It is easy to calculate around 90 % of applications knowing only the following five parameters:
- Mass to be decelerated (weight) m [kg]
- Impact velocity at shock absorber vD [m/s]
- Propelling force F [N]
- Cycles per hour c [1/h]
- Number of absorbers in parallel n
Key to symbols used
| Symbol | Unit | Description | Symbol | Unit | Description |
|---|---|---|---|---|---|
| W1 | Nm | Kinetic energy per cycle | 3ST | 1 to 3 | tall torque factor (normally 2.5) |
| W2 | Nm | Propelling force energy per cycle | M | Nm | Propelling torque |
| W3 | Nm | Total energy per cycle (W1 + W2) | I | kgm2 | Moment of Inertia |
| 1W4 | Nm/hr | Total energy per hour (W3 · c) | g | m/s2 | Acceleration due to gravity = 9.81 |
| me | kg | Effective weight | h | m | Drop height excl. shock absorber stroke |
| m | kg | Mass to be decelerated | s | m | Shock absorber stroke |
| n | Number of shock absorbers (in parallel) | L/R/r | m | Radius | |
| 2v | m/s | Velocity at impact | Q | N | Reaction force |
| 2vD | m/s | Impact velocity at shock absorber | μ | Coefficient of friction | |
| ω | rad/s | Angular velocity at impact | t | s | Deceleration time |
| F | N | Propelling force | a | m/s2 | Deceleration |
| c | 1/hr | Cycles per hour | α | ° | Side load angle |
| P | kW | Motor power | β | ° | Angle of incline |
Note:
When using several shock absorbers in parallel, the values (W3), (W4) and (me) are divided according to the number of units used.
Reaction force Q [N] Q = (1.5 · W3) / s | Stopping time t [s] t = (2.6 · s) / vD | Deceleration a [m/s2] a = (0.75 · vD2) / s |
Applications
| Application | Formulae | Example | |
|---|---|---|---|
1. Mass without propelling force
| W1 = m · v2 · 0.5 W2 = 0 W3 = W1 + W2 W4 = W3 · c vD = v me = m | m = 100 kg W1 = 100 · 1.52 · 0.5 = 113 Nm | |
2. Mass with propelling force
| W1 = m · v2 · 0.5 W2 = F · s W3 = W1 + W2 W4 = W3 · c vD = v me = (2 · W3) / vD2 2.1 for vertical motion upwards W2 = (F – m · g) · s 2.2 for vertical motion downwards W2 = (F + m · g) · s | m = 36 kg 1v = 1.5 m/s F = 400 N c = 1000 1/hr s = 0.025 m (chosen) W1 = 36 · 1.52 · 0.5 = 41 Nm W2 = 400 · 0.025 = 10 Nm W3 = 41 + 10 = 51 Nm W4 = 51 · 1000 = 51000 Nm/hr me = 2 · 51 : 1.52 = 45 kg 1 v is the fi nal impact velocity of the mass: With pneumaticallypropelled systems this can be 1.5 to 2 times the average velocity. Please take this into account when calculating energy. | |
3. Mass with motor drive
| W1 = m · v2 · 0.5 W2 = (1000 · P ·ST · s) / v W3 = W1 + W2 W4 = W3 · c vD = v me = (2 · W3) / vD2 | m = 800 kg W1 = 800 · 1.22 · 0.5 = 576 Nm motor, coupling and gearbox into calculation for W1. | |
4. Mass on driven rollers
| W1 = m · v2 · 0.5 W2 = m · μ · g · s W3 = W1 + W2 W4 = W3 · c vD = v me = (2 · W3) / vD2 | m = 250 kg W1 = 250 · 1.52 · 0.5 = 281 Nm | |
5. Swinging mass with propelling force
| W1 = m · v2 · 0.5 = 0.5 · I · ω2 W2 = (M · s) / R W3 = W1 + W2 W4 = W3 · c vD = (v · R) / L = ω · R me = (2 · W3) / vD2 | m = 20 kg W1 = 20 · 12 · 0.5 = 10 Nm | |
6. Free falling mass
| W1 = m · g · h W2 = m · g · s W3 = W1 + W2 W4 = W3 · c vD = √2 · g · h me = (2 · W3) / vD2 | m = 30 kg W1 = 30 · 0.5 · 9.81 = 147 Nm | |
6.1 Mass rolling/sliding down incline
| W1 = m · g · h = m · vD2 · 0.5 W2 = m · g · sinβ · s W3 = W1 + W2 W4 = W3 · c vD = √2 · g · h me = (2 · W3) / vD2 6.1a propelling force up incline W2 = (F – m · g· sinβ) · s 6.1b propelling force down incline W2 = (F + m · g· sinβ) · s | m = 500 kg W1 = 500 · 9.81 · 0.1 = 490.5 Nm | |
6.2 Mass free falling about a pivot point
| W1 = m · g · h W2 = 0 W3 = W1 + W2 W4 = W3 · c vD = √2 · g · h · (R / L) me = (2 · W3) / vD2 tan α = s / R | m = 50 kg W1 = 50 · 9.81 · 1 = 490.5 Nm | |
7. Rotary index table with propelling torque
| W1 = m · v2 · 0.25 = 0.5 · I · ω2 W2 = (M · s) / R W3 = W1 + W2 W4 = W3 · c vD = (v · R) / L = ω · R me = (2 · W3) / vD2 | m = 1000 kg W1 = 1000 · 1.12 · 0.25 = 303 Nm | |
8. Swinging arm with propelling torque | W1 = m · v2 · 0.17 = 0,5 · I · ω2 W2 = (M · s) / R W3 = W1 + W2 W4 = W3 · c vD = (v · R) / L = ω · R me = (2 · W3) / vD2 | I = 56 kgm2 W1 = 0.5 · 56 · 12 = 28 Nm | |
9. Swinging arm with propelling force | W1 = m · v2 · 0.17 = 0.5 · I · ω2 W2 = (F · r · s) / R = (M · s) / R W3 = W1 + W2 W4 = W3 · c vD = (v · R) / L = ω · R me = (2 · W3) / vD2 | m = 1000 kg W1 = 1000 · 22 · 0.17 = 680 Nm | |
10. Mass lowered at controlled speed
| W1 =m · v2 · 0.5 W2 = m · g · s W3 = W1 + W2 W4 = W3 · c vD = v me = (2 · W3) / vD2 | m = 6000 kg W1 = 6000 · 1.52 · 0.5 = 6750 Nm | |
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Effective Weight (me)
The effective weight (me) can either be the same as the actual weight (examples Aand C), or it can be an imaginary weight representing a combination of the propelling force or lever action plus the actual weight (examples B and D).
| Einsatzfall | Beispiel |
|---|---|
A Mass without propelling force
| m = 100 kg vD = v = 2 m/s W1 = W3 = 200 Nm me = (2 · 200) / 4 = 100 kg Formula: me = m |
B Mass with propelling force
| m = 100 kg F = 2000 N vD = v = 2 m/s s = 0.1 m W1 = 200 Nm W2 = 200 Nm W3 = 400 Nm me = (2 · 400) / 4 = 200 kg Formula: me = (2 · W3) / vD2 |
C Mass without propelling force direct | m = 20 kg vD = v = 2 m/s W1 = W3 = 40 Nm me = (2 · 40) / 22 = 20 kg Formula: me = m |
D Mass without propelling force with | m = 20 kg v = 2 m/s vD = 0.5 m/s s = 0.1 m W1 = W3 = 40 Nm me = (2 · 40) / 0.52 = 320 kg Formula: me = (2 · W3) / vD2 |
