Two terms, and only one of them cares about the material twice
Friction acts on everything that has to be dragged along the length: the belt on the carry run, the belt on the return run, the rotating idlers under both, and the material. Lift acts only on the material. At the defaults — 400 ft, 500 tons an hour at 350 ft per minute — there are 47.6 lb of material on every foot of belt, so friction acts on 2 x 8 + 12 + 47.6 = 75.6 lb per foot and produces 0.022 x 400 x 75.6 = 665 lbf.
Lift is 40 ft x 47.6 lb per foot, which is 1,905 lbf. Add the 120 lbf of skirt and cleaner drag and the effective pull is 2,690 lbf. At 350 ft per minute that is 28.5 hp at the belt and, through a 90 percent drivetrain, 31.7 hp at the motor.
The belt never appears in the lift term. It goes up loaded and comes back down empty, and the rise gives back on the return run exactly what it took on the carry run. Only the material stays at the top.
A decline is negative lift, not a special case
Enter minus 40 ft for the rise and the lift term becomes minus 1,905 lbf. Against 665 lbf of friction and 120 lbf of other resistance, the sum is minus 1,119 lbf: the loaded belt is pulling the drive rather than the drive pulling the belt. The page prints that with the sign it has and calls the power what it is, because rounding it to zero or taking the absolute value hides the entire point.
What has to be done about it is not an arithmetic question. Something absorbs that energy, and whether that is a regenerative drive, a brake, a backstop or a different layout altogether is for the conveyor maker and the drive engineer. It is also not a saving. Plenty of drives turn returned energy straight into heat, and assuming otherwise puts a credit in a spreadsheet that never appears on a meter.
Loading changes the answer more on a decline than on an incline
Because material appears in both terms with opposite signs on a decline, tonnage moves the pull in a way it does not on the flat. More material on a declining belt increases friction a little and increases the downhill pull a lot, so the net drops, crosses zero, and keeps going. A belt that draws power empty and returns it loaded is an ordinary result and it catches people out.
What this sum is not
It is running power at a steady load. Starting a loaded incline is the harder problem and it depends on the drive, the soft start if there is one, and the inertia of the pulleys and material. It is also not a belt rating: effective pull and belt tension are different quantities, and the tension the carcass actually sees depends on where the take-up is, what the wrap is, and what the slack side has to hold. Those come from the belt supplier and the conveyor maker, not from here.
Questions people ask
How do you calculate conveyor horsepower?
Work out the effective belt pull in pounds of force, multiply by belt speed in feet per minute, and divide by 33,000. That gives horsepower at the belt; dividing by the drivetrain efficiency gives what the motor shaft has to produce. At 2,690 lbf and 350 ft per minute that is 28.5 hp at the belt and 31.7 hp at the motor through a 90 percent drive.
Why does belt weight not appear in the lift part of the calculation?
Because the belt comes back. Whatever work the rise takes out of the belt on the carry run is returned on the return run, so it nets to nothing over a full loop. The material does not come back, which is why it is the only thing lift is proportional to. Belt weight still matters in the friction term, and it is counted twice there because there is belt on both runs.
Can a downhill conveyor generate power?
The arithmetic certainly goes negative: once the lift term exceeds the friction, the loaded belt is driving the drive. Whether any of that energy ends up anywhere useful is a property of the drive, and many drives just dissipate it as heat. What absorbs it, and what stops the belt running away, is a question for the conveyor maker and the drive engineer, and this page names the sign and stops there.
What friction factor should I use?
One that came from your installation, not from this page. The number depends on idler condition and seal drag, on temperature, on how well the belt is trained, and on whether the frame is straight, and it varies enough between a new clean conveyor and a tired dirty one to change the answer materially. Your conveyor maker will give you one, or you can back it out of a coast-down test on your own belt.
Is this the motor size I need?
No, and it is not offered as one. This is steady running power on a belt that is already moving at a constant load. Starting a loaded conveyor asks for more, service factors and duty cycles are separate matters, and motor selection also has to survive the worst case rather than the ordinary one. Take the figure to whoever is specifying the drive as an input, not as an answer.