i am a college student currently taking physics, for my mechanical engineering degree. My professor asked a very difficult question.

he said
"the purpose of a sports car is to go faster and faster. but as the car increases speed it takes longer and longer to go from speed interval to speed interval.
example: if it takes a sports car 1 second to get from 20mph to 30 mph...how come it takes 1.2seconds to get from 30mph to 40mph, and so on and so on..?

i said gearing...but he said there is more to it...can anyone help me out? this is gunna be a test question...

 

I would also say that the ammount of drag (air pushing on the car) would increase with speed also. Would also depend on engine combination also. (how fast it pulls at certain speeds)

 

in real world conditions wind resistance increases as speed increases, from a mechanical stand point its becase of gearing, the gear ratios in a car change as speed changes, each gear the transmission shifts into is larger then the previous gear. A larger gear takes more time to turn then a smaller gear. So my answer would be the combination of the two.

 

resistance in the air would be a big factor, but not untill the speed was over 100, however i don't think you can not consider it at any speed. rolling resistance of the tires would be a factor but i think that would be a constant. the performance band of the engine would be a factor, but i doubt he is considering that.

 

It's all about the drag and HP. Pretty much any car on the road will do 100 MPH, but that number gets increasingly smaller as you go from there. My GTA get to triple digits in a quick way, but the time from 100-130 is almost as long as 0-100. In fact, my brother did a little test. He took his Vette out to a "state maintained test facility", by that I mean highway, and ran it up to it's top speed. Then he returned, put tape over all the body lines and did it again, and not only did it go faster, it got there much quicker.

 

https://www.thirdgen.org/techbb2/sho...hreadid=212997

Read thorugh my posts in this thread...some good info there on exterior vehicular resistances.

Also consider the available power, gearing, internal frictions of engine and rotating parts, etc.

Pertinent physics and power related equations are in the above link in my posts.


Also, I am an Engineer...

 

It's all about gearing and where your making power. If you had a CVT always running at the rpm where you made max torque, the car would have a constant acceleration rate. Meaning that the time required to accelertate from 0-10mph=30-40mph. That is assuming ideal no drag, fricton...

Now drag increases by the square of the velocity so as you go faster it takes exponentially more power.

 

FWIW, I have a Master's in Geophysics, and I would make an energy arguement. Neglecting potential energy (you're not going up or down a hill) the total energy of the car is 1/2*m*v^2 where m is the mass and v is the velocity. So, every time you double the velocity, you quadruple the (kinetic) energy. You need to dump in 4 times as much energy from the engine every time you double the speed. So, since you're car can't quadruple it's energy output every time it doubles it's speed (at least my LO3 can't), you will accelerate slower. No doubt that wind resistance and friction are a factor, but even in a vacuum you would not be able to maintain constant acceleration due to the increased energy needs.

Are you by chance studying kinetic energy in this class right now?

 

This is good. I don't know either, but it's a test question in a physics class. It can only be related to drag, hp and torque, and it's got to apply generally to ALL cars. So I doubt it'll be as complex as gearing or other things that are specific to only certain cars. Everything relates to basic fundamentals in physics. So from that, I don't have a clue, lol. But I'd bet that's the place to start.

And the above energy situation probably has something to do with it, too. It sounds plausible.

 

gearing is applicable to all cars unless its a transmissionless car with one direct drive gear. It really is a major factor, drive a manual tranny car and notice the difference from WOT 0-40 in first gear, then try the same thing while starting off in 3rd gear There's one hell of a difference.

 

G-whiz is right.

What he is describing almost sounds like the Law of the Inverse Square. It says that say for example light energy from our sun can heat our planet to a certain temp. If you double the distance from the sun it would take 4x the light energy from the sun to heat the planet to the same temp. If you are 4x the distance then 16x the light energy...and so on...and so on.

Einstein's theory of Relativity also says that an object will never reach the speed of light because it gets heavier the closer it gets. And once it gets heavier it needs more power. Then as it approaches again with the more power it gets heavier and needs more power. And this cycle will continue.

Now I don't think any 3rd gens are going to approach the speed of light. Certainly not my 17.1 sec 1/4 mile LO3. But I think the principle might be the same that G-whiz was describing. The faster you go, the more wind resistance; tire resistance; engine friction and tire heat and friction will require more HP from the engine.

Let's see some Mustang people come up with answers like that!!!

 

I'm no mathematician (I was a liberal arts major), but I think you can boil it down to the power required to perform the work.

Drag is part of it, but it takes exponentially more power to continue to accelerate.

Case in point. 200 hp v8 can accelerate the typical Fbody to a 15 second 1/4mile ET. If its a linear relationship 400hp should get you a 7.5 sec et, but it doesn't, it probably takes 900+ hp to run 7's and a 400hp car might only do low 12s.

Now as to why its exponential? That's the mathematics question.

Maybe I just exposed my own stupidity but I'll now go back and read the replies

-=visit www.burnoutbox.com =-

 

I like Azvolfan's and Gwhiz's responses, I think they most accurately describe why it takes exponentially more energy the faster you go. Drag, gearing, etc. are only part of the equation in acceleration.

Now if you are just talking about maintaining a steady speed w/o accleration, that's all drag and gearing, yeah...

 

Hint:

Stopping from 30 mph takes x feet.

Stopping from 60 mph takes more than 2x feet.

 

G-whiz is right!!!

Ek=.5mv^2 and Ek=max where x is distance.

Also v=at where t is time

So has to = .5mv^2 = max in the end.

Its because when you are going faster you cover the same amount of distance in a shorter time. Or you can go a farther distance in the same amount of time. So you have to apply more power to accelerate in the same distance and time.

It took me a little while but i did figure it out.

BTW i'm a Mechanical Engineering student too. U of Idaho!!

 

Yup, it has to do with the fact that the engine is adding energy to the car at a more or less constant rate, resulting in a more or less linear increase in the car's kinetic energy. But since the kinetic energy is equal to 1/2 * mass * (velocity squared), then as the energy is added linearly, the velocity increases only as the square root of time..... in other words, from 0 to 60 takes 4 times as long as 0 to 30; 0 to 90 takes 9 times as long as 0 to 30; etc. assuming perfect gearing and all that.

I know nothing of math or physics though; I'm just an idiot with no marketable skills, forced (by my unsuitability for gainful employment and otherwise general uselessness) to exist in the world of broadcasting.

 

The last few posts have it correct. The question doesn't even really involve the car at all, it's just a simple definition in physics.

You need a constant increase in acceleration to maintain a constant time between "speed intervals" as your professor put it. His example is describing a single, set rate of acceleration which would increase the amount of time between speed intervals as time progresses.

If you want to impress, tell him a sports car sometimes even has shorter time intervals as it speeds up, because it is impossible to go from 0 acceleration to a set acceleration instantaneously. When you first romp on it, time intervals from 0-5mph are LONGER than 5-10mph, 10-15mph, etc, because your rate of acceleration is increasing.

 

Well I should be able to put some stuff to rest here since I am a Mechanical Engineer and have a focus in IC engines and vehicle analysis.

Your car accelerates slower as speed increases for numerous reasons.

As speed increases air drag increases. Drag increases tremendously based on the shape of your vehilce and at what speed it is going. Sometimes drag is atuall less at high speeds than at lower speeds if the vehicle experiences eddying wake.

As speed increases the rolling resistance of the tires increases as well. Radial tires are more prone to this than pliased by tires. That is why 90% of cop cars bun based ply tires.

Cars do not accelerate or brake linearly. The way highway cops determine your speed after a wreck based on skid marks is WRONG!!!!! They are really off base because all of their little formulae are for constant decelleration and acceleration. Don't be afraid to take a PE to court with you.

Also all the fluids in your system create drag the faster they spin. This is viscous friction and it increases as gear and engine speed increase. Dry or coulumb friction does not increase with speed. Viscous friction is also the reason for air drag increase at speed. Ait is a fluid and behaves just like oil.

This is just a quick run down. If anyone needs clarification on this please let me know.

 

Good thought provoking topic.

Don't forget that energy has a relationship to Work...which is measured in terms of Horsepower.

Go to the math calc formulas for ET. What do you enter...hp and weight.

 

Work is calculated more than just HP. You only need to worry about work if you are displacing something. If you are pushing an object with your car than you would worry about it. Just traveling through a viscous fluid negates work to an extent.

 

Don't forget about torque multiplication. In first gear, your torque is spread over a very short distance of total gearing. in second, the torque is spread over a bit longer spread of gearing, and hence the multiplication is decreased, and so on and so on, until you are in top gear where your total gearing is now quite long (the ratio keeps getting closer to 1:1 with every upshift)(remember torque is what puts the car in motion)

-Shaneo

 

I have an Engineering Technology degree and I don't claim to be an expert in physics but sounds really simple to me. Logrithmic energy consumption. As speed increases then the energy required to maintain a specific velocity increases at a non-linear rate (exponentially). Just as others have said with formulas.

 

Torque mulitplication doesn't enter into the picture at all. All that it does, is to allow the engine to produce the maximum possible amount of energy at any given moment by operating under the most efficient possible conditions. So, it helps the car's velocity to approach the theoretical square-root of time curve as closely as possible. The engine is busy releasing energy, and it that energy is neither created nor destroyed by "torque multiplication" or anything else that goes on in a car, that being one of the immutable laws of this universe (since no conversion of mass to energy or vice-versa is involved). If the engine releases energy, then it goes somewhere, and that "somewhere" is either into car motion, or heat in the drive line (from friction), or if the car is moving fast enough, heat in the surrounding atmosphere from friction with the air (drag). You can bet your life on that.

Incidentally, I was a math and physics major in college; except my specialty was modern physics - quantum mechanics and general relativity and the like - not simple mechanics, although I did have to take alot of classical physics courses of one sort and another (optics, mechanics, thermo, E&M, etc.) in addition to "modern". But all of that isn't really worth a whole lot to me out here in meatspace. Without a docotrate, what I have plus $0.99 and tax is worth just about one cup of coffee. However, I don't claim any particular expertise in any of this; not that "expertise" matters anyway, since it's all just numbers, and the numbers don't care about their user's "expertise". But education does at least help one to apply the numbers to physical reality, which is an ability some of the other posts in this thread seem to be sorely lacking.... both the familiarity with the concepts, and the connection of the concepts with the real-world car.

 

doesn't the car get lighter too as it goes faster. means you wouldn't have as much traction, so you might go slower.

 

No the weight stays the same. If there is any aero lift than the car will not have as much downforce and make it feel like it is lighter. Is that what you are thinking of?

 

All this stuff about weight, lift, drag, traction, etc. is just incremental to the effects of the exponential nature of energy required to accelerate a given mass. That forumla is true whether you're talking about a car or a Saturn V rocket in a total vacuum. Drag and mechanical friction only aggravate the situation, but they are not the primary cause of it.

 

Exactly as I said, too.

 

I'd like to clarify this for myself. I'm an Engineer, but this is absolutely not my area of expertise...and I didn't pay attention much in highschool physics, much less in Engineering Physics at College. Corections welcome:

The mass, for the equation, is derived from weight factoring out the force of gravity...which gravity is only involved during grade resistance. The physical resistances Are part of the complete equation when operating outside a vacuum.

Mass = weight divided by 32.2 feet per second for American units.

Velocity can be measured in feet per second. When squared...it is called acceleration,
Mass*velocity squared is the same as Mass*acceleration which is a Force.

The equation I listed in the link I posted is

Force=ma + Rr (rolling) + Ra (aerodynamic) + Rg (grade)

Where Force is tractive effort (rear wheel torque for rpm at relative velocity*gear ratio) all divided by wheel radius in feet.

Note: Aerodynamic resistance is exponential as well. If your speed is doubled, 8 times more power is required to overcome the resistance.

Energy comes from the engine...and typically about 15% of the energy will make it to the rear wheels...in terms of power.

Now...without my Engineering Physics Book in front of me, can someone detail to me the link between Energy and Work...how it moves the wheels...and how the "divided by 2" gets in there...that is in the above posted equation by Gwhiz.

I think I know...being that Force is a product of force and velocity...but not quite sure without a reference.

Thank you

Edit:
Ahhh. Error.
Velocity is ft/second. Acceleration is ft/ second squared...not the whole thing squared.

So,,,I need more clarification on the Energy equation.

 

Radial tires have LESS rolling resistance than bias ply tires, not more, which is why they became popular in the late '70's/early '80's. Cop cars don't run bias ply tires.

 

Energy is the integral of momentum by time ... that is, momentum is the first time derivative of energy.... (as well as energy being a scalar quantity and momentum being a vector, but that's a whole different can of worms).... so if you apply your first semester high school calculus to energy (which is ½mv^2), you end up with mv, which is of course the scalar part of momentum.

 

Although this is only partially related to the original topic, you can bring this up in your class for some lively discussion......the new Subaru WRX's can accelerate from 0-60 in about 5.2 seconds, but 5-60 takes 7.9 seconds.