Brand new at this please help !

[tortoise]

My question - can you copy the the two relevant answers and post them together?.... I am just finishing my coffee (I did not have breakfast this morning, so I'm having an early lunch) and will not be responding until much later....

Taper,

I'll leave fastman to speak for himself, if he thinks you're worth the trouble.

You, however, have not given him the help he asked for.

[taper41]

there have been a few ...... here

Power required goes up at the cube of change in speed, to double your speed you need 8x the power.

cube root(1.25) = 1.07721735, so a 25% increase in power would net you about 7.7% increase in speed.
That assumes you are going fast enough that aerodynamic drag is the dominant limit on your speed.

However if you are a 50cc motorcycle trying for a 33 mph speed record, rolling resistance is as much a factor as aerodynamic drag. Generally speaking aerodrag is insignificant (for our purposes) at speeds below 45-50 mph. On flat ground it only takes about 12-15 hp to cruise at 60 mph in a full sized car, and much of that is mechanical losses in rolling resistance and power train losses.

When you say 25% reduction in aero drag are you talking about a 25% reduction in frontal area, 25% reduction in CD (coefficient of drag) or a 25% reduction in the actual drag force?

The drag force varies directly with either CD or frontal area, reduce either by 25% and you get a 25% reduction in drag force. Or you can get a little of each. A common metric is to multiply the frontal area x the CD, as that number changes your drag force will change in the same proportion.

power = force x distance.

At a given speed (ie the same distance per second) the change in force will be identically the same as an equal change in power. By definition they are interchangable. But since drag increases at the square of the speed (velocity for the precise), drag changes much faster at higher speeds than at lower speeds.

As mentioned above a change in aerodrag (frontal area or CD) will hardly be noticed if the terminal speed of the car is under 50 mph, and will be very important at higher speeds.

To return to your question if by 25% reduction in drag, whether you reduce frontal area by 25% or CD by 25% or "drag force" by 25% through a combination of both you will get the same result.

Due to the definition of the relationship of power and force, there is no difference between changing power or drag force. An equal change in either will result in the same change in speed,  "if your speed is fast enough that aerodrag completely dominates all other forces like rolling resistance".

what you mean by force in this definition has to do with the aerodynamics right ?

correct drag is a force that is acting opposite to the direction you are trying to go.

Quote
your saying there is no difference between changing power or the drag force . and that an equal change would result in the same speed
but the faster you go does that definition change or is changing the power or drag force still going to be equal.

No the definition does not change at any speed. What changes is that at low speed other forces are far greater than the aerodynamic drag, so even large changes in aero drag at low speed will not make any noticeable distance.

If it takes 16 hp to go 60 mph and 14 hp of that is used to over come rolling resistance and transmission loss, cutting aerodrag in 1/2 will hardly make a difference. If on the other hand the same car is going 206 mph and needs 850 hp to go that fast, it is now spending only 40-50 hp to overcome rolling resistance and transmission losses but using up 800 hp to over come air drag. If you cut its airdrag in half, it now has 400 hp extra available to go faster.

Larry

[tortoise]

Fastman, here is my post that taper thought contradicted you. I don't think you'll argue with it.

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.

[taper41]

ok wellyou obviously understand the other reply You  post something like that why not just post why not post what I am not getting in lahmans terms

[Stan Back]

¿Whose theis lahmans?

[Glen]

New York attitude

[interested bystander]

One of the stranger topics in recent memory,

Is the original poster trying to show how smart he is or is he truly searching for knowledge and enlightenment?

Methinks I'll bet everyone (else) agrees that the collective Scientific, Technical and PRACTICAL knowledge of the individuals who post on this website will FAR outstrip his knowledge or abilities.

This forum will tear apart a person's ego and flush it down the toilet if the poster proceeds in certain manners.

Those have  come and gone.

A lot of valuable stuff has come out of it so far, however.

[taper41]

It would be nice if everyone quite thinking i was trying to cause issues within the forum, and just help me understand what i'm clearly not getting.

[taper41]

At a given speed (ie the same distance per second) the change in force will be identically the same as an equal change in power. By definition they are interchangable. But since drag increases at the square of the speed (velocity for the precise), drag changes much faster at higher speeds than at lower speeds.

I guess this is the part i dont understand, instead of continuing to think i'm trying to stir everyone up can your guys explain it instead

[Jon]

Which bit of what you just wrote don't you understand?

I'm finding it difficult to answer your "question" because I'm not sure what your question is.
I'm no expert but I'll give it a shot.

Cheers
jon

[taper41]

My last post mainly,

[tortoise]

At a given speed (ie the same distance per second) the change in force will be identically the same as an equal change in power. By definition they are interchangable. But since drag increases at the square of the speed (velocity for the precise), drag changes much faster at higher speeds than at lower speeds.

I guess this is the part i dont understand, instead of continuing to think i'm trying to stir everyone up can your guys explain it instead

You haven't asked a question, taper.
 
If you're wondering why aero drag varies as the square of speed, take a course in fluid mechanics (after you take the required math courses), and come back and try to explain it to me. I don't know.

[taper41]

Im asking someone to explain the post that I quoted and what he means by the power and drag being interchangeable

[rgn]

I'm bored of this.   And you Taper 

un-subscribe...done

[salt27]

taper41,
 I do not have the ability to express what you are asking for in the terms that you require, however I am aware of an expert on such subjects.
His name is Franklin Ratliff and I do not have contact information for him, but you should be able to do a search and come up with it [try Florida].
He has been very helpful with information in the past and I'm sure he wouldn't mind helping you out.

Good luck on your quest, Don