Watts to Foot-Pounds per Second for Gearbox Design Work

Checking an SI Drive Datasheet in Foot-Pounds per Second

The gearmotor arrives with a datasheet in newton-metres, revolutions per minute and watts. The design file it has to go into is in inches, pounds and feet — belt pulls, shaft loads, hoist duties and a rating summary that has been in customary units since the drawing was first issued. Foot-pounds per second is the customary power unit that keeps the whole chain consistent, and it is one multiplication away from the catalogue watt figure.

Conversion factor: 1 W = 0.7375621756 ft·lb/s. A 1.5 kW motor is 1 500 × 0.7375621756 = 1 106.34 ft·lb/s. Dividing by 550 gives 2.012 hp, which is the same answer as 1 500 ÷ 745.7 — a useful check that nothing slipped.

Where the Two Systems Meet on a Drive

Torque times angular speed gives watts

Every SI drive figure comes from P = T·ω with torque in newton-metres and speed in radians per second. That is the number on the datasheet, and it is the one this page takes.

IEC motors come in kilowatt steps

0.75, 1.1, 1.5, 2.2, 3, 4, 5.5, 7.5, 11 kW — a ladder that deliberately does not line up with customary sizes, so a converted figure rarely lands on a round customary number.

Losses land between the two ratings

Convert the input rating and you have described the motor; convert the output rating and you have described what reaches the machine. A worm set can put 30 % of the difference between those two into heat.

Customary power keeps the chain intact

With the drive expressed in ft·lb/s, a rope pull in pounds and a line speed in feet per second divide straight into it, with no stray conversion factor hiding in the middle of the calculation.

Turning a Catalogue Watt Figure into a Design-Sheet Number

Decide which point on the drive you are describing before you convert anything. Motor shaft, gearbox output and machine input are three different power figures on the same drawing.

Pick the point on the drive

Take the motor's rated output for a shaft-power check, or the gearbox output after efficiency for what the driven machine actually receives. A 2.2 kW motor through a 96 % helical reducer delivers 2 112 W at the slow shaft.

Enter the watt figure

Type 2112 in the left field and 1 557.73 ft·lb/s reads back as you type. Enter a kilowatt figure by choosing kW in the dropdown rather than adding zeros — pasted values with spaces or a comma decimal separator are handled either way.

Carry the number into the calculation sheet

The copy button hands over the value with no unit or spacing attached, so it drops cleanly into the cell where you divide by a rope pull in pounds to get a line speed in feet per second. Ctrl + C inside a field does the same.

Go the other way to write the enquiry

When the customary figure was worked out first and a supplier needs it in SI, the swap button (↔) runs ft·lb/s → W. It is the same page in the direction that turns your design number into something an IEC catalogue can be searched with.

Check the speed the rating belongs to: a datasheet kilowatt figure is quoted at a stated shaft speed. Move the motor to 60 Hz, add a drive, or read the figure at the gearbox output and the speed changes — the torque and the power change with it, so convert the value at the speed you are designing to.

Motor and Gearbox Output Points in Watts and ft·lb/s

Standard IEC motor ratings at a four-pole 1 450 rpm shaft speed, with two gearbox output points added. Torque comes from P ÷ ω, and each ft·lb/s figure is the watt value multiplied by 0.7375621756.

Drive point	Speed (rpm)	Torque (N·m)	Power (W)	Power (ft·lb/s)
0.75 kW motor shaft	1 450	4.94 N·m	750 W	553.17 ft·lb/s
1.5 kW motor shaft	1 450	9.88 N·m	1 500 W	1 106.34 ft·lb/s
2.2 kW motor shaft	1 450	14.49 N·m	2 200 W	1 622.64 ft·lb/s
4 kW motor shaft	1 450	26.34 N·m	4 000 W	2 950.25 ft·lb/s
7.5 kW motor shaft	1 450	49.39 N·m	7 500 W	5 531.72 ft·lb/s
11 kW motor shaft	1 450	72.44 N·m	11 000 W	8 113.18 ft·lb/s
Helical reducer output, 20:1 at 96 %	72.5	278.18 N·m	2 112 W	1 557.73 ft·lb/s
Worm reducer output, 40:1 at 70 %	36.25	276.60 N·m	1 050 W	774.44 ft·lb/s

The last two rows make the point that matters on a drawing. Both reducers deliver almost the same torque — 278 against 277 N·m — yet one passes 1 557.73 ft·lb/s and the other only 774.44, because the worm turns at half the speed and gives away nearly a third of what it is fed. A torque figure alone never settles how much power reaches the machine; the speed and the efficiency have to be in the same sentence.

On the Design-Office Desk

Catalogue watts in, customary power out, as you type

Step through a page of drive ratings and the customary column keeps up without a keystroke wasted, which suits checking a whole gearmotor family against one design sheet.

Turn the pair round for the supplier enquiry

The ↔ button gives ft·lb/s → W, so a duty already worked out in pounds and feet can be handed over in the unit an IEC drive catalogue is indexed by.

Horsepower a dropdown away from the same drive point

Searchable unit lists on both sides mean the same watt figure can be read as hp for the rating summary or ft·lb/min for an older calculation sheet without retyping anything.

Spaced thousands for four-figure work rates

Results run to eight decimals with thousands separated by a space, so 8 113.18 reads at a glance instead of being miscounted as a five-figure number on a busy sheet.

Drivetrain Conversion Questions from the Design Office

The datasheet gives torque and rpm but no power — how do I get to ft·lb/s?

Multiply the torque in newton-metres by the speed in radians per second to get watts, then multiply by 0.7375621756. For 14.49 N·m at 1 450 rpm: ω is 151.84 rad/s, so the power is 2 200 W and the customary figure 1 622.64 ft·lb/s. Staying in customary units throughout works too — torque in lb·ft times speed in radians per second lands directly in ft·lb/s — but the SI route matches the numbers already printed in front of you.

My gearbox is rated in ft·lb and the motor in ft·lb/s — can I compare them?

No, and it is the single most expensive mix-up in drive selection. The gearbox rating is torque, the twisting effort its shaft and housing can take; the motor rating is power, an amount of work per second. A reducer rated for 300 lb·ft is describing a mechanical limit that applies whether the shaft is turning or stalled. Bring speed into the torque figure and only then can the two be set side by side.

How do I turn rpm into radians per second?

Multiply by 2π and divide by 60, which is the same as multiplying by 0.10471976. So 1 450 rpm is 151.84 rad/s, 1 750 rpm is 183.26 rad/s, and a 72.5 rpm gearbox output shaft is 7.59 rad/s. The step exists because torque × speed only produces watts when the angle is counted in radians; leave the speed in rpm and the answer is out by a factor of about 9.55.

Should I convert the reducer's input rating or its output rating?

The output, whenever the number is going to be compared with what the machine demands. Helical and bevel-helical stages typically pass 95–98 % per stage, planetary units similar, but a single-start worm set can drop to 50–70 % at high ratios — 1.5 kW into a 40:1 worm can leave only about 1 050 W, or 774.44 ft·lb/s, at the slow shaft. Convert the input rating and you have flattered the drive by everything the gears turned into heat.

Does the service factor go on before or after the conversion?

It makes no difference arithmetically, since the conversion is a straight multiplication, but it is cleaner to apply it to the load side and keep the equipment rating untouched. A crusher or reciprocating compressor running long hours might carry a factor of 1.75 or more, a smooth conveyor around 1.25, so a 1 106.34 ft·lb/s absorbed duty becomes 1 382.93 ft·lb/s of required capacity at 1.25. Record which number carries the factor, because a design sheet that leaves it ambiguous invites someone to apply it twice.

Watts to Foot-pounds per Second