FINAL CASE STUDY
Final Case Study: Customer Inquiry
You will apply all eight modules to a single, end-to-end customer inquiry—from identifying missing information to the final drive design.
9 exercises in a fixed order · approx. 60 minutes · template for a design report (2 pages)
Getting Started
A customer sends a brief, incomplete inquiry regarding a linear axis. You work through it step by step—from Module 1 to Module 8—designing the motor, gearbox, linear guide, ball screw, coupling, and lubrication system. Each task builds on the results of the previous one.
Calculation exercises with immediate feedback alternate with decision-making questions, for which you can view the worked solution in a collapsible text box. At the end, you’ll find the complete worked solution as well as a summary of the final design.
This case study is an exercise, not a test—no points or progress are saved.
Background: The customer inquiry
“We are looking for a linear axis for our packaging system. It needs to move a workpiece 400 mm horizontally. The mass to be moved is approximately 80 kg. We have estimated the required feed force at 200 N; the maximum travel speed should be 0.5 m/s, and the positioning accuracy should be ±0.05 mm. We would like to use a ball screw. Due to space constraints, the motor cannot be mounted in line with the lead screw but must be connected with a lateral offset via a spur gear stage (center distance fixed at 60 mm by the housing, gear ratio i=2). Please provide a proposal with justification."
| Size | Value |
|---|---|
| Translational motion | 400 mm horizontal |
| Moving mass | ca. 80 kg |
| Feed force F | 200 N (estimated) |
| Travel speed v | 0.5 m/s (maximum) |
| Positioning accuracy | ±0.05 mm |
| Desired element | Ball screw |
| Motor–Spindle Connection | Spur gear stage, laterally offset |
| Center distance a (housing specification) | 60 mm |
| Translation i | 2 |
Reference: Module 1 – Load Profile, Torque, and Inertia
Problem 1: Missing Data and Base Power
What information is missing from the inquiry that you need for a reliable design?
View Worked Solution
Four missing pieces of information: (1) operating hours per day and working days per year, (2) ambient temperature and conditions at the installation site, (3) required service life of the axis, (4) permissible shaft misalignment or alignment accuracy between the motor and the spindle.
Calculation task
You can also calculate the required mechanical power (P_erf) based on the feed force and speed.
- F = 200 N (feed force)
- v = 0.5 m/s (travel speed)
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Hints:
Use the basic formula P = F · v.
Set F = 200 N and v = 0.5 m/s.
Use the basic formula P = F · v.
Set F = 200 N and v = 0.5 m/s.
Show solution
The solution unlocks after your first check.
Worked solution – how to get the result:
P_erf = F · v = 200 N · 0.5 m/s = 100 W
Tolerance: exact (pure multiplication).
Source: Module 1 – Load Profile →Additional information (provided upon request; applies starting with Exercise 2)
Two-shift operation, 16 hours/day, 250 working days/year (= 4,000 hours/year); workshop temperature 25 °C, normal cleanliness; required service life 5 years; radial shaft misalignment due to the spur gear connection approx. 0.1 mm.
Reference: Module 2 – Selecting Motor and Drive
Task 2: Motor Type and Required Power Class
The ball screw has an efficiency of η = 0.90, and the spur gear stage has an efficiency of η = 0.98. Calculate the required motor power, including a reserve factor of 1.3 (start-up/acceleration), as a rough estimate of the power class. Also, verify the required positioning accuracy of ±0.05 mm on the axis against the achievable angular accuracy on the motor shaft: Is a simple induction motor sufficient, or is a controlled drive (servo motor with encoder) required?
Calculation task
Calculate the required motor power with a safety margin, as well as the angle at the motor shaft that corresponds to the required positioning accuracy.
- P_erf = 100 W
- η_KGT = 0.90; η_Stirnrad = 0.98
- 1.3 =
- Positioning accuracy along the axis = ±0.05 mm
- Lead Ph = 20 mm/revolution (already specified, confirmed in Problem 6)
- Translation: Spur gear stage i = 2
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Hints:
Multiply the two efficiencies to obtain the overall efficiency: η_gesamt = η_KGT · η_Stirnrad.
Divide P_erf by η_gesamt and multiply the result by the reserve factor 1.3.
Convert ±0.05 mm, based on the lead Ph = 20 mm/revolution, into a lead angle (one revolution = 360°), and then multiply this angle by the gear ratio i = 2.
Multiply the two efficiencies to obtain the overall efficiency: η_gesamt = η_KGT · η_Stirnrad.
Divide P_erf by η_gesamt and multiply the result by the reserve factor 1.3.
Convert ±0.05 mm, based on the lead Ph = 20 mm/revolution, into a lead angle (one revolution = 360°), and then multiply this angle by the gear ratio i = 2.
Show solution
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Worked solution – how to get the result:
η_gesamt = 0.90 · 0.98 = 0.882
P_Motor,erf = 100 W / 0.882 ≈ 113.4 W
With a margin: 113.4 W · 1.3 ≈ 147.4 W → As a rough estimate of the power class, this is close to the standard market power rating of 0.18 kW (180 W).
Spindle angle: 0.05/20 · 360° = 0.9°
Angle at the motor shaft via the spur gear stage (i=2): 0.9° · 2 = 1.8°
Source: Servo or Induction Motor? – Learning unit, Module 2 →Worked Solution: What type of motor is required?
An induction motor without an encoder achieves only about ±5–10°—3 to 6 times less accurate than the required 1.8° at the motor shaft. A servo motor with an encoder, on the other hand, achieves ±0.01–0.001° and meets the requirement with a large margin. Conclusion: A controlled drive (servo motor with encoder) is required; a simple induction motor is not sufficient.
Reference: Module 3 – Selecting Gearboxes
Problem 3: Justification for a spur gear stage
Why is a spur gear stage (η ≈ 0.98) preferable to a worm gear stage in this case, even though a worm gear stage often allows for a higher gear ratio in a more compact space?
View Worked Solution
The efficiency of a worm gear stage is often significantly lower (typically 50–70%, highly dependent on the angle) than that of a spur gear stage (typically 97–99%). Since each stage of loss increases the required motor power (see Exercise 2), a worm gear stage would necessitate a noticeably larger and more expensive motor in this case. For this application (i=2, moderate installation space), the spur gear stage offers the better overall performance.
This problem does not require a number and has no tolerance.
Reference: Module 4 – Understanding Gearing, Selecting Gears
Problem 4: Design a spur gear set
Design the gear train for the spur gear stage: center distance a = 60 mm (housing specification), gear ratio i = 2. Select a module according to DIN 780 and the number of teeth.
Calculation task
Determine the module m, the number of teeth on the pinion z1, and the number of teeth on the gear z2.
- a = 60 mm (center distance, housing specification)
- i = 2 (translation)
- a = m · (z1 + z2) / 2
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Hints:
Substitute z2 = i · z1 into a = m · (z1 + z2) / 2: a = m · 3 · z1 / 2, so z1 = 2a / (3m).
Select a module according to DIN 780 (e.g., m = 2 mm) and check whether z1 is an integer.
Substitute z2 = i · z1 into a = m · (z1 + z2) / 2: a = m · 3 · z1 / 2, so z1 = 2a / (3m).
Select a module according to DIN 780 (e.g., m = 2 mm) and check whether z1 is an integer.
Show solution
The solution unlocks after your first check.
Worked solution – how to get the result:
a = m·(z1+z2)/2, where z2 = i·z1 = 2·z1 → a = m·3·z1/2 → z1 = 2a/(3m)
With m = 2 mm (DIN 780): z1 = 2·60/(3·2) = 20. Therefore, z2 = 40.
Example: d1 = m·z1 = 40 mm, d2 = m·z2 = 80 mm, a = (d1+d2)/2 = 60 mm ✓ (exact, no rounding necessary).
Tolerance: exact (integer number of teeth; a must match exactly).
Source: Module, Pitch Circle, and Center Distance – Learning unit, Module 4 →Reference: Module 5 – Selecting Linear Guides, Calculating Service Life
Problem 5: Sizing the Guide
The moving mass exerts a lateral force on the guide due to its weight. Calculate this force and the static safety factor S0 for a guide with C0 = 10 kN.
Calculation task
Calculate the weight force F_G and the static safety factor S0.
- m = 80 kg (moving mass)
- g = 9.81 m/s²
- C₀ = 10 kN (dynamic load rating of the guide)
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Hints:
First, calculate F_G = m · g.
S0 = C0 / F_G – pay attention to the units (C0 in kN = 10,000 N).
First, calculate F_G = m · g.
S0 = C0 / F_G – pay attention to the units (C0 in kN = 10,000 N).
Show solution
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Worked solution – how to get the result:
F_G = m·g = 80 kg · 9.81 m/s² = 784.8 N ≈ 785 N
S0 = C0/F_G = 10,000 N / 785 N ≈ 12.7
Evaluation: S0 ≈ 12.7 is significantly above the guideline values for standard design (5–8) and even exceeds the range for overhead installation/high risk potential (12 and above)—with a positioning accuracy of ±0.05 mm, the additional margin is reasonable and does not constitute oversizing.
Source: Load Ratings and Static Safety – Learning unit, Module 5 →Reference: Module 6 – Linear Drives: Ball Screws, Lead Screws, and Actuators
Problem 6: Spindle and Torques
Select a lead screw with a pitch of Ph = 20 mm/revolution from the catalog. Calculate the lead screw speed, the motor speed (via the spur gear stage with i = 2), and the drive torque at the lead screw (η_KGT = 0.90).
Calculation task
Calculate the spindle speed, motor speed, drive torque at the spindle, and required motor torque.
- v = 0.5 m/s = 500 mm/s
- Ph = 20 mm/revolution
- i = 2
- F = 200 N
- η_KGT = 0.90; η_Stirnrad = 0.98
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Hints:
Spindle speed: n_Spindel = v · 60 / Ph.
Motor speed: n_Motor = n_Spindel · i.
Drive torque at the spindle: M_Spindel = F · Ph / (2000 · π · η_KGT).
Required motor torque: M_Motor,erf = M_Spindel / (i · η_Stirnrad).
Spindle speed: n_Spindel = v · 60 / Ph.
Motor speed: n_Motor = n_Spindel · i.
Drive torque at the spindle: M_Spindel = F · Ph / (2000 · π · η_KGT).
Required motor torque: M_Motor,erf = M_Spindel / (i · η_Stirnrad).
Show solution
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Worked solution – how to get the result:
n_Spindel = v·60/Ph = 500 mm/s · 60 / 20 mm = 1,500 min⁻¹
n_Motor = n_Spindel · i = 1,500 · 2 = 3,000 min⁻¹ (The motor rotates faster; the smaller pinion z1=20 is mounted on the motor shaft)
M_Spindel = F·Ph/(2000·π·η) = 200·20/(2000·π·0.90) = 4,000/5,654.9 ≈ 0.71 Nm
M_Motor,erf = M_Spindel/(i·η_Stirnrad) = 0.71/(2·0.98) ≈ 0.36 Nm
Verification: P = M_Motor ·n_Motor/ 9550 = 0.36 ·3,000/ 9550 ≈ 0.113 kW = 113 W – matches Problem 2 (113.4 W before reserve).
Source: Ball Screw Fundamentals – Learning unit, Module 6 →Reference: Module 7 – Connecting and Fitting: Couplings, Magnetic Couplings, Fits, and Standard Parts
Problem 7: Design a coupling
What is the rated torque of the selected motor (0.18 kW, 3,000 min⁻¹)? How much reserve is there relative to the required torque from Problem 6? Design the coupling with an appropriate service factor.
Installation location: The coupling is located between the motor shaft and the pinion shaft of the spur gear stage, not on the spindle—according to the initial conditions, this is where the radial shaft misalignment of 0.1 mm occurs. Only at this point (n = 3,000 min⁻¹) is M_N the correct reference torque for the coupling.
Calculation task
Calculate the rated torque of the motor at the motor/pinion shaft and the rated coupling torque.
- P = 0.18 kW; n = 3,000 min⁻¹ (motor/pinion shaft)
- M_Spindel,erf = 0.36 Nm (Problem 6, for comparing the reserve)
- Service factor S_B = 1.5 (start-stop operation, no continuous shock loading)
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Hints:
Rated torque: M_N = 9550 · P / n.
Clutch torque: M_dim = M_N · S_B.
Rated torque: M_N = 9550 · P / n.
Clutch torque: M_dim = M_N · S_B.
Show solution
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Worked solution – how to get the result:
M_N = 9550·P/n = 9550·0.18/3,000 ≈ 0.57 Nm (on the motor/pinion shaft, n =3,000 min⁻¹)
Reserve = 0.57/ 0.36 ≈ 1.6 times (sufficient, in part because the reserve factor 1.3 from Problem 2 is already factored into the motor selection)
M_dim = M_N·S_B = 0.57·1.5 ≈ 0.86 Nm → Select the next larger catalog coupling (e.g., size 1 Nm)
Source: Coupling Types and Selection – Learning unit, Module 7 →Worked Solution: Coupling Type and Shaft-Hub Fit
The shaft misalignment of 0.1 mm between the motor shaft and the pinion shaft requires a flexible coupling (with compensation and damping), not a rigid coupling. Shaft-hub connection of the coupling: H7/k6 transition fit (tight fit for reliable torque transmission), not H7/g6 (which would be a clearance fit for rotating parts).
Reference: Module 8 – Lubrication, Maintenance, and Total Cost
Problem 8a: Relubrication interval
Calculate the relubrication interval for the spindle bearing (n = 1,500 min⁻¹ spindle speed, bore d = 20 mm, k_f = 10, mild conditions f_T = f_L = f_U = 1) and compare it with the required service life of 5 years (at 4,000 operating hours/year).
For simplicity, it is assumed that the axis moves throughout the entire operating time (4,000 hours/year)—cycle pauses and downtime are not deducted separately here.
Calculation task
Calculate the relubrication interval t_f in hours and in years.
- n = 1,500 min⁻¹ (spindle speed)
- d = 20 mm (bearing bore)
- k_f = 10
- f_T = f_L = f_U = 1 (mild conditions)
- t_f = k_f · (14·10⁶/(n·√d) − 4·d) · f_T · f_L · f_U
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Hints:
First, calculate 14·10⁶/(n·√d) and subtract 4·d.
Multiply the result by k_f (the factors f_T, f_L, and f_U are all 1 here).
Divide t_f by 4,000 operating hours per year to obtain the service life in years.
First, calculate 14·10⁶/(n·√d) and subtract 4·d.
Multiply the result by k_f (the factors f_T, f_L, and f_U are all 1 here).
Divide t_f by 4,000 operating hours per year to obtain the service life in years.
Show solution
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Worked solution – how to get the result:
14·10⁶/(1,500·√20) = 14,000,000/(1,500·4.472) = 14,000,000/6,708 ≈ 2,087. Minus 4·20=80 → 2,007.
t_f = 10 · 2,007 = 20,070 h. In years: 20,070 h / 4,000 h/year ≈ 5.0 years.
Evaluation: The relubrication interval barely covers the required service life—the design report should note that either relubrication must be performed in a timely manner or a bearing with lifetime lubrication or a larger reserve should be selected.
Source: Lubrication Intervals and Relubrication Periods – Learning unit, Module 8 →Problem 8b: Energy Comparison: Ball Screw vs. Trapezoidal Screw
A colleague suggests using a trapezoidal lead screw instead of a ball screw for cost reasons (the less favorable case of the 25–50% efficiency range mentioned in Module 6; here, η is assumed to be 0.35). Compare the annual energy consumption of both variants under otherwise identical conditions.
For simplicity’s sake: As in Problem 8a, it is assumed that the drive operates at the calculated power for the entire operating time (4,000 h/year)—cycle pauses with lower or no power consumption are not deducted separately here; both variants are affected equally; the comparison between them remains unbiased.
Calculation task
Calculate the annual energy consumption for the ball screw and for the trapezoidal screw.
- P_erf = 100 W; η_Stirnrad = 0.98
- η_KGT = 0.90
- η_Trapez = 0.35 (worst-case scenario)
- 4,000 Operating hours per year
- P_Motor = P_erf/(η_Spindel·η_Stirnrad); E_Jahr = P_Motor·4,000 h/year
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Hints:
For each variant, first calculate P_Motor = P_erf/(η_Spindel·η_Stirnrad).
Multiply P_Motor by 4,000 operating hours per year to calculate the annual consumption in kWh.
For each variant, first calculate P_Motor = P_erf/(η_Spindel·η_Stirnrad).
Multiply P_Motor by 4,000 operating hours per year to calculate the annual consumption in kWh.
Show solution
The solution unlocks after your first check.
Worked solution – how to get the result:
Ball screw (η =0.90): P = 100/(0.90 ·0.98) ≈113.4 W → E ≈453.5 kWh/year
Trapezoidal lead screw (η =0.35): P = 100/(0.35 ·0.98) ≈291.5 W → E ≈1,166.2 kWh/year
Difference ≈ 712.7 kWh/year (factor ≈2.6, approximately 157% more) – over the required 5-year service life, ≈3,560 kWh more.
Evaluation: The lower initial cost comes at the expense of operating energy consumption that is more than twice as high—recommendation in the design report: stick with the ball screw.
Source: Calculate the TCO of the powertrain →Worked Solution (Complete)
A summary of all tasks in context—for comparison with your own design report.
View Worked Solution
“Missing data: operating hours per year, ambient temperature, required service life, permissible shaft misalignment. Estimated power requirement: 100 W, with efficiency chain (0.90×0.98) and a reserve of 147 W → power class approximately 0.18 kW; the required accuracy (1.8° at the motor shaft) necessitates a controlled drive with a servo motor and encoder; a simple induction motor (only ±5–10°) is insufficient. A spur gear stage (m=2, z1=20, z2=40) precisely achieves the required center distance of 60 mm at i=2—significantly more efficient than a worm gear stage. A linear guide with C0=10 kN achieves S0≈12.7, which is adequate for the required accuracy. A ball screw with Ph=20 mm operates at 1,500 min⁻¹ (motor 3,000 min⁻¹) with a drive torque of 0.71 Nm; the motor requires 0.36 Nm for this—a cross-check of the power confirms 113 W. The coupling is mounted on the motor/pinion shaft, where the motor (M_N≈0.57 Nm) has sufficient reserve; it is designed with a service factor of 1.5 for 0.86 Nm, with elastic compensation due to shaft misalignment, H7/k6 fit. The relubrication interval of approximately 5 years barely covers the required service life—plan for timely maintenance. A less expensive trapezoidal lead screw would more than double energy consumption (from 454 to 1,166 kWh/year)—recommendation: stick with the ball screw."
Summary of the Design
| Assembly | Design value | From the assignment |
|---|---|---|
| Motor | Servo motor with encoder, power class approx. 0.18 kW (180 W) | 2 |
| Gearbox (spur gear stage) | m = 2 mm, z1 = 20, z2 = 40 (i = 2) | 4 |
| Linear Guides | C₀ = 10 kN, S₀ ≈ 12.7 | 5 |
| Ball screw | Ph = 20 mm, n_Spindel = 1,500 min⁻¹, M_Spindel ≈ 0.71 Nm | 6 |
| Coupling | elastic, M_dim ≈ 0.86 Nm, fit H7/k6 | 7 |
| Lubrication | Relubrication intervals ≈ 5.0 years (barely covers the required service life) | 8a |
| Energy Comparison | Ball screw ≈ 454 kWh/year, trapezoidal screw ≈ 1,166 kWh/year – Recommendation: stick with the ball screw | 8b |