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82 Cards in this Set
- Front
- Back
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Formula of tooth thickness of (14.5°) |
Tooth thickness = 1.5708/DP(diametral pitch) |
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Speed and teeth relation formula |
T1N1 = T2N2 |
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Formula of circular pitch(spur gears) |
Pc = πD/Tooth(total) Pc = π ( Module) |
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Formula of lead triple |
Lead = 3(Pc) |
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Formula of center distance? |
C = T / 2Dₚ |
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Formula for Dₚ ( diametral pitch), tooth/mm |
Dₚ = T / D Dₚ = 25.4 / M Dₚ = 1 / M Dₚ = π / Pc |
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Formula for Module |
M = D(pitch diameter) / T M = (25.4) / Dₚ |
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Formula for when two gears turning in opposite direction ( spur gears) |
C = ( D₁+D₂)/2 |
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Formula for when two gears turning in the same direction ( spur gears) |
C = (D₂ - D₁)/2 |
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Formula for pitch line velocity |
V = π D N |
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Formula for total load, tangential load and separation load relation. |
Fₙ = sqrt. (Fₜ²-Fᵣ²) Fₙ = Fₜ / cos θ |
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GEAR TOOTH PROPORTION TABLE |
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Formula for normal diametral pitch |
Pₙ = P / cos x where: Pₙ = diametral pitch P = diametral pitch |
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Formula for normal pressure angle(spur gears) |
Tanθₙ = Tanθ tanx |
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Formula for axial load or end trust(spur gears) |
Fₐ = Fₜ tanx |
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Formula for virtual number of teeth |
Nᵥ = N / cos³ x |
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Formula for strength of helical gears |
F = (SwfY)/P = (78) / [78 + sqrt.V] |
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Dynamic load on helical gears |
Fd = Ft + [ 0.05V(Cf cos²x + Ft) cos x ] / [0.05V +(Cf cos²x + Ft) ^ 1/2 |
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Formula for worm gear Nomenclature |
P = πD / T |
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Formula for velocity ratio |
( no. of threads on gear) / ( no. of threads on worm gear) |
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formula for strength of worm gear |
F = SwPfY ( 1200 / (1200 + V ) where: Sw = safe stress P = circular pitch f = face width Y = form factor |
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Formula for efficiency of worm gear |
e = [tan x (cosθₙ - f tan x )] / cosθₙ tanx + f where : f = coefficient of friction |
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Formula for clearance |
c = d - a c = W - Wf c = 0.25/ diametral pitch
where: a = addendum W = whole depth Wf = working depth |
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Formula for addendum |
a = 1/Pd |
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formula for Dedendum A. For 14.5° and 22.5 ° B. For 20° and 25° |
A. 1.157/Pd B. 1.25/Pd |
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Formula for outside diameter(spur gears) A. For Pinion B. For Gear |
A. Dₒ = Dₚ + 2a = (Nₚ + 2)/Pd B. Dₒ = [(Np + 2)/Pd] = Dg + 2a Dₚ = pinion diameter Np = no. of teeth in pinion Dg = gear diametral |
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Formula for Root diameter( spur gears) A. for pinion B. For gear |
A. D = Dₚ - 2d B. D = Dg - 2d
where: Dₚ = pinion diameter a = addendum d = dedendum |
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Formula for whole depth ( spur gears) |
W = a +d W = Wr + c W = 2.25 / diametral pitch
where: Wᵣ = Working depth c =clearance |
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Formula for working depth ( spur gears) |
Wᵣ = 2a = 2 / diametral pitch |
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Formula for Tooth thickness A. for backlash is zero B. for backlash is not zero |
T = Pc/2 T = ( Pc - B )/2 |
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Formula for Tooth space (S) |
S = T(tooth thickness) + B |
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Formula for Base circle diameter A. For Pinion B. For Gear |
A. D = Dₚ cosθ B. D = D cosθ |
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Formula for Backlash |
B = 0.03 / Pd to 0.04/Pd |
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Formula for face width ( spur gears) |
2.5 Pc < b < 4pc (8/Pd) < b < (12.5)/Pd |
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Formula for beam fatigue strength of the teeth (spur gears) |
Fₛ = (S b Y) / Kf Pd where: S = endurance strength b = face width Kf = stress reduction factor Pd = diametral pitch |
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Formula for dynamic load( continuous service) |
Fd = Fₜ +[ (0.05 V (bC'+Fₜ) ] 0.05V + sqrt.(bC'+Fₜ) where: Fd = dynamic load, lb Fₜ = tangential load b= face width C' = factors which depends upon the elasticity of material |
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Formula for dynamic load( intermittent service) A. For commercial cut V ≤ 2000 fpm B. For carefully cut 200 < V ≤ 4000 fpm C. For precision cut V > 4000 fpm |
A. Fd = [(600 +V) / 1200] Fₜ B. Fd = [ (1200 +V)/1200] Fₜ C. Fd = [50+sqrt.(V)] / 50 |
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Formula for wear load |
Fw = Dp b Q kg Dp = pinion diameter b =face width Q = ratio factor kg = wear factor |
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Formula for checking failure based on fatigue |
Nsf = Fs / Fd where: Nsf = service factor Fs = beam fatigue strength Fd = dynamic load |
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Note for service factor |
Nsf = 1.0 to 1.25 for uniform load without shock Nsf = 1.25 to 1.50 for medium shock Nsf = 1.50 to 1.75 for moderately heavy shock Nsf = 1.75 to 2.0 for heavy shock |
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Formula for tangential force, radial force, axial force, normal pressure angle & axial pitch ( helical gears) |
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Formula for normal diametral pitch, normal circular pitch, lead & beam fatigue strength of the teeth(helical gears) |
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Formula for dynamic load, wear load & formative number of teeth(helical gears) |
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Formula for lead angle (worm gears) |
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Formula for seprating force & tangential force on the worm( worm gears) |
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Formula for efficiency of the worm gear, face width , outside diameter and worm diameter |
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Formula for worm gears A. Beam fatigue strength of the teeth B. Dynamic loaf for worm gear teeth C. Wear load D. Thermal capacity E. Normal pressure |
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A. What is bevel gears B. Typed of bevel gears |
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Formula for lead ( worm gears) |
L = Nt Pa Where : Nt = number of threads Pa = axial pitch |
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Formulas for bevel gears A. Cutting or root angle B. Face angle C. Pitch angle D. Face width |
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Formulas for bevel gears A. Length of cone B. Working depth C. Total depth |
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Formulas for bevel gears A. Strength of straight bevel gears B. Dynamic load |
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(prob.1) Compute the tooth thickness of 14.5° spur gear with diameter pitch of 5. |
0.31416 inch |
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(prob.2) A 36 tooth pinion with a turning speed of 300 rpm drives 120 tooth gear of 14.5° involute full depth pressure angle. What would be the speed of the driven gear. |
90 rpm |
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(prob.3) A triple thread worm has a pitch diameter of 3 inches. The wheel has 25 teeth and a pitch diameter of 5 inches. Material for both the worm and the wheel is of phosphor bronze. Compute the helix angle ( tan x). |
0.20 |
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(prob.4) A pair of gear/pinion of 42 tooth and 18 tooth with a diametral pitch of 0.7874 teeth/cm and the addendum is 0.80/p and the addendum 1/p. The gear pressure angle is 20°. Compute the center distance of the pair of gears in meters. |
0.381 m |
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(prob.5) A helical hear having a 14.5° normal pressure angle and transverse diametral pitch of 2.3622 per cm. The helix angle is at 45° and has 8 teeth. Compute the transverse pressure angle in degrees. |
20.1° |
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(prob.6) Two idlers of 28T and 26T are introduced between the 24T pinion with a turning speed of 400 rpm driving a final 96T gear. What would be the final speed of the driven and its direction relative to the driving gear rotation? |
100 rpm ( opposite direction) |
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(prob.7) A spur pinion rotates at 1800 rpm and transmits to a mating gear 30 HP. The pitch diameter is 4" and the pressure angle 14.5°. Determine the tangential load in lbs. |
525 lbs. |
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(prob.8) Compute the pitch angle of a bevel gear given the pinion's number of teeth of 14 and 42 teeth on the gear. |
18.4° |
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(prob.9) Compute the circular pitch of a pair having a ratio of 4 and a center distance of 10.23. Each gear has 72 teeth and pinion has 18 teeth. |
0.7854 inch |
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(prob.10) The tooth thickness of a gear is 0.5 inch and its circular pitch is 1.0 inch. Calculate the dedendum of the gear. |
0.3979" |
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(prob.11) An internal gear is set up with a 5 inch diameter pinion and center distance of 18 inches. Find the diameter of the internal gear. |
41" |
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(prob.12) The minimum whole depth of spur gear of 14.5° involute type with diameter pitch of 24 and circular pitch of 0.1309. |
0.08987 |
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(prob.13) Determine the pitch diameter of a gear with 28 teeth, 4 diametral pitch. |
7 inch |
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(prob.14) The parallel shafts have a center distance of 15 inch. One of the shaft carries a 40 tooth, 2 diametral pitch gear which drives a gear on the other shaft at a speed of 150 rpm. How fast is the 40-tooth turning? |
75 rpm |
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(prob.15) A pair of meshing gears has a diametral pitch of 10, a center distance of 2.6 inches, and velocity ratio 1.6. Determine the number of teeth of smaller gear. |
20 teeth |
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(prob.16) A spur gear 20 degrees full-depth involute has an outside diameter of 195 mm and a module of 6.5. Determine the number of teeth |
28 Teeth |
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(prob.17) What is the pitch diameter of a 40-tootj gear having a circular pitch of 1.5708 inch? |
20 inch |
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(prob.18) How many revolutions per minute is a spur gear turning if it has 28 teeth, a circular pitch of 0.7854 inch and a pitch line velocity of 12 ft/sec? |
392.88 rpm |
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(prob.19) A gear turning 300 rpm has a diametral pitch of 8. If there are 40 teeth on the gear, find the pitch line velocity of gear. |
6.54 ft/s |
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(prob.20) Two gears meshing each other has a center distance of 200 mm. The pinion has 20 teeth and gear has 60 teeth. Find the module of meshing gears. |
M = 5 |
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(prob.21) A gear has a tooth thickness of 0.1308 inch. Find the addendum of the gear at 14.5° involute. |
2.12 mm |
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(prob.22) A gear has an addendum of 0.10 inch. What is the dedendum of the gear at 20° pressure involute. |
0.125 inch |
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(prob.23) A gear is use to transmit 20 kW at 600 rpm to a driven gear. The pinion has a pitch diameter of 200 mm. If pressure angle is 14.5° , find the load that tends to separate the two gears. |
0.823 kN |
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(prob.24) The pressure angle of the 12 inch diameter gear is 20°. The total load of the gear is 5 kN and is turning at 750 rpm. Find the power delivered by the gear. |
56.23 kW |
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(prob.25) Three gears meshing each other has a driving power of 30 kW with 900 rpm. The speed ratio is 1:3:5 and each meshing gear has an efficiency of 96%. FInd the torque developed of the driven gear, kn.m |
4.40 kN-m |
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formula for gear formulas from machineries handbook 1. Gear set center distance 2. Circular pitch for given center distance and ratio |
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formula for circular thickness of tooth when outside diameter has been enlarged |
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formula for chordal thickness of tooth when outside diameter is special |
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formula for chordal addendum |
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Formula for chordal thickness of tooth when outside diameter is standard |
t = D sin(90°/Number of teeth) |
