train help
train help
langjahr@comcast.net said:
The following probably is completely false since I know squat about jet engines but, since we are having fun, a = F/m. Let's call a 10 second quarter mile acceptable for a train.
Now let's pick a train, say a SD90MAC-H at 6000HP, 425000lbs (192777_kg), 83:16 gear, top speed of 75MPH. Assume we can fix the gearing, traction, constant acceleration, etc, and have perfect conditions.
With rough physics, a=(2*D-2*V0*t)/t^2 = (2*1320-2*0*10)/(10^2) = 26.4_ft/s^2. The final velocity is V = V0 + a*t = 0 + 26.4_ft/s^2*10_s = 264_ft/2 = 180_mph = 80.4672_m/s.
The energy change is 0.5*m*V^2 = 0.5*192777_kg*(80.4672_m/s)^2 = 624111887_kg*m^2/s^2(J) = 591544_Btu. This expended over 10 seconds yields a power of 83,695_HP
Since we have 6,000_HP already, we need an additional 77,695_HP. So, we need 92.831% of 62411887_N*m or a thrust of 34,873_lbf for 1320 feet.
One engine from an F-15 can produce 25-29000 pounds of thrust while an F-22 engine can produce 35000 lbs of thrust (1/2 of a 747 engine). A dragster jet engine might run 5-10000 lbs of thrust.
So, we could make a decent showing with one or two fighter jet engines, or about three to seven smaller jet engines.
If I completely messed this math up, feel free to run it again as I only went through it once and am too tired to do it again.
