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**Summary:**Calculating engine power required for semi trucks under different circumstances.

Hello everyone,

I would like to ask you for help if that is allowed.

For a current project I need to calculate engine power needed in watts under different circumstances.

This is something that I will have do more often, so I would like to make this into an excell sheet where I can simply input the prevailing conditions, and have it calculate the right value(s).

However, I underestimated the complexity of the formula due to all the different variables.

Can someone help me get to the right formula?

**The value(s) are as follows:**

Calculate the engine power needed for the following semi trucks in the described situations:

**Semi Truck data:**

Dimensions: L: 14m W: 2,55m H: 2,74

Front face of truck in m2: 7m2

Gross train weight: 40.000 kg

Overall drive train efficiency: 95%

CW value air resistance: 0,6

**Situation 1:**

Engine power (in kW), needed for the truck above to:

Drive up a slope of 3% with 80k/ph, while experiencing a headwind of 10kph.

Road Condition: Good asphalt with a rolling friction coefficient of 0.007.

**Situation 2:**

Engine power (in kW), needed for the truck above to:

Drive up a slope of 3% with 80k/ph, while experiencing a headwind of 15kph.

Road Condition: Sandy roads with a rolling friction coefficient of 0.2.

So far I have come up with for situation 1:

**Formula for rolling resistance in Newtons:**

9.81 * friction coefficient * vehicle weight in kilo’s * cos (angle of slope in degrees)

9.81 * 0.007 * 40000 * cos 1.72

= 2746 N

**Formula for slope resistance in Newtons:**

9.81 * vehicle weight in kilo’s * sin 1.72

9.81 * 40000 * sin 1.72

= 11778 N

**Formula for air resistance in Newtons:**

0.6 * 0.6 * 7 * 22,22^2

= 1244 N

**Total resistance in Newton:**

2746 N+ 11778 N + 1244 N = 15768 N

**Formula for Required power without taking into account the driveline efficiency:**

Total resistance * velocity in meters per second

15768 * 22,22 = 350364 watts (350,364kW)

**Formula for Required power taking into account the driveline efficiency:**

Power Required / driveline efficiency

350,364 / 0,95 = 368,804kW

This is how far I’ve come on my own, and I know I haven’t taken into account the headwind yet. I would like it if anyone could help me by looking over the calculations so far, and help me perfect them.

Kind Regards,

Wim