest precision piece of input information.

Larry

Offline taper41

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Re: Brand new at this please help !
« Reply #191 on: June 05, 2012, 06:34:52 PM »
What I am saying is, by the results of your calculations, you have said that if the drag coefficient was more important than power to the result of top speed, then the result should have came out as less than 75%.

there for isn't the fact that this is showing a result of 77.24% (2.9%) above the expected outcome, inferring that power is superior than aero drag?

In which both these test with the opposite outcome show that there is a small statistical error involved in these calculations, there for helping us conclude that the if the test was exactly precise the outcome of changing either drag or power by the same percent would lead to identical results

Offline WOODY@DDLLC

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Re: Brand new at this please help !
« Reply #192 on: June 05, 2012, 06:35:30 PM »
Taper41, here is all the help you need for now: http://www.landracing.com/forum/index.php/topic,9378.msg153450.html#msg153450

You need a better overall grasp of the fundamentals before you consume 13 more pages of this forum!  :-o
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Offline taper41

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Re: Brand new at this please help !
« Reply #193 on: June 05, 2012, 06:50:45 PM »
this is definitely showing that aerodrag has a superior effect, now would there be an error with this graph aswell?

why would this graph show a much more definitive result

Offline tortoise

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Re: Brand new at this please help !
« Reply #194 on: June 05, 2012, 06:56:09 PM »
You're all missing something I mentioned umpteen posts ago, and several posters have linked to, but nobody seems to have absorbed.

Assuming CdA and power were the only factors (we know they're not, but leave that aside for now), two cars with the same ratio of power to CdA would go the same speed.  Agreed?

Now, let's start with a car with a car with 400 horsepower and a CdA of 4 square feet.
Increase power by 25%, power to CdA is 500/4 = 125
Instead, decrease CdA by 25%, power to CdA is 400/3 = 133.3

The decreased CdA car will be a bit faster. Each horse is pushing a little less air.

What the Bonneville Pro run showed is how much running at the salt is a drag race, which the veteran racers repeat over and over.
« Last Edit: June 05, 2012, 07:00:13 PM by tortoise »