Ford's Electric F-150 pulled a 1 million pound train, but Ford's stunt was way easier than it looked. This isn't a competitive advantage, instead, it's a very clever, very cool, marketing exercise.
Towing a million pounds sounds impossible. On an open road, it pretty much is. On a railroad, however, things are different. With a steel wheel riding on a steel track, the rolling resistance of the train is extremely low. The result? A relatively tiny force is required to move the train in comparison to the weight of the train.
What makes it easy? Well it all comes down to the coefficient of rolling resistance. This is the ratio of force required to pull a certain rotational mass. For example, a coefficient of rolling resistance of 0.1 means you only need a 1 pound-force to move a 10 pound ball. With steel wheels on a steel track, that ratio is far, far lower. Plenty of trucks out there could have done the same thing Ford's electric F-150 did, if not family crossovers and SUVs. Check out the video for a full breakdown of how it all works.
Important Note: This video was filmed prior to my article being published in Road & Track (link below). Upon release of the article, Ford reached out with clarification. Here is additional information with regards to the power required for accelerating the 1.25 million pound train:
1. We want to know how much power it takes to move 1.25 million pounds at a speed of 4.5 mph in just 1,000 ft.
2. We’re going to assume the truck is accelerating for those entire 1,000 ft, and reaches 4.5 mph right when it crosses the 1,000 ft. line.
3. Power = Work / Time. If we calculate work & time, we determine power.
4. Time can be calculated based on average speed over 1,000 ft. Top speed 4.5 mph, starting at 0 mph. Average 2.25 mph. 2.25 mph = 3.3 ft/s. 1000 ft / 3.3 ft/s = 303 seconds (5 minutes, 3 seconds, to go from 0 to 4.5 mph). T = 303 sec.
5. Work = change in kinetic energy. We start at 0, so kinetic energy is zero. All we need to know is final kinetic energy. Kinetic energy is equal to one-half mass times velocity squared (k=0.5mv^2). 12.5M lb = 566,990 kg. 4.5 mph = 2.01 m/s. KE = (0.5)*(566,990 kg)*(2.01168 m/s)^2 = 1,147,263.561 N-m.
6. Plug in variables to Power equation. Power = Work/Time. P = (1,147,263.561 N-m)/(303 seconds) = 3.786 kW or ~ 5 Horsepower.
7. It’s important to note that 5 HP is the power required to to accelerate this mass assuming zero losses. Heat, friction, aerodynamics, etc. There are also smaller assumptions like not accelerating any rotational masses (the train has wheels, etc). A sanity check using the same math shows a 3,300 lb car requires 240 HP to go 0-60 in 3.0 seconds. We know it’s generally double that in the real world. Even giving the truck a factor of 10 advantage, 50 HP is nothing crazy.
8. The train is very heavy, this is the impressive part. Unfortunately, the duration is very long (5 minutes), and the speed is very low (5 mph). Both time and speed variables work heavily against the power required. If the train were to accelerate to 60 mph in the same 1,000 ft distance, it would take 12,000 HP without losses. Quite a difference, just from changing speed!
Road & Track article covering the stunt:
www.roadandtra...
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28 авг 2024
