Brand new at this please help !

[McRat]

Sombody correct me if I'm wrong.  If the coefficient of friction of rubber on salt is .25, and you have 10 sq ft of frontal area, weigh 6000lb, and have a Cd of 0.1, you could go >700mph before traction becomes an issue?

[taper41]

well i apologize for annoying everyone with my questioning and my hypothesis. you do not have to answer me if you would not like to.

i have one last question and i wont bother asking for anymore information, though i do appreciate all that has been given to me.

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?

[manta22]

You have not been reading the many answers on this very subject.

[hotrod]

Depends on the situation.

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".

Larry

[jl222]

Sombody correct me if I'm wrong.  If the coefficient of friction of rubber on salt is .25, and you have 10 sq ft of frontal area, weigh 6000lb, and have a Cd of 0.1, you could go >700mph before traction becomes an issue?

   I inputed '' 10 sq ft--cd .1--6000lbs and 16'' tire width and increased the hp in steps to reach 700 mph at 10075 hp.

  Program does not have coefficient of friction but a traction index 8 for good salt 11 for loose. inputed 11

  Then I decreased the tire width in steps down to 12'' when a note popped up saying that there was something seriously
wrong with the inputs. When I clicked to close the note, Bville Pro program shut down and went back to here.
  Happened twice

   I suspect at the 12'' tire width it blew the tires off. Streamliner would have to be 2 wheel drive with 14'' tires or 4 wheel
drive with 7'' tires or 6 wheel drive at 4.6'' wide which is close to tire widths available.

  Or more weight or downforce for smaller tires and probably more hp.

  Summers Bros with a tandem rear axle?

              JL222

   Also program said 10 sq ft was small for body type but let me continue inputs.

   Input taper41s cd and hp next on Camaro style.

                

[taper41]

Depends on the situation.

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".

Larry

thank you larry there is a lot of information that i do not know about, ao what your saying is that power and aerodynamics are interchangeable and you would get the same result even at higher speeds using the formulas you stated in your answer,

if thats the case then what do you mean by if your going faster enough aero drag dominates all other forces , does that mean at that point a aerodynamic change is more effective than a change of power or no, they are still equal factors

[Stan Back]

Come on now -- why do you all try to answer these hypothetical questions for someone who isn't, nor probably will never, build something.  It's an exercise in jacking-off, and you're all at hand.

(So to speak.)

[hotrod]

No! -- It means that at low speeds aero changes have almost no effect but power changes would make a difference.

For example at the drag strip where terminal speeds are near 100 mph for street cars, aero changes have almost no impact on trap speed. The car is simply not going fast enough long enough for the difference to be noticeable.

Same as the example I gave, for cars and bikes that are going for records under 100 mph aero changes will not make significant changes in top speed because most of their power requirement is other forms of drag like rolling resistance and power train losses. On the other hand aero changes are very noticeable for cars going over 200 mph, and helpful for cars in between those two speeds.

The problem is you are looking for a simple answer for a very complex question. There is no simple answer to most of the questions you are asking.

Each car will have a weak spot, some issue where it is farther from ideal performance than another. That is the issue that car owner needs to deal with as they will get the most return on investment. The same modification made to another car might make no noticeable change at all because you would be trying to fix something that was not a significant problem.

answer to Stan -- because others will have the same questions.

Larry

[rgn]

a 5% increase in aero is equivalent to having a 16% increase in power, well what about traction

Lol, I said this was entertaining on many levels, this has been a great post.  

Taper, I think the principles-physics at play here are fairly obvious.  Real world insights have also been generously provided by some of the folk here who engage in the science of doing.

As hotrod and Dr J have said above, the power required to push an object through air (or any other fluid) increases as the cube of the velocity, so what you say above is pretty much on the mark.  

With drag quadrupling as speed increases, so does the amount of work that need to be done for it to be achieved.  The work required on a wheel driven vehicle is transmitted through the wheels, so without traction work cannot be done, and drag cannot be overcome.  

[taper41]

No! -- It means that at low speeds aero changes have almost no effect but power changes would make a difference.

For example at the drag strip where terminal speeds are near 100 mph for street cars, aero changes have almost no impact on trap speed. The car is simply not going fast enough long enough for the difference to be noticeable.

Same as the example I gave, for cars and bikes that are going for records under 100 mph aero changes will not make significant changes in top speed because most of their power requirement is other forms of drag like rolling resistance and power train losses. On the other hand aero changes are very noticeable for cars going over 200 mph, and helpful for cars in between those two speeds.

The problem is you are looking for a simple answer for a very complex question. There is no simple answer to most of the questions you are asking.

Each car will have a weak spot, some issue where it is farther from ideal performance than another. That is the issue that car owner needs to deal with as they will get the most return on investment. The same modification made to another car might make no noticeable change at all because you would be trying to fix something that was not a significant problem.

answer to Stan -- because others will have the same questions.

Larry

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 ?
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.

im sorry im taking forever to understand exactly the concept please be patient

[jl222]

Come on now -- why do you all try to answer these hypothetical questions for someone who isn't, nor probably will never, build something.  It's an exercise in jacking-off, and you're all at hand.

(So to speak.)

 Yeah Stan I know were your coming from. I expected a comment from you about the small i, but I'm not so good either.

 But some questions might help others.

  As I have inputs for our Camaro in Bville Pro, I took the hp down to 1228 hp and with CD at .3 program shows 259 mph
                                                                
                                                                                                        25% less CD at .225  shows        283 mph

                                                                                                        25% more hp at 1535 shows        288 mph  
  As I change CD and Hp I have to change the gearing for best speed

                   JL222

[hotrod]

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.

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

[jl222]

  Taper41---You might look at bullet ballistics. A 270 caliber is faster than a 306 because it has the same
cartridge but is smaller diameter.

  You could make a chart with amount of powder and size of bullet. Sporting good stores have them and probably the internet.

           JL222

[taper41]

ok larry so i think im getting it,

so the car would then have 400 left over for top end speed but if you cut that in half then you would be right back where you started from .

the changes and aero increase as the drag increases, which is due to going faster, but no matter how fast you go whether you decrease the aero drag by 50% or increase the horsepower by 50% you will still always get the same result.

[fastman614]

Come on now -- why do you all try to answer these hypothetical questions for someone who isn't, nor probably will never, build something.  It's an exercise in jacking-off, and you're all at hand.

(So to speak.)

Stan,
Sometimes this stuff just plain gets interesting.

And.... you never know about someone like taper41..... he may have been helped in ways that none of us fully understand (perhaps not even he)..... If he comes out of engineering college with a degree in this stuff, there is a huge chance that he takes a career opportunity in the field of aerodynamics....it will pay him good... and, if he remembers this, that a lot of real world hotrodders who don't and never will have the "book learning" ar actively engaged in "seat of the pants" modification to vehicle aerodynamics, there is a chance albeit smaller that if or when he has spare time, he may get involved in some design of bodies for use in land speed racing. (I can hope and dream, can't I)