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Conveyor Belt Speed and Motor Power Calculator
Work out how fast a conveyor belt runs from its drive, and how much motor power that belt needs to move the load. Enter the drive pulley diameter, the motor speed, and the gear reduction to get the belt speed, or type the belt speed directly. Then add the material throughput, the conveyor length, the incline, the belt and moving parts mass, the friction factor, and the drive efficiency. The tool returns the belt speed in m/s and fpm, the effective belt tension, the motor power split into the part that fights friction and the part that lifts the load, and the belt loading in mass per metre.
It uses the simplified main resistance and lift model from ISO 5048 and DIN 22101, the same physics behind a full CEMA design, so you can size a drive, sanity check a supplier quote, or see why a steep incline needs a much bigger motor than a flat run of the same length. Free, no sign-up, and your numbers stay in your browser.
In short: belt speed is v = pi x D x N, where N is the drive rpm divided by the gear reduction ratio. The motor power is the effective belt tension times the belt speed, divided by the drive efficiency, and that tension has two parts: friction along the belt and the weight of the load lifted up the incline. A 0.32 m pulley at 1,450 rpm through a 20:1 reduction gives a belt speed of 1.21 m/s (239 fpm). Carrying 120 t/h up a 30 m belt at 10 degrees, with 16 kg/m of belt and moving parts, a 0.03 friction factor, and 90% drive efficiency, the drive needs about 2.59 kW (3.47 hp), the effective tension is 1,919 N (431 lbf), and the lift term (1.70 kW) is more than double the friction term (0.63 kW).
Conveyor result
2.59 kWmotor / drive power needed
- Motor power
- 2.59 kW (3.5 hp)
- Belt speed
- 1.21 m/s (239 fpm)
- Effective belt tension
- 1919 N (431 lbf)
- Power to overcome friction
- 0.63 kW
- Power to lift the load
- 1.70 kW
- Belt loading
- 27.4 kg/m
- Material throughput
- 120.0 t/h
The drive needs about 2.59 kW (3.5 hp) at 1.21 m/s (239 fpm). The lift up the incline is the bigger draw. Add a service factor and pick the next standard motor up.
How the calculator works
The tool answers two linked questions in one pass. First, how fast does the belt run. Second, how much power does the motor have to deliver to keep that belt moving under its load. The belt speed comes straight from the drive geometry, and the power comes from the tension the drive has to pull against, which is set by friction along the whole belt plus the weight of the material carried up any incline. Enter the drive, the load, and the belt, and the panel reports the belt speed, the effective tension, the motor power, and how that power splits between friction and lift.
Belt speed is v = pi x D x N, where D is the drive pulley diameter and N is the pulley speed in revolutions per second. The pulley turns slower than the motor because a gearbox sits between them, so N equals the drive motor rpm divided by the gear reduction ratio, then divided by 60 to reach revolutions per second. With D in metres and N in rev/s the speed comes out in m/s; with D in feet and N in rpm it comes out in feet per minute (fpm). One fpm equals 0.00508 m/s, so the tool reports both side by side. If you already know the belt speed, switch the source to enter it directly and skip the pulley math.
Belt loading is the mass of material sitting on each metre of belt, mm = Q / v, where Q is the mass flow rate in kg/s and v is the belt speed in m/s. A throughput of 120 t/h is 33.33 kg/s, and at 1.21 m/s that spreads to about 27.4 kg per metre of belt. Loading matters because it is the mass that has to be dragged along and lifted, so it feeds directly into the tension.
The effective belt tension is the force the drive pulley has to apply at the belt line to keep everything moving. In the simplified ISO 5048 and DIN 22101 form the tool uses, it is Te = f x g x L x (2 x mb + mm) x cos(a) + mm x g x H. The first term is the main resistance: the friction factor f times gravity g (9.81 m/s squared) times the conveyor length L times the moving mass per metre, where 2 x mb counts the belt and moving parts on both the carrying and return runs and mm is the material. The cos(a) trims that term slightly on an incline. The second term is the lift: the material mass per metre times gravity times the lift height H, where H = L x sin(a). Friction acts along the belt whether it goes up, down, or flat. Lift only appears when the belt climbs.
Power follows from tension and speed. The belt power is Te x v, the tension times the speed at which the belt moves. The motor has to supply a little more than that because the gearbox and drive lose some of it to their own friction, so the motor power is the belt power divided by the drive efficiency. At 90% efficiency the motor delivers about 1.11 times the belt power. To convert, 1 hp equals 0.7457 kW, so the tool prints kW and hp together. This is the steady running power. Starting a fully loaded belt draws more, which is what the service factor covers.
Belt speed from the drive
Every belt speed on a driven conveyor traces back to three numbers: the drive pulley diameter, the motor speed, and the gear reduction between them. The motor spins fast, often near 1,450 rpm on a four-pole machine or 1,750 rpm on a US 60 Hz motor, and the gearbox slows that down to the pulley. Divide the motor rpm by the reduction ratio to get the pulley rpm. A 1,450 rpm motor through a 20:1 gearbox turns the pulley at 72.5 rpm, or about 1.208 rev/s.
The belt then travels at the rim speed of that pulley, which is its circumference times how many times it turns each second. The circumference is pi x D, so a 0.32 m pulley has a circumference of about 1.005 m, and at 1.208 rev/s the belt moves about 1.21 m/s. In imperial terms that same belt runs at 239 fpm. Change any of the three inputs and the speed moves with it: a bigger pulley, a faster motor, or a smaller reduction all raise the belt speed, and the reverse lowers it.
Speed is not a free choice. It is a design lever that trades against the cross-section of material on the belt, the wear on the belt and idlers, the spillage at transfer points, and the noise the line makes. That is why the tool lets you enter a target belt speed directly and work backward to the drive, or start from a drive you already have and see what speed it produces. Either way, the speed you land on drives both the capacity the belt can carry and the power the motor has to find.
Where the power goes
The single most useful thing this calculator shows is not the total power but how that total splits. Every watt the motor delivers goes to one of two jobs: dragging the belt and its load along against friction, or lifting the material up the incline. The panel reports each separately, and the balance between them tells you what kind of conveyor you are really dealing with.
On the default inclined feeder, the friction term is about 0.63 kW and the lift term is about 1.70 kW. Lift is more than double friction, which means most of the motor exists to raise 120 tonnes an hour up the slope, not to overcome rolling resistance. On a flat belt the lift term is exactly zero, and every watt goes to friction, so length and the friction factor become everything. On a decline the lift term goes negative, because gravity now helps rather than hinders, and if the slope is steep enough the whole power number turns negative and the belt tries to drive the motor.
Reading the split changes how you act. When lift dominates, the way to cut power is to reduce the height the material has to climb or to move less of it, not to chase a slicker idler. When friction dominates, a lower friction factor from better idlers and alignment, or a shorter run, is where the savings live. The chart in the result panel draws the two shares so the dominant term is obvious at a glance.
Effective belt tension and why it sets the motor
Power on a conveyor is really a tension problem in disguise. The motor turns the drive pulley, the pulley grips the belt, and the belt has to be pulled tight enough to move the whole loaded run. That pulling force at the belt line is the effective belt tension, Te, and once you know it, the power is just Te times the belt speed. So sizing the motor comes down to getting the tension right.
The tension gathers up two resistances. The main resistance is the rolling and rubbing friction of the belt on its idlers and the material on the belt, spread over the full length of the conveyor, which is why longer belts pull harder even on the flat. The lift resistance is the plain weight of the material multiplied by the vertical height it is raised, which is why a short steep belt can pull harder than a long gentle one. In the default case the tension works out to 1,919 N, or 431 lbf, and multiplying that by the 1.21 m/s belt speed gives the belt power that the efficiency then converts to motor power.
This effective tension is not the same as the total tension in the belt, which is higher because the belt also carries a takeup tension to stop it slipping on the drive pulley and sagging between idlers. The simple power calculation uses the effective tension only. A full belt selection, checking the belt strength and the takeup, works with the peak tension, which is the effective tension plus the slack-side tension. The tool reports effective tension because that is what sets the motor. Belt strength is a separate check you carry out once the drive is sized.
Incline, lift, and lift height
The incline is the input that changes a conveyor the most, because it turns on the lift term. On a flat belt, the material is carried but never raised, so the lift power is zero and the motor only fights friction. Tilt the same belt up a few degrees and the motor suddenly has to raise every tonne through a real height, and that height grows with both the length of the belt and the sine of the angle.
The lift height is H = L x sin(a). At 30 m and 10 degrees, sin(10 degrees) is about 0.1736, so the material rises about 5.21 m over the run. The lift power is the mass flow times gravity times that height, which for 33.33 kg/s over 5.21 m is about 1.70 kW. Steepen the angle or lengthen the belt and the height climbs, and the lift power climbs with it in direct proportion. This is why the tool lets you enter the incline either as an angle or as a vertical rise: sometimes you know the slope, and sometimes you only know how high the material has to go.
The main resistance term shrinks very slightly on an incline because it carries a cos(a) factor, and cos(10 degrees) is about 0.985, so the friction is trimmed by around 1.5%. That trim is small and never offsets the lift it comes with. The practical takeaway is that on anything but a gentle slope, the lift term runs the show. If you want to move the same tonnage up the same height for less power, the honest levers are a shallower path or a shorter climb, not a marginal friction improvement.
Belt loading and the material on the belt
Belt loading is how much material rides on each metre of belt, and it links throughput to power. It is the mass flow divided by the belt speed, mm = Q / v. Two conveyors moving the same tonnes per hour can carry very different loads per metre if they run at different speeds: the faster belt spreads the same tonnage thinner, so it carries less mass on each metre, while the slower belt piles it deeper.
That loading feeds both resistance terms. In the main resistance it adds to the moving mass the friction acts on, and in the lift term it is the mass being raised. In the default case, 120 t/h at 1.21 m/s gives about 27.4 kg per metre of belt. Speed the belt up and the loading falls, which lowers the tension per metre but does nothing to the total tonnage or the total lift, since the same mass still has to be moved and raised each second. Loading is most useful as a design check: a very high loading warns that the belt cross-section is nearly full and spillage is likely, while a very low loading suggests the belt is faster than it needs to be for the tonnage.
Choosing a belt speed for the material
Belt speed is a tradeoff, and the right value depends on what the belt carries. A faster belt moves the same tonnage with a smaller cross-section of material, so the belt can be narrower and the load per metre lighter, which tends to lower the tension and the motor size per metre of belt. The cost of speed is more wear on the belt and idlers, more spillage at loading and transfer points, more dust thrown into the air, and more noise. Fine or dusty materials, or fragile products, usually want a slower belt to keep them on the belt and intact.
Typical bulk material belts run somewhere between 1 and 3.5 m/s, roughly 200 to 700 fpm, with heavier and coarser materials at the lower end and free-running granular materials able to go faster. Package handling lines are often slower still, set by the rate the downstream process can take rather than by the belt. The tool makes this easy to explore: set the target speed you have in mind, or read the speed a given drive produces, and watch how the loading and power respond. Match the speed to the material and the process, then size the drive to the speed you chose.
CEMA and the simplified resistance model
Two standards frame conveyor power. In the United States, CEMA’s “Belt Conveyors for Bulk Materials”, now in its 7th edition, is the rigorous reference, building the belt tension from a detailed set of resistances and then selecting the belt, idlers, and takeup from it. In Europe and much of the rest of the world, ISO 5048 and its close relative DIN 22101 describe an equivalent resistance-based method. This calculator uses the simplified main-resistance plus lift form of that method, the same core physics both standards agree on.
What “simplified” means here is honest to state. The tool captures the two largest resistances, the main friction along the belt and the lift up the incline, which together dominate the power on most conveyors. It does not add the secondary resistances that a full design includes: the resistance of the material and belt at the loading point, the scrapers and skirtboards, the pulley bearings, the belt wrap and flexure, and the tilt of the idlers. On a long or complex conveyor those secondary terms can add a meaningful slice, often 10 to 20% of the main resistance, so treat the tool’s power as a sound first estimate rather than the final selection figure.
A full CEMA or ISO design also does work this tool does not attempt. It selects a specific belt carcass and cover from the peak tension, sizes the takeup to hold the slack-side tension, checks the belt against startup and braking loads, and confirms the idler spacing and the transition geometry. Use this calculator to get the belt speed, the effective tension, and the drive power quickly, to compare options, and to understand the split between friction and lift. Take the result into a full design, or to a conveyor manufacturer, for the belt and takeup selection that a real installation needs.
Metric and imperial units
The tool accepts inputs in either metric or imperial and always reports the two headline results, belt speed and motor power, in both systems. In metric mode you enter the pulley diameter in metres, the length and rise in metres, the throughput in tonnes per hour, and the belt mass in kg per metre, and the power comes out in kW with the hp shown alongside. In imperial mode you enter the pulley in inches, the length and rise in feet, the throughput in tons per hour, and the belt mass in pounds per foot, and the belt speed reads in fpm with m/s alongside.
The conversions the tool uses are exact. One horsepower is 0.7457 kW, one fpm is 0.00508 m/s, and force converts between newtons and pounds-force at 4.448 N per lbf. Because the two systems are shown together on every result, a plan drawn up in one set of units reads cleanly to a colleague or supplier working in the other. Enter your numbers in whatever system your data arrives in, and read the answer in whichever system you report in.
Five worked examples
Example 1: inclined bulk feeder (the default)
This is the case the tool opens on. A 0.32 m drive pulley turns at 1,450 rpm through a 20:1 reduction, so the pulley runs at 72.5 rpm and the belt moves at 1.21 m/s (239 fpm). The belt carries 120 t/h over a 30 m run at a 10 degree incline, with 16 kg/m of belt and moving parts, a friction factor of 0.03, and 90% drive efficiency. The lift height is 30 x sin(10 degrees), about 5.21 m. The effective tension works out to 1,919 N (431 lbf), and the motor power is 2.59 kW (3.47 hp). Of that, only 0.63 kW fights friction while 1.70 kW lifts the load. The lesson: on an incline the lift term usually dominates, so the height the material climbs, not the friction, sets the motor.
Example 2: long horizontal belt
Now a flat belt. A 0.40 m pulley at 1,200 rpm through a 15:1 reduction turns at 80 rpm, giving a belt speed of 1.68 m/s (330 fpm). It moves 200 t/h over a 60 m run at 0 degrees, with 20 kg/m of belt and moving parts, a 0.03 friction factor, and 90% efficiency. Because the belt is level, the lift term is exactly zero and every watt goes to friction. The effective tension is 1,292 N and the motor power is 2.40 kW (3.23 hp), with a belt loading of 33.2 kg/m. The lesson: on the flat, length and friction are everything, so a lower friction factor from good idlers and alignment, or a shorter run, is the only way to cut the power.
Example 3: steep incline conveyor
A short, steep climb. A 0.25 m pulley at 1,750 rpm through a 25:1 reduction turns at 70 rpm, giving a belt speed of 0.92 m/s (180 fpm). It carries 80 t/h over 40 m at 18 degrees, with 14 kg/m of belt and moving parts, a friction factor of 0.033, and 88% efficiency. The lift height is 40 x sin(18 degrees), about 12.36 m. The effective tension climbs to 3,584 N (806 lbf) and the motor power is 3.73 kW (5.00 hp), with 2.69 kW going to lift and only 0.59 kW to friction. The lesson: a steep belt runs at high tension, so check the belt strength and fit a holdback, because both the motor and the belt are working hard against gravity.
Example 4: decline that drives itself
A downhill belt that gives power back. Here the belt speed is entered directly at 1.5 m/s (295 fpm). It moves 150 t/h over 50 m at minus 12 degrees, with 18 kg/m of belt and moving parts, a 0.03 friction factor, and 90% efficiency. The slope is negative, so the lift term is negative: gravity pulls the load down the belt and feeds energy in rather than taking it out. The lift share is about minus 4.25 kW, which overwhelms the 1.38 kW of friction, so the motor power comes out negative, about minus 3.19 kW (minus 4.28 hp). The lesson: the load pushes the belt, so you size a brake or a regenerative drive and a holdback, not a bigger motor. A decline conveyor is a braking problem, not a driving one.
Example 5: imperial packaging line
The same physics in US units. Working in imperial, a 16 in pulley at 1,750 rpm through a 20:1 reduction gives a belt speed of 367 fpm (1.86 m/s). It carries 300 TPH over a 120 ft run at 8 degrees, with a belt and moving parts mass of 12 lb/ft, a 0.03 friction factor, and 90% efficiency. The effective tension is 2,841 N (639 lbf) and the motor power is 5.88 kW (7.88 hp). The lesson: the tool takes imperial inputs and still reports kW and hp side by side, so a line specified in feet, inches, and tons per hour hands off cleanly to a motor catalog quoted in either system.
Three expert tips
Size to the worst case, not the average
The motor is set by the hardest moment the conveyor ever faces, not by a typical shift. A fully loaded belt that has to start from rest on an incline on the coldest morning, when the grease is stiff and the friction is highest, draws far more than the steady running number this tool reports. That is what the service factor is for. Take the calculated power, multiply by a service factor, commonly 1.2 to 1.5 depending on how heavy the starts are and how variable the load is, and then pick the next standard motor size up from that. It is cheaper to fit slightly more motor than to burn one out on a hard start six months in.
Belt speed is a tradeoff, so match it to the material
A faster belt carries the same tonnage in a smaller cross-section, so it can be narrower and lighter loaded, and the motor per metre of belt is smaller. The price is more spillage, more wear on the belt and idlers, and more noise, and fragile or dusty materials suffer at speed. Typical bulk belts run 1 to 3.5 m/s, about 200 to 700 fpm, with coarse or heavy material at the low end and free-running granular material able to go higher. Pick the speed from the material and the process first, then size the drive to it, rather than accepting whatever speed a stock gearbox happens to give.
Watch the decline and the starting tension
Two things the plain power number does not capture can still catch you out. A downhill conveyor can overhaul, meaning the load drives the belt faster than the motor intends, so it needs a brake or a holdback to stay controlled, and on a steep decline the drive becomes a generator that has to dissipate or return energy. Separately, at startup the takeup tension has to be high enough to keep the belt from slipping on the drive pulley while it accelerates the loaded run, and that peak tension can exceed the running tension by a wide margin. Size the motor from the power, but confirm the brake, the holdback, and the takeup as their own checks.
Limits of the method
This calculator gives a first estimate of belt speed, tension, and drive power, not a finished conveyor design. It models the two resistances that dominate most conveyors, the main friction along the belt and the lift up the incline, and it reports the effective tension and the motor power those produce. It is accurate enough to size a drive to within a service factor, to compare layouts, and to see where the power is going, which covers most planning and teaching needs.
What it leaves out is the secondary resistances and the selection work a full design carries. It does not add the resistance at the loading point where material is accelerated onto the belt, the drag of skirtboards and scrapers, the bearing and belt-flexure losses, or the effect of tilted idlers, which together can add roughly 10 to 20% to the main resistance on a long belt. It does not select the belt carcass and cover from the peak tension, size the takeup, or check the belt against startup, braking, and holdback loads. It does not size the gearbox or the coupling, and it treats efficiency as a single number rather than a chain of drive-train losses. Use the result as a sound starting point, then confirm the belt, the takeup, and the braking against a full CEMA or ISO 5048 design and with a conveyor manufacturer before you build or buy.
Where this calculator fits
It suits anyone who needs a defensible power number for a belt conveyor quickly. A plant or project engineer scoping a new line can get the belt speed and drive size before the detailed design starts. A maintenance engineer replacing a burned-out motor can check what the drive actually needs rather than copying the old nameplate. A buyer comparing supplier quotes can see whether a proposed motor is sensible for the duty. A student learning conveyor design can watch how speed, incline, and load move the tension and the power, and see the friction-versus-lift split that a formula alone hides.
Because it shows the working, it also builds intuition. You can watch a flat belt put all its power into friction, then tilt it and see the lift term take over. You can lengthen the run and see the friction grow while the lift stays put, or steepen the slope and see the reverse. You can flip to a decline and watch the power go negative, which is the moment the design problem changes from driving to braking. Use it to test a layout against a power budget, to compare a fast narrow belt against a slow wide one, or to sanity-check a number someone else produced, then take the winning option into a full design. These conveyor drives also tie into the wider plant: the motor power feeds the load studies in the Energy Management and Facility Infrastructure work, and the throughput links to the line-balancing and takt-time tools in Lean Production.
Common mistakes to avoid
The first mistake is sizing the motor on the running power alone and skipping the service factor, so the drive that is fine at steady speed stalls or overheats on a loaded start. The second is ignoring the incline, treating a climbing belt as if it were flat, which can miss the lift term that is often the largest part of the power. The third is the mirror image, forgetting that a decline needs a brake and a holdback rather than a bigger motor, and being surprised when the calculated power comes out negative.
A fourth mistake is confusing the effective tension with the belt working tension and trying to select the belt from the effective number, which is too low because it omits the takeup and slack-side tension. A fifth is mixing units, reading a pulley in inches against a length in metres, or a belt mass in lb/ft against a throughput in t/h, which throws the whole result off; enter one consistent system and let the tool show both. A sixth is treating this simplified figure as a final design and skipping the secondary resistances, the belt and takeup selection, and the startup and braking checks that a real conveyor needs. Apply a service factor, respect the incline sign, keep the units consistent, and confirm the belt and braking against a full design, and the drive you size here will hold up.
Frequently asked questions
What does this conveyor calculator do?
It works out how fast a conveyor belt runs and how much motor power that belt needs to move its load. You enter the drive pulley diameter, the motor speed, and the gear reduction to get the belt speed, or type the belt speed directly. Then you add the material throughput, the conveyor length, the incline, the belt and moving parts mass, the friction factor, and the drive efficiency. The tool returns the belt speed in m/s and fpm, the effective belt tension, the motor power split into a friction part and a lift part, and the belt loading in mass per metre. In the default case a 0.32 m pulley at 1,450 rpm through a 20:1 reduction gives 1.21 m/s (239 fpm), and moving 120 t/h up a 30 m belt at 10 degrees needs about 2.59 kW (3.47 hp).
What is the conveyor belt speed formula?
Belt speed is v = pi x D x N, where D is the drive pulley diameter and N is the pulley speed in revolutions per second. The pulley turns slower than the motor because a gearbox sits between them, so N equals the motor rpm divided by the gear reduction ratio, then divided by 60 to reach rev/s. With D in metres the speed is in m/s; with D in feet and N in rpm it comes out in fpm, and one fpm equals 0.00508 m/s. For example, a 0.32 m pulley driven at 1,450 rpm through a 20:1 reduction turns at 72.5 rpm, or about 1.208 rev/s, so the belt moves about 1.21 m/s, which is 239 fpm. If you already know the belt speed, you can enter it directly and skip the pulley math.
How do I calculate conveyor motor power?
Motor power comes from the effective belt tension and the belt speed. First find the effective tension, Te = f x g x L x (2 x mb + mm) x cos(a) + mm x g x H, which adds the main friction along the belt to the weight of the material lifted up the incline. The belt power is then Te times the belt speed, and the motor power is that belt power divided by the drive efficiency, since the gearbox and drive lose a little of it. To convert, 1 hp equals 0.7457 kW. In the default case the tension is 1,919 N, the belt runs at 1.21 m/s, and at 90% efficiency the motor power works out to about 2.59 kW (3.47 hp). This is the steady running power, so add a service factor before you pick the motor.
How does an incline change the motor power?
An incline turns on the lift term, which is often the biggest part of the power. On a flat belt the lift is zero and the motor only fights friction. Tilt the belt up and the motor has to raise every tonne through a real height, H = L x sin(a), and the lift power is the mass flow times gravity times that height. In the default case, 30 m at 10 degrees lifts the material about 5.21 m, and the lift power is about 1.70 kW against only 0.63 kW of friction, so lift is more than double friction. The main friction term shrinks very slightly on a slope because of a cos(a) factor, but that trim is tiny. On anything but a gentle grade, the height the material climbs sets the motor.
What happens on a decline, and what is regeneration?
On a downhill belt the lift term goes negative, because gravity now helps move the load instead of resisting it. If the slope is steep enough, the negative lift outweighs the positive friction and the total power comes out negative, which means the load is driving the belt rather than the motor driving the load. In the decline example, a belt running down a 12 degree slope has a lift share of about minus 4.25 kW against 1.38 kW of friction, so the net power is about minus 3.19 kW. A negative power number is the signal to size a brake or a regenerative drive that can absorb or return the energy, plus a holdback to stop the belt running away. A decline conveyor is a braking problem, not a bigger-motor problem.
What friction factor should I use?
The friction factor f is the artificial coefficient that rolls up the rolling resistance of the belt on its idlers and the internal flexing of the belt and material into one number. For a well-aligned belt on good idlers in normal conditions, a value around 0.02 is typical, and many designs use about 0.03 as a sound default that allows for real-world alignment and maintenance. Harder conditions, poor alignment, sticky material, cold temperatures, or worn idlers push it higher, toward 0.033 or more. The tool defaults to 0.03 and lets you change it. Because friction acts along the whole length of the belt, the factor matters most on long flat conveyors where friction dominates the power, and less on short steep ones where lift runs the show.
What is the belt plus moving parts mass?
It is the mass per metre of everything that moves with the belt except the material being carried: the belt itself on both the carrying and return runs, plus the rotating mass of the idler rolls that turn as the belt passes. It is entered as kg per metre in metric or lb per foot in imperial. In the effective tension formula it appears as 2 x mb, the factor of two counting the belt on both the top and return strands. A heavier belt and heavier idlers raise the main friction resistance, so a wide heavy-duty belt with steel-cord carcass has a much larger moving mass than a light fabric belt. If you do not know the exact figure, a manufacturer’s belt and idler data sheet gives the mass per metre for the belt width and construction you are using.
What service factor should I apply?
The service factor is the multiplier you put on the calculated running power before choosing a motor, to cover the harder conditions the drive meets in service. The biggest of these is starting a fully loaded belt from rest, especially on an incline and in cold weather, which draws well above the steady running power. A common range is 1.2 to 1.5, with the higher end for frequent heavy starts or a load that varies a lot. Take the tool’s power, multiply by the service factor, and then round up to the next standard motor size. In the default case, 2.59 kW with a 1.25 factor is about 3.24 kW, so you would fit the next standard motor above that. Fitting a little extra motor is cheaper than replacing one that fails on a hard start.
Can I use metric and imperial units?
Yes. In metric mode you enter the pulley diameter in metres, the length and rise in metres, the throughput in tonnes per hour, and the belt mass in kg per metre, and the power reads in kW with hp alongside. In imperial mode you enter the pulley in inches, the length and rise in feet, the throughput in tons per hour, and the belt mass in pounds per foot, and the belt speed reads in fpm with m/s alongside. The two headline results, belt speed and motor power, are always shown in both systems. The conversions are exact: 1 hp is 0.7457 kW, 1 fpm is 0.00508 m/s, and force converts at 4.448 N per lbf. Enter your numbers in whichever system your data arrives in, and read the answer in whichever system you report in.
How does capacity relate to belt speed?
Capacity, or throughput, and belt speed are linked through the belt loading, the mass of material on each metre of belt, mm = Q / v. For a fixed throughput, a faster belt spreads the same tonnage thinner, so it carries less mass per metre, while a slower belt piles it deeper. That means you can move a given tonnage with a narrower, faster belt or a wider, slower one. Speed does not change the total tonnage moved per second or the total lift, since the same mass still has to be raised and dragged along, but it does change the cross-section on the belt and the loading per metre. In the default case, 120 t/h at 1.21 m/s gives about 27.4 kg per metre. Very high loading warns the belt is nearly full and likely to spill.
Why does the lift term dominate on an incline?
Because lifting mass against gravity is expensive, and friction is comparatively cheap. The lift power is the mass flow times gravity times the height climbed, and gravity is a large, constant force, so raising 120 tonnes an hour even a few metres takes real power. The friction power, by contrast, is only the small friction factor times the weight of the moving parts, spread over the length, so it is a fraction of the full weight rather than the whole of it. In the default case the lift is 1.70 kW against 0.63 kW of friction, more than double, even at a modest 10 degrees. The steeper the belt, the more the lift term grows, because the height climbed rises with the sine of the angle, while the friction barely changes. This is why cutting the climb saves far more power than chasing a slicker idler.
What is effective belt tension?
The effective belt tension, Te, is the net force the drive pulley has to apply at the belt line to keep the loaded belt moving. It is the sum of the main friction resistance along the belt and the lift resistance of the material up the incline, and it is what you multiply by the belt speed to get the belt power. In the default case it is 1,919 N, or 431 lbf. It is not the same as the total working tension in the belt, which is higher because the belt also carries a takeup tension to stop it slipping on the drive pulley and sagging between idlers. The power calculation uses the effective tension only. Selecting the belt carcass and the takeup uses the peak tension, which is the effective tension plus the slack-side tension, so belt strength is a separate check after the drive is sized.
How does this compare with a full CEMA design?
This tool uses the simplified main-resistance plus lift model from ISO 5048 and DIN 22101, which shares its core physics with CEMA’s “Belt Conveyors for Bulk Materials” (7th edition). It captures the two largest resistances, the belt friction and the lift, which dominate the power on most conveyors, so it sizes a drive well to within a service factor. A full CEMA or ISO design goes further: it adds the secondary resistances at the loading point, the skirtboards and scrapers, the pulley bearings, and the belt flexure and idler tilt, which can add roughly 10 to 20% on a long belt. It also selects the belt carcass and cover from the peak tension, sizes the takeup, and checks startup, braking, and holdback loads. Use this tool for a fast, sound estimate, then confirm the belt and takeup selection with a full design.
Is the calculator free, and does it store my data?
Yes, the tool is free with no sign-up, and every calculation runs in your browser. The numbers you enter are never sent to a server, stored, or shared. You can download a clean PDF or export a CSV of the result, and share a summary on WhatsApp, all from the numbers computed on your own device. The calculator is for planning and education, so confirm any figure that informs a capital purchase, a motor selection, or a safety decision with a qualified engineer and a full conveyor design against CEMA or ISO 5048 for your site.
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Sources, disclaimer, and editorial transparency
The belt speed relation, the material loading formula, the simplified effective-tension model with its main-resistance and lift terms, and the conversion between belt power and motor power described here follow recognized conveyor engineering references, including the resistance-based method of ISO 5048 and DIN 22101 and the rigorous US method in CEMA’s “Belt Conveyors for Bulk Materials” (7th edition). Belt speed is treated as v = pi x D x N, the effective tension as f x g x L x (2 x mb + mm) x cos(a) + mm x g x H, and the motor power as Te x v divided by the drive efficiency. This calculator uses the simplified main-resistance plus lift form and does not add the secondary resistances, belt and takeup selection, or startup and braking checks that a full design carries. This tool and guide are built and reviewed by the OpsCalculators team; see our Editorial Policy for how each tool is researched, built, and tested.
Results are accurate estimates for planning and education, not a substitute for a full conveyor design or an engineering review. The method models only the main friction and the lift, treats efficiency as a single figure, and reports the effective tension rather than the peak belt tension, so it does not size the belt, the takeup, the brake, or the holdback, any of which can change the equipment required. Confirm the drive, the belt selection, and the braking against a full CEMA or ISO 5048 design and with a qualified engineer or a conveyor manufacturer before you build or buy. See our full Disclaimer. OpsCalculators.com is operated by MAFHH INTERNATIONAL LTD. Your inputs are processed in your browser and are never stored; see our Privacy Policy.