Brand new at this please help !

[Glen]

14 pages and 209 replies and still he doesn't understand.

[jl222]

  Taper.. these drag power formulas work more for objects not trying to put power to the ground through tires
with a limited track and fuel load.

                      JL222

[tortoise]

14 pages and 209 replies and still he doesn't understand.

I gave up on taper many pages ago. Simple algebra is beyond him. I'm just waiting for someone, anyone, to tell me they understand what I'm saying.

[Interested Observer]

To perhaps clarify a few things:

Re: Hotrod’s reply # 187
The 1544.8 horsepower result that yields the “77.24%” is in error.  Re-doing the calculation will give the expected 1500 hp and 75%.  So, all the ensuing discussion about this number is pointless.

Re: Hotrod’s reply # 199
The units for rho, the air density, should be “slugs/cu. ft.” and at sea level the value would be on the order of 0.00238.  This misapplication resulted in the ludicrous results of 20,565 pounds of drag at 200 mph and the requirement of only 182 hp to achieve that.

Also--   Power = Force x VELOCITY, not distance.


Many pages ago, Tortoise, in reply # 126, correctly and succinctly, albeit somewhat obscurely, answered the question.  Drag reduction by a given percentage produces very slightly better results than power increase by the same percentage.

And, in deference to reality, it is an academic answer to a contrived academic question.

[hotrod]

Crap you are right mass should be in slugs not pounds --- serves me right for trying to pound out that complex answer while at work and up to my eyeballs in other things.

Although the magnitude of the answer is screwed by using that improper unit for Rho the ratio of the results should be correct and they are identical if you keep the terminal speed the same. If you do figure it as stated by tortise there is a small 2% difference again, but I strongly suspect that is due to rounding errors in the cube root function but have not had time to figure out if there is some other cause.
I also used the per minute conversion to hp rather than the per second conversion --- in short I totally screwed it up.

Got a ton of work to do so back to that this will have to wait until later.

Good catch !

Larry

[tortoise]

Many pages ago, Tortoise, in reply # 126, correctly and succinctly, albeit somewhat obscurely, answered the question.  Drag reduction by a given percentage produces very slightly better results than power increase by the same percentage.

Thank you!

And, in deference to reality, it is an academic answer to a contrived academic question.

Reality is just a crutch for people who can't handle drugs.

[tortoise]

If you do figure it as stated by tortise

As asked by taper41

there is a small 2% difference again, but I strongly suspect that is due to rounding errors in the cube root function but have not had time to figure out if there is some other cause.

I show  the drag-reduced car as 2.17459098580708087981673712439% faster, per the Microsoft Windows calculator. Unix numbers are surely better, though, so I must be wrong.

[fastman614]

  It was Experts not expert, [and I did think I was among them] showing your contempt and disrespect for everyone.

  Next time you refer to anyone as an ''armchair expert'' I WILL repost your threat as an inspector to fail a fellow

posters car.

             JL222

               

I perhaps should post my personal message to you on this page, right?  Or, perhaps I should go through this thread and collect your posts and then repost them all in one.....

Whatever you say Mr Langlo.....

  Been in Santa Barbara and just catching up

  Yeah, do both, and your post also. The only person I give a hard time is Taper [for lack of punctuation skills and claiming to be a college student] skills taught by 4th grade, and you for disrespecting everyone calling them ''armchair experts'' saying some that post here think they have a God given right to be right. Yeah post them side by side and your personal message.

  AND don't forget to post my personal message

  Leave out your utopia B.S.about how everyone gets along at your work.

             JL222

Mr Langlo.... WHATEVER!.... maybe you ought to go out and meet a few more high school and college students....

In case you haven't noticed, this thread got interesting in a positive way in the last little bit....

Continuing to wish you nothing but good posts, I remain....
Dave

[hotrod]

Ok lets try this again I think I have it all corrected for my original calc.
Ratio at fixed terminal speed again is exactly the same (with in the rounding limits of the calculation) if you have a fixed terminal speed, if you lower your drag by 25% your power requirement to go that speed also drops 25% a 1:1 relationship.

When I get the time to do the other version I will

==============

density of air .002373 slugs/ft^3


force = 1/2 Rho (v)^2 CD Fa

Rho = .002373 slugs/ft^3
CD=.30
Fa=20
mph = (ft/sec)/1.466667     or ft/sec = mph*1.466667

At 200 mph this car has an air drag of:
(.002373/2) *.30 * 20 * ((200 * 1.466667)^2)= 612.55067843 pounds drag

Distance covered per second is 200 * 1.466667 = 293.333400 ft/sec

Power = force x distance/second

hp = (force x distance/second )/550

hp = (612.55067843 * 293.333400)/550 = 326.69376941 hp




reduce frontal area by 25%

Rho = .002373 slugs/ft^3
CD=.30
Fa=15
mph = (ft/sec)/1.466667     or ft/sec = mph*1.466667

At 200 mph this car has an air drag of:

(.002373/2) *.30 * 15 * ((200*1.466667)^2) = 459.41300882 pounds drag

Distance covered per second is 200*1.466667 = 293.333400 ft/sec

Power = force x distance

hp = (force x distance/sec )/550

hp = (459.41300882 * 293.333400 )/550 = 245.02032705


The new power required is exactly 75% of the original power required to go the same speed
245.02032705/326.69376941 = .74999999


======================

Insert Quote
Quote from: hotrod on Today at 07:48:45 PM
If you do figure it as stated by tortise
As asked by taper41
Quote
there is a small 2% difference again, but I strongly suspect that is due to rounding errors in the cube root function but have not had time to figure out if there is some other cause.
I show  the drag-reduced car as 2.17459098580708087981673712439% faster, per the Microsoft Windows calculator. Unix numbers are surely better, though, so I must be wrong.

Nope the windows calculator is actually a pretty good high precision calculator, it allows you to get the same useless precision I did in Unix. Since the lowest precision input is the CD and frontal area at one decimal place it is a difference of about 2%, any more precision is only useful for showing how sensitive the calculation is to rounding errors. In both cases we got the same number out to 8 decimals (ie the same answer).

Calculated to 8 digit precision in bc 1.02174590

It is interesting to see where that 2% difference is coming from when the fixed terminal speed calc shows they are interchangeable variables.

Larry

[fastman614]

If you do figure it as stated by tortise

As asked by taper41

there is a small 2% difference again, but I strongly suspect that is due to rounding errors in the cube root function but have not had time to figure out if there is some other cause.

I show  the drag-reduced car as 2.17459098580708087981673712439% faster, per the Microsoft Windows calculator. Unix numbers are surely better, though, so I must be wrong.

I think I am going to go back and find that post that stated to the effect that where the engineers (or possibly the theoreticians) end up in situations where, when reality does not match the theory, reality is always right!

To Taper41.... there comes a time when the theory has to be proven in a practical manner!.... the "arguing" over percentages of about 2%(or even a bit more) is, in this kind of "racing", not statistically imporant until ALL OTHER AVENUES OF PERFORMANCE ENHANCEMENT HAVE BEEN EXHAUSTED!....That is about when our team may revisit the theory in search of that 2%.... I said, as did others, early on that most of us are using "seat of the pants" comparison (sorry, I do not recall who first used the expression -NO PLAGIARISM INTENDED!) .... In industry, specifically the mining and ore extraction industry (with which I have experience), 2% is NOT enough of a margin for which a "go/no-go" decision would be made - mostly because of the "fact" that the numbers cannot be relied upon to be that exact....

Back to this topic, it is a likely possibility that, when the differences reach less than 5%, "real world" evaluation begins to take place.... scale models and wind tunnel testing will begin.... an automobile with more or less exact Cd numbers and frontal area is procured or created and then actual testing takes place.... etc....

[tortoise]

Hotrod, I got my result by dividing the cube root of 4/3 by the cube root of 5/4.

[fastman614]

Whoa!..... My MS Word file is getting rather large!....

Hotrod- You have posted a lot of what I would call interesting information... So have several others of you... I have copied it all!....

It will probably take me "years" to digest it all but I am going to be slowly working on doing so....

Thanx Again!

[taper41]

Ok lets try this again I think I have it all corrected for my original calc.
Ratio at fixed terminal speed again is exactly the same (with in the rounding limits of the calculation) if you have a fixed terminal speed, if you lower your drag by 25% your power requirement to go that speed also drops 25% a 1:1 relationship.

When I get the time to do the other version I will

==============

density of air .002373 slugs/ft^3


force = 1/2 Rho (v)^2 CD Fa

Rho = .002373 slugs/ft^3
CD=.30
Fa=20
mph = (ft/sec)/1.466667     or ft/sec = mph*1.466667

At 200 mph this car has an air drag of:
(.002373/2) *.30 * 20 * ((200 * 1.466667)^2)= 612.55067843 pounds drag

Distance covered per second is 200 * 1.466667 = 293.333400 ft/sec

Power = force x distance/second

hp = (force x distance/second )/550

hp = (612.55067843 * 293.333400)/550 = 326.69376941 hp




reduce frontal area by 25%

Rho = .002373 slugs/ft^3
CD=.30
Fa=15
mph = (ft/sec)/1.466667     or ft/sec = mph*1.466667

At 200 mph this car has an air drag of:

(.002373/2) *.30 * 15 * ((200*1.466667)^2) = 459.41300882 pounds drag

Distance covered per second is 200*1.466667 = 293.333400 ft/sec

Power = force x distance

hp = (force x distance/sec )/550

hp = (459.41300882 * 293.333400 )/550 = 245.02032705


The new power required is exactly 75% of the original power required to go the same speed
245.02032705/326.69376941 = .74999999


======================

Insert Quote
Quote from: hotrod on Today at 07:48:45 PM
If you do figure it as stated by tortise
As asked by taper41
Quote
there is a small 2% difference again, but I strongly suspect that is due to rounding errors in the cube root function but have not had time to figure out if there is some other cause.
I show  the drag-reduced car as 2.17459098580708087981673712439% faster, per the Microsoft Windows calculator. Unix numbers are surely better, though, so I must be wrong.

Nope the windows calculator is actually a pretty good high precision calculator, it allows you to get the same useless precision I did in Unix. Since the lowest precision input is the CD and frontal area at one decimal place it is a difference of about 2%, any more precision is only useful for showing how sensitive the calculation is to rounding errors. In both cases we got the same number out to 8 decimals (ie the same answer).

Calculated to 8 digit precision in bc 1.02174590

It is interesting to see where that 2% difference is coming from when the fixed terminal speed calc shows they are interchangeable variables.

Larry

There is a lot of great information here! I'm interested to see the how the other calculations come out in the other version ! On the contingency that all the little math errors are worked out the numbers should end up the same shouldn't they

[jl222]

I know that some of these question i have might be elementary to you folks that are veterans at this but please help to inform me of some of the simple questions due to a study that i have been doing for physics....

in a perfect world where traction was not a problem, when trying to reach top end speeds, is it more advantageous to have the streamliner equipped with 4 wheels like a tradition car or 2 wheels like a tradition motorcycle, lets keep the wheels all at the same with for either vehicle for a controlled variable and say that both liners have the same weight and power along with same dimensions. theoretically wouldn't the motorcycle style have a lower drag thus  enabling it to reach higher speeds.

from the outside looking in i dont understand how the 4 wheel streamliners have been able to reach higher top speeds
wouldnt the 2 wheel design be more efficient for top speed

as silly as this might be i thank whomever helps me with this

  First it's a perfect world were traction is not a problem, then you refer to the current record speeds set on a track were distance and traction is a problem.

  Thats why 4 wheels and hp rule.

  Inputing and refining data in the Bville Pro program shows why.

           Starting data 1208 hp CD .300  gearing  2.3   mph after 1 mile 213   2 miles 245  3 miles 257  4 m 263  5 m 265

          -25% CD        1208 hp CD .225    ''        2.15   ''             ''    218       ''      253     ''       270   ''    280   ''  284

          +25% HP        1510 HP CD .300    ''        2.1    '''            ''     231       ''       263     ''      279    ''   285   ''  289

        At the finish the lower Cd is catching up but Bville Pro only goes 5 miles

       Also the program is optimistic on acceleration as it won't input the 4'' wide tires used by most.
       But we have been ahead of program inputs at El Mirage [which has better traction than Bville] with the 222 camaro


                    JL222    
 

This is what i don't get, you torture me for a few days then come up with data which explains what i was asking.  That's all i wanted in the first place.
I see at the end that the vehicle with the 25% drag increase is catching up, but will it eventually  top out at a certain speed , equal to the car with the power change , or would it just keep going with no end

  I had more time to post the mile times than my other posted BvillePro results just showing top speed.

  After looking closer at the print outs tonight [been away]  

                                                              -25% CD car at 4.5 miles 73.67 sec-- 282.5 mph --acceleration .02--7250 rpm
                                                                                5   miles  80 sec       284.4 mph      acl           .01    7290 rpm
                                                             little weird     5    mi     80.2 sec    284.4 mph       acl          .01     7290 ''

                                                              +25% HP car 4.5 mi  70.29 sec      287.5  ''           acl          .01     7200 rpm
                                                                                4.88 mi 75 sec          288.4  ''            ''             .01     7220 ''
                                                                                5 mi      76.54 sec      288.6  ''            ''            .01     7230  ''

       Looks like the low drag car was catching up fast but hit the wall and it looks like the hp cars tongue was hanging out also.  

      Like another poster said [to tired to look] a different cd and frontal area might give a different answer.

  I have no clue what formulas Bville Pro uses but I did input the know CD- frontal area and hp for the Summers Bros streamliner and came up with 425 mph.

  Bville Pro has been fun and it is really useful in figuring gearing.
  And you don't have to be a math wizard

                  JL222
                                                            
                                                                                
 
 

 

[hotrod]

Okay let's try this again -- with the restated problem I get the same result about a 2% difference.
Hopefully I have fixed all the typo's and brain farts here too.

Here is the problem as stated.

Tortoise statement of the problem:
Increase power by 25%. How fast then?
Return to original power, then decrease frontal area by 25%. How fast then?

Original poster's statement of the problem:
May 30, 2012, 05:23:12 PM »
my question is if you had the same vehicle for a top speed run what would be more effective in raising top speed,
 a 25% increase in power, or a 25% decrease in aerodynamic drag?


To figure that you would need to rearrange the formula a bit to be able input horsepower directly

(2 * hp * 550)/ (Rho * Cd * Fa) = ft/sec^3

ft/sec^3 = ((2 * hp * 550)/ (Rho * Cd * Fa))

ft/sec = ((2 * hp * 550)/ (Rho * Cd * Fa))^-.3333
(400 * 1100)/ (.002373 * .30 * 20) = 30903216.74392470
cube root(30903216.74392470) = 313.810807 ft/sec = 213.96186523 mph


(500 * 1100)/ (.002373 * .30 * 20) = 38629020.92990588
cube root(38629020.92990588) = 338.042445 ft/sec = 230.48343284 mph
 

(400 * 1100)/ (.002373 * .30 * 15 ) = 41204288.99189961
cube root(41204288.99189961) = 345.393485 ft/sec = 235.49550443 mph


235.49550443/230.48343284 = 1.02174590 apparent advantage for drag improvement over power improvement.

That 2% and a fudge difference is consistent but I have not been able to figure out the mechanics of why it exists when stating the problem this way but when stating the problem on a fixed terminal speed basis you come out with no difference.
As above it is useless precision because we don't know the frontal area or CD to a 1/50 accuracy in the first place and the engine power is probably no better than +/- 5% unless the dyno operator was very very careful and you have made corrections for all sorts of esoteric issues like oil temp, coolent temp intake air temp , barometer etc.

Like I said, my gut instinct is it is an artifact of the calculation process (cube root calculations are just an approximation), and all calculators tend to accumulate rounding errors with large numbers, but it has got my curiosity peeked to see if I can figure out why it shows up in one variation of the problem and goes away in another variation of the problem.

Larry