Three calculators for the drive arithmetic a millwright does every week and the 433A exam tests under mechanical power transmission. The first takes tooth counts and an input speed and gives the ratio, the output speed, whether the train is a reduction or an overdrive, the ideal torque multiplication and, with a second pair of gears, the total ratio of a compound train; it works for chain sprockets as well. The second solves D1 × N1 = D2 × N2 for any one of the four pulley values and gives the belt speed in ft/min and m/s. The third finds the pitch length of an open belt from two pulley diameters and a centre distance, or works backwards from a belt length to the centre distance. Defaults are loaded (20 and 60 teeth at 1,750 RPM gives 583.3 RPM; a 4 in sheave driving a 12 in sheave gives 583.3 RPM; 4 and 12 in pulleys on 24 in centres take a 73.80 in belt), and the working is printed under each result in the form an exam answer takes.
Gear ratio and output speed
Pulley (sheave) speed
Belt length and centre distance
01Gear ratio basics: teeth, speed and torque
When two gears mesh, every tooth on one passes a tooth on the other, so in one full turn of a 20-tooth pinion exactly 20 teeth of the mating gear go by. If that gear has 60 teeth it has made a third of a turn. That is the whole of gear ratio arithmetic: the ratio is the driven tooth count divided by the driver tooth count, and speed goes the other way. In symbols, with N for speed in RPM and T for teeth,
Ndriver × Tdriver = Ndriven × Tdriven, so Ndriven = Ndriver × Tdriver / Tdriven.
A 20-tooth pinion turning 1,750 RPM drives a 60-tooth gear at 1,750 × 20 / 60 = 583.3 RPM. The ratio is 60 / 20 = 3, written 3:1. Because the ratio is greater than 1 the output is slower: a reduction. When the driver has more teeth than the driven gear the ratio is less than 1 and the output is faster: an overdrive or step-up. Ratios of 1 (equal tooth counts) change direction and shaft position without changing speed.
Torque follows the ratio in the opposite direction to speed. Power in equals power out, and power is torque times angular speed, so if the speed comes down by a factor of 3 the torque goes up by 3. A 3:1 reduction fed with 50 N·m delivers 150 N·m at the output shaft, less whatever the meshes and bearings lose, which for a well-made spur or helical stage is a few per cent. The calculator reports the ideal figure and says so.
Two things that do not affect the ratio: the tooth size (pitch), and any idlers between the gears. Meshing gears must share the same pitch, so the tooth counts alone set the ratio. An idler is a gear placed between the driver and the driven gear; it turns at its own speed set by its own tooth count, but it passes the driver's tooth rate straight through, so the ratio between the first and last gears is unchanged. What the idler does change is direction. Two external gears in mesh turn in opposite directions; add one idler and the driven gear turns the same way as the driver. Count the meshes: an odd number reverses, an even number does not. Internal (ring) gears are the exception, since a pinion inside a ring gear turns the same way as the ring.
The other quantity the exam uses is the pitch diameter. Gears of the same pitch have pitch diameters proportional to their tooth counts, so the ratio can be found from pitch diameters as well as from teeth. That is also why the pulley formula in the next calculator has the same shape: a belt drive is a pair of pitch circles with a belt instead of teeth.
02Compound trains: ratios multiply
A single pair of gears is limited in how much reduction it can give before the big gear becomes impractical. A compound train fixes that by putting two gears on one intermediate shaft: the first driven gear and the second driver turn together at the same speed. Each stage has its own ratio, and the total ratio is the product of the stage ratios, not the sum.
Take an 18-tooth pinion driving a 54-tooth gear (stage 1, ratio 3.0:1). On the same shaft as the 54-tooth gear is a 20-tooth gear driving a 50-tooth gear (stage 2, ratio 2.5:1). The total is 3.0 × 2.5 = 7.5:1, so a 1,200 RPM input comes out at 1,200 / 7.5 = 160 RPM. Written as one fraction, the total ratio is the product of all the driven tooth counts over the product of all the driver tooth counts: (54 × 50) / (18 × 20) = 2,700 / 360 = 7.5.
The compound rule is the one that catches people. Two stages of 3:1 give 9:1, not 6:1. Three stages of 4:1 give 64:1. A gearbox nameplate ratio is the product of everything inside it. Torque multiplies the same way, so the ideal output torque of the train above is 7.5 times the input torque, and the intermediate shaft carries 3 times the input torque, which is why intermediate shafts and their bearings are heavier than the input side.
Direction in a compound train is counted mesh by mesh, exactly as for a simple train. The two meshes in the example above give two reversals, so the output turns the same way as the input. Gears that share a shaft do not count as a mesh.
To use the calculator for a compound train, enter the stage 2 tooth counts in the two optional fields. Leave both blank for a single pair. The working shows each stage ratio and the product.
03Chains and sprockets: the same rules
A roller chain drive behaves like a pair of gears with a very long idler between them. The chain engages one tooth on each sprocket per pitch of chain, so the ratio is the driven sprocket teeth over the driver sprocket teeth, speed is inverse to teeth, and torque is proportional to the ratio. The tooth-count formula above works unchanged, which is why the gear calculator on this page is labelled for sprockets too.
One difference is direction. Gears in direct mesh reverse; a chain, like a belt, carries the driver's motion round to the far side of the driven sprocket, so both sprockets turn the same way. A crossed chain is not a thing, so if the driven shaft has to turn the other way the answer is a gear, not a sprocket. The other difference is that chain does not slip, so the calculated speed is the real speed, unlike a V-belt. A 17-tooth driver sprocket at 1,160 RPM driving a 51-tooth sprocket gives 3:1 and 1,160 × 17 / 51 = 386.7 RPM, and the driven shaft really does turn at 386.7 RPM.
Chain drives have their own arithmetic for chain length in pitches and for centre distance, which uses the sprocket tooth counts rather than diameters and rounds to an even number of pitches to avoid an offset link. The belt-length calculator on this page is not the tool for that; it is the open-belt formula for pulleys.
04Pulley and belt drives: diameter ratio, pitch diameter, slip and belt speed
A belt runs at one speed round the whole drive, and that speed is the surface speed of each pulley at the point where the belt is. Surface speed is π × diameter × RPM, and π cancels, so D1 × N1 = D2 × N2: driver diameter times driver speed equals driven diameter times driven speed. Any one of the four can be found from the other three, which is what the second calculator does. Leave the value you want blank and it solves for it.
The default is a 4 in motor sheave at 1,750 RPM driving a 12 in sheave: N2 = 4 × 1,750 / 12 = 583.3 RPM, a 3:1 reduction. The ratio is D2 / D1, and a larger driven pulley means a slower driven shaft, exactly as more teeth on a driven gear does. Torque scales with the ratio in the same way.
The diameter that belongs in the formula is the pitch diameter, the diameter of the circle the belt's neutral axis runs on. For a flat belt that is close to the outside diameter of the pulley plus the belt thickness. For a V-belt sheave it is noticeably less than the outside diameter, because the belt sits down in the groove: on a small sheave the difference can be a quarter of an inch or more, which is enough to change the answer by a few per cent. Catalogues list the pitch (or datum) diameter, and that is the number to use. Using the outside diameter is one of the standard exam traps.
Belt speed matters on its own.In inches, belt speed in ft/min = π × D × N / 12; in millimetres, belt speed in m/s = π × D × N / 60,000. The default drive runs at π × 4 × 1,750 / 12 = 1,833 ft/min, which is 9.31 m/s. This is also the surface-speed formula on the 433A formula sheet, and the reason the calculator prints both units.
Slip is the last correction. A V-belt transmits power by friction, and under load it creeps back on the driver and forward on the driven pulley, so the driven shaft turns slightly slower than D1 × N1 / D2 predicts. One to three per cent is typical for a properly tensioned drive; a loose or glazed belt slips more. A timing (synchronous) belt has teeth and does not slip, so for it, as for chain, the formula is exact using the pitch diameters or the tooth counts. Exam questions ignore slip unless they state it, in which case reduce the driven speed by the stated percentage.
05Belt length and centre distance: the formula, forwards and backwards
An open belt (both pulleys turning the same way) wraps a little more than half of the large pulley and a little less than half of the small one, with two straight runs between. The exact length needs the wrap angles, but the standard approximation, which is what catalogues, textbooks and this calculator use, is
L = 2C + π(D1 + D2) / 2 + (D2 − D1)² / (4C)
where L is the belt pitch length, C is the centre distance, and D1 and D2 are the pitch diameters of the small and large pulleys. The first term is the two straight runs, the second is the two half-circumferences, and the third corrects for the runs being slanted and the wrap being unequal. The third term is small whenever C is comfortably larger than the pulleys, which is the normal case, and the approximation is well within the tolerance of a catalogue belt size.
Worked forwards with the calculator's defaults, D1 = 4 in, D2 = 12 in and C = 24 in: 2 × 24 = 48; π × (4 + 12) / 2 = 25.133; (12 − 4)² / (4 × 24) = 64 / 96 = 0.667. Added together, L = 73.80 in. That is the pitch length. Belt catalogues list nominal or standard sizes, so in practice you choose the nearest listed belt and let the motor base or take-up absorb the difference in C.
Going backwards, from a known belt to the centre distance it needs, takes a little algebra. Multiply the formula by 4C and gather terms:
4CL = 8C² + 2πC(D1 + D2) + (D2 − D1)², so 8C² − bC + (D2 − D1)² = 0 where b = 4L − 2π(D1 + D2).
That is a quadratic in C, and the quadratic formula gives
C = [ b + √( b² − 32(D2 − D1)² ) ] / 16
The plus sign is the root that matters; the minus root is the centre distance a crossed belt of the same length would need, or is negative. If the quantity under the root is negative, no centre distance works: the belt is shorter than the pulleys, and the calculator says so. Some references write the same result as C = [b′ + √(b′² − 2(D2 − D1)²)] / 4 with b′ = L − π(D1 + D2) / 2. That is the same formula with the constants moved around (b′ is b / 4); check by putting the result back into the forward formula, which is what the calculator's check tile does. Example 4 below works a full reverse case.
Two rules of thumb sit alongside the formula. The centre distance must be more than half the sum of the diameters or the pulleys touch, and for V-belts it is usually kept between the large pulley diameter and about three times the sum of the two diameters, so that the small pulley gets enough wrap and the belt does not flap on a long run. The calculator flags a C below either lower bound.
06Worked examples in exam style
Each of these is written the way a Red Seal question is: a situation, the numbers, and one thing to find. Work them on paper first. The 433A exam gives about 1.8 minutes a question, and questions like these should take less.
Example 1: simple gear train, speed and torque
Question. A 24-tooth pinion on a motor shaft turning 1,750 RPM drives a 96-tooth spur gear. The motor delivers 20 N·m. What is the ratio, the output speed and the ideal output torque?
Solution. Ratio = driven / driver = 96 / 24 = 4:1, a reduction. Output speed = 1,750 × 24 / 96 = 437.5 RPM. Output torque = 20 × 4 = 80 N·m, ignoring losses. The driven gear turns opposite to the pinion because there is one external mesh.
Example 2: compound train
Question. A gearbox has an 18-tooth input pinion meshing with a 54-tooth gear. On the same shaft as the 54-tooth gear, a 20-tooth gear drives a 50-tooth output gear. The input shaft turns 1,200 RPM. What is the total ratio and the output speed?
Solution. Stage 1: 54 / 18 = 3.0:1. Stage 2: 50 / 20 = 2.5:1. Total = 3.0 × 2.5 = 7.5:1. Output speed = 1,200 / 7.5 = 160 RPM. If you added the stages instead (3.0 + 2.5 = 5.5) you would get 218.2 RPM, which is one of the wrong options a question writer would offer.
Example 3: sizing a driven sheave and checking belt speed
Question. A fan must run at 700 RPM. The motor turns 1,750 RPM and carries a 6 in pitch diameter sheave. What pitch diameter is needed on the fan sheave, and what is the belt speed in ft/min and m/s? Ignore slip.
Solution. D2 = D1 × N1 / N2 = 6 × 1,750 / 700 = 15 in, a 2.5:1 reduction. Belt speed = π × 6 × 1,750 / 12 = 2,749 ft/min, which is 2,749 × 0.3048 / 60 = 14.0 m/s. With 2% slip the fan would run at about 686 RPM, which is why a fan drive is often sized a little fast and trimmed with a variable-pitch motor sheave.
Example 4: centre distance from a known belt (the reverse solve)
Question. A drive uses a 5 in and a 10 in pitch diameter sheave and a belt with a 60 in pitch length. At what centre distance should the motor be set?
Solution. b = 4L − 2π(D1 + D2) = 4 × 60 − 2π × 15 = 240 − 94.248 = 145.752. Then b² − 32(D2 − D1)² = 21,243.71 − 32 × 25 = 20,443.71, and its square root is 142.982. C = (145.752 + 142.982) / 16 = 18.05 in. Check it forwards: 2 × 18.05 + π × 15 / 2 + 25 / (4 × 18.05) = 36.09 + 23.562 + 0.346 = 60.00 in, which matches. On the job you would set the base a little short of 18.0 in, fit the belt, then tension it out to about 18.0 in.
Example 5: chain drive
Question. A 17-tooth drive sprocket on a 1,160 RPM motor drives a 51-tooth sprocket on a conveyor head shaft. Find the ratio and the head shaft speed.
Solution. Ratio = 51 / 17 = 3:1. Head shaft speed = 1,160 × 17 / 51 = 386.7 RPM. Both sprockets turn the same way, and there is no slip to allow for.
Enter any of these into the calculators to see the same working printed step by step.
07Common mistakes
- Inverting the ratio. Ratio is driven over driver, and speed is driver over driven. If a 20-tooth pinion drives a 60-tooth gear the ratio is 3:1 and the speed comes down; getting 5,250 RPM out of a 1,750 RPM motor through that pair is the sign the fraction is upside down. Sanity-check: more teeth on the driven side means slower.
- Using outside diameter on V-belt sheaves. The belt runs on the pitch diameter, which for a V-belt sheave is smaller than the outside diameter. On a small sheave the difference is a few per cent of the speed, enough to separate two multiple-choice options.
- Adding stage ratios instead of multiplying them. Two stages of 3:1 and 2.5:1 make 7.5:1, not 5.5:1. The total ratio of a compound train is the product of the driven tooth counts over the product of the driver tooth counts.
- Mixing RPM and rad/s. Ratio formulas work in any speed unit as long as both sides use the same one. But if a question gives angular speed in rad/s and asks for RPM, or gives power and asks for torque, convert: 1 RPM = 2π/60 = 0.1047 rad/s, and P (W) = T (N·m) × ω (rad/s). Dropping the 2π/60 changes the answer by a factor of about 9.5.
- Expecting an idler to change the ratio. An idler gear changes direction and fills space; the ratio between first and last gear depends only on those two tooth counts. Idler pulleys and chain tensioners likewise change wrap and tension, not speed.
- Forgetting direction. Two gears in external mesh turn opposite ways; belts and chains turn the driven pulley the same way as the driver. A question that describes the motor turning clockwise and asks which way the output turns is a direction question, not a ratio question.
- Calling the belt formula exact. It is an approximation that is good when the centre distance is comfortably larger than the pulleys. Catalogue belts come in nominal sizes anyway, so pick the nearest size and adjust C with the take-up.
- Applying slip where there is none. V-belts slip one to three per cent under load; timing belts, chains and gears do not. Do not deduct slip from a chain drive speed, and do not ignore it if a V-belt question states a slip figure.
08What the 433A exam expects about drives
Everything on this page sits in the largest block of the Industrial Mechanic (Millwright) exam. Our 433A exam guide sets out the Red Seal Program's breakdown: Block C, mechanical power transmission components and systems, is 32 of the 135 questions, and within it the 2017 standard weights gear systems at 16 per cent of the block and chain and belt drive systems at 15 per cent, with couplings, clutches and brakes, prime movers, shaft alignment, and shafts, bearings and seals making up the rest. Each of those equipment tasks has the same four sub-tasks: installs, diagnoses, maintains and repairs. Ratio and speed arithmetic belongs to installs (selecting and sizing components) and to diagnoses (a driven shaft running at the wrong speed).
The program describes the 433A question mix as 30 to 40 per cent knowledge and recall, 40 to 50 per cent procedural and application, and 20 to 30 per cent critical thinking. A drive question can be any of the three: a recall question on what pitch diameter or an idler is, a procedural question like examples 1 to 5 above, or a critical thinking question that gives you a symptom, say a fan delivering less air than the sheet calls for, and expects you to work back to a slipping belt or a sheave of the wrong size.
On the formula sheet, the guide's summary of the one-page sheet provided with the exam lists rpm from cutting speed and diameter, surface speed, and gear pass frequency, alongside the rigging, area, volume and expansion formulas. The surface-speed formula is the belt-speed formula on this page. The guide's list does not include a ratio formula or the belt-length formula, so treat those as ones you carry in your head: driven over driver, stages multiply, D1 × N1 = D2 × N2, and L = 2C + π(D1 + D2)/2 + (D2 − D1)²/(4C). A calculator is provided if the exam needs one; you cannot bring your own.
Gear pass frequency being on the sheet is a hint about how Block F (vibration analysis) connects to Block C: gear mesh frequency is shaft speed times tooth count, which is the same tooth arithmetic in a different coat. If you can find the output speed of a compound train, you can find the mesh frequency of each stage in it.
Practise until the sequence is automatic: identify driver and driven, write the ratio as driven over driver, multiply stages, invert for speed, and check the direction and the units. The free 433A questions include power transmission items in the same format as the exam.
09Sources
- Red Seal Program: Industrial Mechanic (Millwright) trade page the trade the 433A exam certifies, the 2017 and 2026 occupational standards, and the link to the exam information page with the block breakdown, formula sheet and acronym list.
- The Engineering ToolBox: Belt Transmissions, Speed and Length of Belts the open-belt length formula L = πd1/2 + πd2/2 + 2C + (d2 − d1)²/(4C) and the pulley speed relation used by the calculators.
- TicketPrep: 433A Industrial Mechanic (Millwright) Red Seal exam guide the block and task weightings, the question-type mix, the formula sheet contents and what is provided in the exam room.
10Questions people ask
- How do you calculate a gear ratio?
- Divide the number of teeth on the driven gear by the number of teeth on the driver. A 20-tooth pinion driving a 60-tooth gear is 60 / 20 = 3, written 3:1. Output speed is the input speed divided by the ratio, so 1,750 RPM in gives 583.3 RPM out, and the ideal output torque is the input torque multiplied by the ratio. Chain sprockets use the same rule.
- How do you find the total ratio of a compound gear train?
- Multiply the stage ratios together. If the first pair is 3:1 and the second pair is 2.5:1, the total is 3 x 2.5 = 7.5:1. Equivalently, multiply all the driven tooth counts together and divide by the product of all the driver tooth counts. Adding the stage ratios is the common mistake.
- What is the formula for pulley speed?
- D1 x N1 = D2 x N2, where D is pitch diameter and N is RPM. Driven speed = driver diameter x driver speed / driven diameter. A 4 in sheave at 1,750 RPM driving a 12 in sheave gives 4 x 1,750 / 12 = 583.3 RPM. Use pitch diameters for V-belt sheaves, and remember a V-belt slips about 1 to 3% under load, so the real driven speed is slightly lower.
- How do you calculate belt length from pulley sizes and centre distance?
- Use the open-belt approximation L = 2C + pi(D1 + D2)/2 + (D2 - D1)^2/(4C), with C the centre distance and D1 and D2 the pitch diameters. For 4 in and 12 in pulleys on 24 in centres, L = 48 + 25.13 + 0.67 = 73.80 in. This is the pitch length; belt catalogues list nominal sizes, so choose the nearest and adjust the centre distance with the take-up.
- Does an idler gear change the gear ratio?
- No. An idler between the driver and the driven gear changes the direction of rotation and bridges the gap, but the ratio between the first and last gears depends only on their two tooth counts. Two external gears in mesh turn opposite ways; adding one idler makes the driven gear turn the same way as the driver.
11Practice for this exam
Drive ratios and speeds are a few questions among the 32 on mechanical power transmission in the 433A exam, and the same gear arithmetic turns up in the automotive trades. TicketPrep 433A and 310S practice tests follow the Red Seal Program's block-by-block breakdown, and every answer is explained, including why each wrong option is wrong. You can try free sample questions with no account.
Ten free questions per trade, no account: free practice questions. How our questions are written and checked: how our questions are made.