>Hi Matthew
>The problem is that I know the acceleration rate is parabolic shaped.
How do you know this? Sounds like you are looking at one particular curve.
No. It takes time for the velocity to increase, not the accel.
Accel is not a fcn of distance. Well, only in so far as you are looking at
one continuous curve. Velocity is the result of time and accel history.
Distance is the result of the velocity history. Thus yes, if you look at one
curve, there is a relationship betweeen distance and accel, but that is only
because they were both fcns of time. Well, the distance is a fcn of accel,
accel is NOT a fcn of distance.
Hmmm. I'd say "yes way". Acceleration can be instantaneous (though I'm not
saying 5 km/h/m is the correct value).
Let's do the math...
You say you start with knowing the velocities for two combos of "tractive
efforts" (force at the wheel), grades (1%), speeds. From these two, you have
to calculate the retarding forces which vary with speed & grade:
aero drag = 0.5*rho*FA*V^2
(FA=frontal area, rho=air density, V=velocity. If there is a wind, it is the
vector sum of the wind and velocity normal to the
frontal area).
rolling resistance= Cr*M*g
(Cr=rolling coef usually expressed as KN(force)/KN(Weight), M=mass, g=grav
constant). In reality, there is often a velocity-dependent term, but I showed
the simplest, and often used, form.
grade force = sin(arctan(grade %)*M*g
You have to do the sin/arctan thing since grade is expressed as a percent
relative to horizontal distance, not in degrees in the direction of travel.
At a given velocity and grade, the sum of those forces is calculable if you
know all the constants.
You then use a = [F(tractive)-F(retarding)]/m to calculate instantaneous
accel.
Velocity = Vo + integral (a*dt)
distance = So + integral (velocity*dt)
For your case, you have two data points, the only thing changing is the
velocity and tractive force, so you can determine the aero drag constant
(0.5*rho*FA). Thus you can determine the combination of the other two terms.
ie, your problem is solvable. If OTOH, you wanted to now calculate the result
at another grade, you'd need another data pt to differentiate the rolling term
from the drag term.
I probably made this almost impenetrable. But the info is there. If the
problem is as you state, once the extra tractive force is applied, the train
will instantly accelerate (NOT instantly increase its velocity!). The
acceleration will continuously *decrease* with time due to the buildup of the
aero resistance. At the terminal velocity, the aero drag has increased such
that the sum of the retarding forces equals the tractive force, accel = 0,
velocity = steady state.
With all the above, you can solve your problem. And move on to more esoteric
concepts like hp vs torque and why they are the same, but different :-)
