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Gear Ratio worksheet

Grades 6–7 · CCSS 6.RP.A.3, 6.RP.A.1

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Gear set: Two-stage compound train
- Driving gear teeth (input): 12
- Driven gear teeth: 38
- Second-stage driving gear teeth: 12
- Second-stage driven gear teeth (output): 30
- Input speed: 1,100 rpm
- Input torque: 7 N·m
- Efficiency per mesh: 90%
Find: Gear ratio (driven ÷ driving)
Answer
2.
Problem 2.
- Gear set: Two-stage compound train
- Driving gear teeth (input): 13
- Driven gear teeth: 45
- Second-stage driving gear teeth: 8
- Second-stage driven gear teeth (output): 39
- Input speed: 1,000 rpm
- Input torque: 12 N·m
- Efficiency per mesh: 85%
Find: Gear ratio (driven ÷ driving)
Answer
3.
Problem 3.
- Gear set: Two-stage compound train
- Driving gear teeth (input): 15
- Driven gear teeth: 40
- Second-stage driving gear teeth: 7
- Second-stage driven gear teeth (output): 40
- Input speed: 1,300 rpm
- Input torque: 10 N·m
- Efficiency per mesh: 95%
Find: Gear ratio (driven ÷ driving)
Answer
4.
Problem 4.
- Gear set: One gear pair
- Driving gear teeth (input): 8
- Driven gear teeth: 34
- Input speed: 1,100 rpm
- Input torque: 11 N·m
- Efficiency per mesh: 90%
Find: Gear ratio (driven ÷ driving)
Answer
5.
Problem 5.
- Gear set: Two-stage compound train
- Driving gear teeth (input): 11
- Driven gear teeth: 31
- Second-stage driving gear teeth: 9
- Second-stage driven gear teeth (output): 50
- Input speed: 1,100 rpm
- Input torque: 10 N·m
- Efficiency per mesh: 69%
Find: Gear ratio (driven ÷ driving)
Answer
6.
Problem 6.
- Gear set: Two-stage compound train
- Driving gear teeth (input): 9
- Driven gear teeth: 40
- Second-stage driving gear teeth: 9
- Second-stage driven gear teeth (output): 42
- Input speed: 1,100 rpm
- Input torque: 12 N·m
- Efficiency per mesh: 80%
Find: Gear ratio (driven ÷ driving)
Answer
7.
Problem 7.
- Gear set: Two-stage compound train
- Driving gear teeth (input): 11
- Driven gear teeth: 38
- Second-stage driving gear teeth: 10
- Second-stage driven gear teeth (output): 24
- Input speed: 1,000 rpm
- Input torque: 9 N·m
- Efficiency per mesh: 85%
Find: Gear ratio (driven ÷ driving)
Answer
8.
Problem 8.
- Gear set: Two-stage compound train
- Driving gear teeth (input): 13
- Driven gear teeth: 40
- Second-stage driving gear teeth: 8
- Second-stage driven gear teeth (output): 39
- Input speed: 800 rpm
- Input torque: 12 N·m
- Efficiency per mesh: 100%
Find: Gear ratio (driven ÷ driving)
Answer
9.
Problem 9.
- Gear set: One gear pair
- Driving gear teeth (input): 10
- Driven gear teeth: 30
- Input speed: 1,500 rpm
- Input torque: 9 N·m
- Efficiency per mesh: 90%
Find: Gear ratio (driven ÷ driving)
Answer
10.
Problem 10.
- Gear set: One gear pair
- Driving gear teeth (input): 15
- Driven gear teeth: 40
- Input speed: 900 rpm
- Input torque: 9 N·m
- Efficiency per mesh: 90%
Find: Gear ratio (driven ÷ driving)
Answer

Answer key: Gear Ratio worksheet

1. Gear ratio (driven ÷ driving): 7.92
1. First pair: divide the driven teeth by the driving teeth
  $\frac{38}{12}$ $= 3.1666666667$
2. Second pair: divide its driven teeth by its driving teeth
  $\frac{30}{12}$ $= 2.5$
3. Stages on a shared shaft multiply, so the overall ratio is their product
  $ratio$ $= 3.1666666667 \times 2.5$ $= 7.9166666667$
4. Divide the input speed by the ratio to find how fast the output turns
  $\frac{1,100 rpm}{7.9166666667}$ $= 138.9473684211 rpm$
5. Multiply the input torque by the ratio, then by the efficiency to allow for friction at each of the 2 meshes
  $7 N·m \times 7.9166666667 \times 81%$ $= 44.8875 N·m$
2. Gear ratio (driven ÷ driving): 16.88
1. First pair: divide the driven teeth by the driving teeth
  $\frac{45}{13}$ $= 3.4615384615$
2. Second pair: divide its driven teeth by its driving teeth
  $\frac{39}{8}$ $= 4.875$
3. Stages on a shared shaft multiply, so the overall ratio is their product
  $ratio$ $= 3.4615384615 \times 4.875$ $= 16.875$
4. Divide the input speed by the ratio to find how fast the output turns
  $\frac{1,000 rpm}{16.875}$ $= 59.2592592593 rpm$
5. Multiply the input torque by the ratio, then by the efficiency to allow for friction at each of the 2 meshes
  $12 N·m \times 16.875 \times 72.25%$ $= 146.30625 N·m$
3. Gear ratio (driven ÷ driving): 15.24
1. First pair: divide the driven teeth by the driving teeth
  $\frac{40}{15}$ $= 2.6666666667$
2. Second pair: divide its driven teeth by its driving teeth
  $\frac{40}{7}$ $= 5.7142857143$
3. Stages on a shared shaft multiply, so the overall ratio is their product
  $ratio$ $= 2.6666666667 \times 5.7142857143$ $= 15.2380952381$
4. Divide the input speed by the ratio to find how fast the output turns
  $\frac{1,300 rpm}{15.2380952381}$ $= 85.3125 rpm$
5. Multiply the input torque by the ratio, then by the efficiency to allow for friction at each of the 2 meshes
  $10 N·m \times 15.2380952381 \times 90.25%$ $= 137.5238095238 N·m$
4. Gear ratio (driven ÷ driving): 4.25
1. Divide the driven teeth by the driving teeth to get the gear ratio
  $ratio$ $= \frac{34}{8}$ $= 4.25$
2. Divide the input speed by the ratio to find how fast the output turns
  $\frac{1,100 rpm}{4.25}$ $= 258.8235294118 rpm$
3. Multiply the input torque by the ratio, then by the efficiency to allow for friction
  $11 N·m \times 4.25 \times 90%$ $= 42.075 N·m$
5. Gear ratio (driven ÷ driving): 15.66
1. First pair: divide the driven teeth by the driving teeth
  $\frac{31}{11}$ $= 2.8181818182$
2. Second pair: divide its driven teeth by its driving teeth
  $\frac{50}{9}$ $= 5.5555555556$
3. Stages on a shared shaft multiply, so the overall ratio is their product
  $ratio$ $= 2.8181818182 \times 5.5555555556$ $= 15.6565656566$
4. Divide the input speed by the ratio to find how fast the output turns
  $\frac{1,100 rpm}{15.6565656566}$ $= 70.2580645161 rpm$
5. Multiply the input torque by the ratio, then by the efficiency to allow for friction at each of the 2 meshes
  $10 N·m \times 15.6565656566 \times 47.61%$ $= 74.5409090909 N·m$
6. Gear ratio (driven ÷ driving): 20.74
1. First pair: divide the driven teeth by the driving teeth
  $\frac{40}{9}$ $= 4.4444444444$
2. Second pair: divide its driven teeth by its driving teeth
  $\frac{42}{9}$ $= 4.6666666667$
3. Stages on a shared shaft multiply, so the overall ratio is their product
  $ratio$ $= 4.4444444444 \times 4.6666666667$ $= 20.7407407407$
4. Divide the input speed by the ratio to find how fast the output turns
  $\frac{1,100 rpm}{20.7407407407}$ $= 53.0357142857 rpm$
5. Multiply the input torque by the ratio, then by the efficiency to allow for friction at each of the 2 meshes
  $12 N·m \times 20.7407407407 \times 64%$ $= 159.2888888889 N·m$
7. Gear ratio (driven ÷ driving): 8.29
1. First pair: divide the driven teeth by the driving teeth
  $\frac{38}{11}$ $= 3.4545454545$
2. Second pair: divide its driven teeth by its driving teeth
  $\frac{24}{10}$ $= 2.4$
3. Stages on a shared shaft multiply, so the overall ratio is their product
  $ratio$ $= 3.4545454545 \times 2.4$ $= 8.2909090909$
4. Divide the input speed by the ratio to find how fast the output turns
  $\frac{1,000 rpm}{8.2909090909}$ $= 120.6140350877 rpm$
5. Multiply the input torque by the ratio, then by the efficiency to allow for friction at each of the 2 meshes
  $9 N·m \times 8.2909090909 \times 72.25%$ $= 53.9116363636 N·m$
8. Gear ratio (driven ÷ driving): 15
1. First pair: divide the driven teeth by the driving teeth
  $\frac{40}{13}$ $= 3.0769230769$
2. Second pair: divide its driven teeth by its driving teeth
  $\frac{39}{8}$ $= 4.875$
3. Stages on a shared shaft multiply, so the overall ratio is their product
  $ratio$ $= 3.0769230769 \times 4.875$ $= 15$
4. Divide the input speed by the ratio to find how fast the output turns
  $\frac{800 rpm}{15}$ $= 53.3333333333 rpm$
5. Multiply the input torque by the ratio, because gears trade speed for force (no losses)
  $12 N·m \times 15$ $= 180 N·m$
9. Gear ratio (driven ÷ driving): 3
1. Divide the driven teeth by the driving teeth to get the gear ratio
  $ratio$ $= \frac{30}{10}$ $= 3$
2. Divide the input speed by the ratio to find how fast the output turns
  $\frac{1,500 rpm}{3}$ $= 500 rpm$
3. Multiply the input torque by the ratio, then by the efficiency to allow for friction
  $9 N·m \times 3 \times 90%$ $= 24.3 N·m$
10. Gear ratio (driven ÷ driving): 2.67
1. Divide the driven teeth by the driving teeth to get the gear ratio
  $ratio$ $= \frac{40}{15}$ $= 2.6666666667$
2. Divide the input speed by the ratio to find how fast the output turns
  $\frac{900 rpm}{2.6666666667}$ $= 337.5 rpm$
3. Multiply the input torque by the ratio, then by the efficiency to allow for friction
  $9 N·m \times 2.6666666667 \times 90%$ $= 21.6 N·m$