Bruce2
Member
Posts: 293
From:Vancouver, Canada
Registered: Oct 2001
posted 05-31-2002 12:56 AM
--------------------------------------------------------------------------------
Shad, you must've skipped out that day of high school physics class.
The relationship is that power requirement goes up by the square of the speed, not the cube.
The formula to use is K.E.=1/2*m*V^2
Shad Laws
Member
Posts: 978
From:Evanston, IL, USA
Registered: Aug 2001
posted 05-31-2002 02:33 AM
--------------------------------------------------------------------------------
Hello0
Shad, you must've skipped out that day of high school physics class.
The relationship is that power requirement goes up by the square of the speed, not the cube.
The formula to use is K.E.=1/2*m*V^2
Nope... sorry...
I was there back in high school physics. However, high school physics didn't teach me near enough to understand systems like these. The problem is that without some knowledge of dynamics, you may not understand my derivation above.
The equation you listed above is the Newtonian equation for ENERGY as a function of VELOCITY. If a ball of some known mass (say in lb, or lb-m to be proper) is flying through the air at some known velocity (say in mph), we can figure out how much more energy it has then than if it were at zero velocity (say in btu, or J, etc.). We can do the same for a car: we can find out how much energy you have just before you hit a brick wall... or, you can use that number to estimate how much energy your brakes have to dispell to very quickly stop you to zero. This equation is 100% valid for a given specific time (not necessarily time frame!) for speeds far away from light speed.
Now, we are looking for the POWER consumption as a function of VELOCITY. POWER is the time derivative of energy, i.e. ENERGY PER TIME.
Power and energy are VERY, VERY different things. Analogy: power is to energy as velocity is to length as flow is to volume. Power is a rate of energy change!
If you sat in on elementary high school calculus, you should know that the time derivative of E=1/2*m*v^2 is dE/dt=m*v*dv/dt, or P=m*v*a, or P=F*v.
What does this mean? It means that power generation is related to the VELOCITY and the NET FORCE on an object. In true vector form, it's a bit different, but our equation is 100% valid if the directions of force and velocity are parallel. In a car's case, they are anti-parallel: drag forces push you backward while the car moves forward. So, the power generation is P=-F*v. Or, if we flip our perspective, power consumption is P=F*v. Make sense?
Now, I don't want to get into vector notation using ASCII on an online forum, so you'll have to take a leap of faith and trust me on the shortcuts I'm taking to getting to the next step :-). To keep an object (like a car) moving at a constant speed, we need the net force to be zero, right? Well, that means that the net power must be zero. Therefore, you can see that the power generation by the engine must be equal to the power consumption by various drags. Bingo!
Well, what drags do we have? Lots. Okay, so how do we characterize them and make them make sense? If we collect data from various cars, we will find that the drag as a function of velocity can be very well-approximated by: F=A*v + B*v^2 (A and B are some constants that change from car to car). The linear term has to do with tire drag, brake drag, etc. The quadratic term has to do overwhelmingly primarily with aerodynamic drag. The higher the velocity, the more the second term dominates the equation. In other words, the higher the vehicle speed, the more we can essentially ignore the first term. So, at high vehicle speeds: F=B*v^2.
Back to the power consumption equation for high vehicle speeds: P=F*v, or P=B*v^3. Voila, we have our simple approximation.
If we have some reference point (i.e. a standard bug using about 35hp at about 75mph), then we can figure out B, right? Or, we could just use ratios of speeds to ratios of power consumptions, either work well.
(35hp)=B*(75mph)^3
B= (35hp)/(75mph)^3
@some velocity v in mph:
P=(35hp)/(75mph)^3*v^3
P=35*(v/75 mph)^3 hp
So, we can approximate that at 125mph, we need 162hp for a standard bug.
Take care,
------------------
Shad Laws
LN Engineering
http://www.lnengineering.com
