Brand new at this please help !

[taper41]

This is why I keep asking to understand each answer, there are a few people that state that the result should be the same, and there is a few that show the result of drag being superior than power.

[saltwheels262]

the best thing to do is probably to make the most amount of power and the least amount of drag.

[tortoise]

the best thing to do is probably to make the most amount of power and the least amount of drag.

Yes, my son, it is better to be rich and healthy than poor and sick.

[Peter Jack]

That is in fact the ultimate answer. Because the two are in no way linked, neither depends on the other, so we work to optimize both as best we can.

Pete

[hotrod]

Lets go back to the raw formula for aerodynamic drag

The following was calculated to 8 significant digits on a unix system using the bc calculator routine.

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

Rho = 0.07967 lbs/ft^3   (Rho is the air density at the temperature and pressure of the test conditions, this is the value for standard temperature and pressure at sea level)

CD= .30
Fa = 20 (ft^2)
mph = (ft/sec)/1.466667     or ft/sec = mph*1.466667

At 200 mph this car has an air drag of:
(0.07967/2) *.30 * 20 * ((200*1.466667)^2)= 20565.49201461 pounds drag

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

Power = force x distance

hp = (force x distance )/33000


20565.49201461 * 293.333400 = 6032545.69531840 ft pounds / sec

6032545.69531840 / 33000 = 182.80441500   Hp required to go 200 mph
 
=========================

reduce frontal area by 25%

Rho = 0.07967 lbs/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:

(0.07967/2) *.30 * 15 * ((200*1.466667)^2) = 15424.11901096 pounds drag

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

Power = force x distance

hp = (force x distance )/33000

15424.11901096 * 293.333400 = 4524409.27148953

4524409.27148953/33000 = 137.10331125 hp required to go 200 mph with new 25% lower frontal area.




The new power required is exactly 75% of the original power required to go the same speed
137.10331125/182.80441500 = .75000000

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

You reduce the frontal area by 25% and the power required to go the same speed drops by 25%
drag force at 200 mph reduced by 25%
15424.11901096/20565.49201461 = .75000000


Larry

[taper41]

So because i can't read all of that lol what exactly is it showing in the conclusion, larry

[Glen]

Maybe you have a problem

[hotrod]

If you can't read and understand all that, you are not ready to be an engineer!

It is telling you exactly what I told you in the beginning, power and aero drag are interchangeable.
If you increase one, the other changes by an equal amount in the appropriate direction to go the same speed.

Increase drag force and power required to go the same speed goes up by the same amount.
reduce drag force and the power required to go the same speed drops by the same amount.

You have to be careful with on line calculators many of them use rule of thumb approximations that are "close enough" but not mathematically correct.

Larry

[taper41]

Ok, I was just looking at the chart that woody had put on the forum

[tortoise]

That is in fact the ultimate answer. Because the two are in no way linked, neither depends on the other, so we work to optimize both as best we can.

Pete

When Mickey Thompson put the top-mounted blowers on Challenger, it increased power and drag.

[tortoise]

Lets go back to the raw formula for aerodynamic drag

The following was calculated to 8 significant digits on a unix system using the bc calculator routine.

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

Rho = 0.07967 lbs/ft^3   (Rho is the air density at the temperature and pressure of the test conditions, this is the value for standard temperature and pressure at sea level)

CD= .30
Fa = 20 (ft^2)
mph = (ft/sec)/1.466667     or ft/sec = mph*1.466667

At 200 mph this car has an air drag of:
(0.07967/2) *.30 * 20 * ((200*1.466667)^2)= 20565.49201461 pounds drag

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

Power = force x distance

hp = (force x distance )/33000


20565.49201461 * 293.333400 = 6032545.69531840 ft pounds / sec

6032545.69531840 / 33000 = 182.80441500   Hp required to go 200 mph
  
=========================

reduce frontal area by 25%

Rho = 0.07967 lbs/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:

(0.07967/2) *.30 * 15 * ((200*1.466667)^2) = 15424.11901096 pounds drag

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

Power = force x distance

hp = (force x distance )/33000

15424.11901096 * 293.333400 = 4524409.27148953

4524409.27148953/33000 = 137.10331125 hp required to go 200 mph with new 25% lower frontal area.




The new power required is exactly 75% of the original power required to go the same speed
137.10331125/182.80441500 = .75000000

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

You reduce the frontal area by 25% and the power required to go the same speed drops by 25%
drag force at 200 mph reduced by 25%
15424.11901096/20565.49201461 = .75000000


Larry

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?

All quite correct, but you have answered a different question than was asked.

Here is the problem as stated.

Decrease CdA by 25%. How fast then?

Return to original conditions, then increase power by 25%. How fast then?

[taper41]

With that said isn't it showing the same thing?

[tortoise]

With that said isn't it showing the same thing?

NO!

If you reduced drag 100%, you'd go infinitely fast. (You'd need a very tall final drive.) If you increased power 100% (the same amount), you'd go 26% faster.

[taper41]

How is it not according to his work if decreasing drag was a mote important factor to the top speed then the car would have needed less than the 75 percent of the original power needed to achieve the same top speed

[jl222]

  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