Calculating the top speed of a train (almost there)

[Diresu]

I'm trying to calculate the top speed of a train and I think I'm almost there but I've hit a wall.

I started off by using the equation found at the link below and solved it for V (velocity).
D=.5*PV^2ACd
http://en.wikipedia.org/wiki/Drag_equation

That gives me

V=SQRT((2D)/(PACd))

When I had finished I realized that this only accounts for air drag. Doesn't the mass of the train also play a factor in it's top speed? I'm not saying that drag isn't the biggest problem it faces but I have to believe mass is also a factor.

I'm using the metric system. Could I add the mass of the train in Kilograms to the right side of the original equation and then solve for V again? More or less that would give me...
The force needed=air resistance at velocity + train mass

Thanks,

Al

 

[Chi Meson]

Top speed depends most importantly on the power of the engine (specifically, how quickly the engine delivers energy). At top speed, the engine must put energy into the train at the same rate as energy is taken from the train by friction and air drag.

Power = (energy transfer)/time = work/time = (force*distance)/time = force * (distance/time) = force * speed

power = force * speed

speed = power/force

 

[Diresu]

Thanks Chi!

I already know the power (or in this case the force) created by the theoretical engine.

In the Drag equation D=.5PV^2ACd

D = Force of drag (the force the engine needs to create for forward motion?) = 10,000 Newtons of engine thrust/force
P = Density of the air fluid the train is traveling in= .1 Kg per cubic meter
V = Velocity of the train in meters per second = what I'm solving for
A = frontal reference area of the train = 2 square meters
Cd = Drag coefficient = .15

When I solved for V to determine it's top speed I got
V=Sqrt((2*D)/(PACd))

or

816.5 Meters per second = Sqrt((2*10,000)/(.1*2*.15))

But that equation doesn't include the mass of the train. That only takes into account the negative drag force created by air.

If I added the mass of the train (5,000 Kilograms) onto the right side of the original equation and then solved for V would I be accounting for mass correctly?

D=.5PV^2ACd + Mass of 5,000 Kg

http://en.wikipedia.org/wiki/Drag_equation

thanks a billion,

Al

 

[Chi Meson]

No, mass does not directly figure into the top speed of a train unless you're going uphill. Mass would be a factor in how much energy is required to get it to top speed, but as it approaches top speed, all that really matters is the frictional and drag forces holding the engine back. Mass will be part of what determine the frictional drag in the wheel bearings, but this will be negligible compared to air drag.

 

[Diresu]

How is that possible?

How can mass not be a factor in top speed? The other day I was pulling my youngest son around the backyard in a wagon as fast as I could run. After a few minutes his older brother got jelous and climbed in. I couldn't run nearly as fast with both of them in the wagon. There has to be a really basic equation to account for mass as well as drag.

<----Force = Drag + Mass ---->

I could make up my own screwed up equation but it's bound to be wrong. I know that slightly "off" stuff like this can be hard even for experts but isn't there somebody in your group who would know which equation I'm talking about? It's got to be simple. I'm not smart enough to think up hard questions.

Thanks,

Al

 

[Chi Meson]

The reason you got tired more quickly has to do with two things. It took more than twice the amount of energy to get your two sons to the same speed, because more mass means more kinetic energy at the same speed.

Second, mass is directly proportional to the frictional force (that's the resistance to rolling int the wheels, in this case). For a little wagon, this friction is very large and is the most significant resistive force. FOr a train that is moving in the hundreds of miles per hour, this frictional force (which does not depend on speed) will be overwhelmed by the air drag which increases with speed.

If you added in the frictinal force, you will need the effective "coefficient of friction" for the wheels of a train. Multiply this number by the trains weight (kilograms times 9.8), and add this frictional force to the drag force. You will find that you will be adding "thousand of Newtons" of frictional force, to "hundreds of thousand" (or even millions) of Newtons of air drag force.