chriserx
Well-known member
I'd say this was about right, at 100 mph I glanced at carscanner and motor output was roughly 90 kw, which puts it at 120 hp, wasn't willing to take my eyes off the road or do a longer term test other than that.I ran the numbers and came up with higher hp requirements than did Gemini, but either way the amount of power to keep a truck rolling is substantial. I think Gemini underestimated the energy losses. When using the power to run an EREV you will have both losses charging the battery and discharging.
Anyway here’s what Gemini says is the power to keep a F150 rolling at various speeds:
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To maintain a constant speed of 60 mph on a level road, a typical gasoline-powered Ford F-150 requires approximately 25 to 35 horsepower (hp).
While these trucks have engines capable of 325 to over 700 hp, that massive power is only used for acceleration or towing. At a steady cruise, the engine only needs to produce enough power to overcome two main forces: Aerodynamic Drag and Rolling Resistance.
The Physics Breakdown
To calculate this, we look at the power required at the wheels (\bm{P_{wheels}}) and then factor in the efficiency of the drivetrain.
1. Aerodynamic Drag (\bm{F_d})
For a large vehicle like an F-150, air resistance is the biggest hurdle at 60 mph.
• Formula: \bm{F_d = \frac{1}{2} \cdot \rho \cdot v^2 \cdot C_d \cdot A}
• Drag Coefficient (\bm{C_d}): Approx 0.44 to 0.46 (Trucks are "blocky" and less aerodynamic than cars).
• Frontal Area (\bm{A}): Approx 31.5 to 35 sq ft.
• Speed (\bm{v}): 60 mph (88 feet per second).
• Result: At 60 mph, aerodynamic drag consumes about 18–22 hp.
2. Rolling Resistance (\bm{F_{rr}})
This is the friction between the tires and the road.
• Formula: \bm{F_{rr} = C_{rr} \cdot m \cdot g}
• Weight (\bm{m}): A typical F-150 weighs about 5,000 lbs.
• Rolling Coefficient (\bm{C_{rr}}): Approx 0.010 to 0.015 for standard highway tires.
• Result: Rolling resistance consumes about 8–10 hp.
3. Total Power at the Wheels
• Sum: \bm{20\text{ hp (Air)} + 9\text{ hp (Tires)} = \mathbf{29\text{ hp}}}
Why the Engine Works Harder
The number above is the power needed at the wheels. To provide 29 hp to the ground, the engine must produce more because of:
• Drivetrain Loss: About 15–20% of power is lost through the transmission, driveshaft, and differentials.
• Accessories: The alternator, water pump, and air conditioning also draw a few horsepower.
Estimated Engine Output at 60 mph: 34–40 hp.
Comparison Table: Power Required vs. Speed
As speed increases, the power required rises exponentially because aerodynamic drag increases with the cube of velocity (\bm{P \propto v^3}).
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