Sorry about the off topic post but this problem has been driving me
nuts for ages and a few people here know a thing or two about physics.
Is it possible to calculate the acceleration curve of a train if the
tractive effort and total train resistance (grade, rolling, wind,
curve) are known? For example:
A train with a mass of 1410 tonne requires a TE of 195 KN to maintain
65 Km/h on a 1 in 100 grade. If I go to full throttle at that speed
the TE will be 230KN. Eventually the train will reach a balance speed
of 74km/h with a tractive effort of 202 KN.
How in earth do I calculate the acceleration rate at different
speeds/times/distance travelled? Obviously I can't use the
acceleration formula f=m*a. Any help would be great.
Jeff
[email hidden]
Jeff wrote:
Use a = f / m
instead ;)
Seriously.
Sum the forces, divide by the mass, the result is the acceleration.
At the "balance" speed mentioned above, the sum of all the forces
is zero and the train's acceleration is zero.
Quoted text (19 lines)
> > Sorry about the off topic post but this problem has been driving me > nuts for ages and a few people here know a thing or two about physics. > > Is it possible to calculate the acceleration curve of a train if the > tractive effort and total train resistance (grade, rolling, wind, > curve) are known? For example: > > A train with a mass of 1410 tonne requires a TE of 195 KN to maintain > 65 Km/h on a 1 in 100 grade. If I go to full throttle at that speed > the TE will be 230KN. Eventually the train will reach a balance speed > of 74km/h with a tractive effort of 202 KN. > > How in earth do I calculate the acceleration rate at different > speeds/times/distance travelled? Obviously I can't use the > acceleration formula f=m*a. Any help would be great. > > Jeff > [email hidden]Hi Matthew
The problem is that I know the acceleration rate is parabolic shaped.
It takes time and distance for the acceleration rate to increase.
Using F=M*A (I made a typo with F=M*S) is saying that the train will
instantly increase it's acceleration to 5 km/h/m then decrease as it
reaches the balance speed of 74 km/h. No way.
The trouble with calculating it using the 2 known speeds at different
times (average acceleration between 2 points) is that you already have
to know the answer to the problem to solve the problem. (Got a
headache coming up with that sentence). It's no problem calculating
the resistance and TE at different speeds. It's the
time/distance/acceleration rate that I can't crack.
Jeff
In article <[email hidden]>, [email hidden] (Jeff) wrote:
>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.
>It takes time and distance for the acceleration rate to increase.
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.
>Using F=M*A (I made a typo with F=M*S) is saying that the train will
>instantly increase it's acceleration to 5 km/h/m then decrease as it
>reaches the balance speed of 74 km/h. No way.
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 :-)
