Calculation Bases

Design of Industrial Shock Absorbers

ACE shock absorbers provide linear deceleration and are therefore superior to other kinds of damping element. With the help of the overview of the calculation bases for the design of a shock absorber you will be able to select the right industrial shock absorber for your project from the ACE web catalogue.

Key to symbols used

Applications

Effective Weight (me)

It is easy to calculate around 90 % of applications knowing only the following five parameters:

Abzubremsende Masse (Gewicht) m [kg]
Aufprall- oder Auffahrgeschwindigkeit v_D [m/s]
Evtl. vorhandene zusätzliche Antriebskraft F [N]
Anzahl der Hübe oder Takte pro Stunde x [1/h]
Anzahl Stoßdämpfer parallel n

Calculate the correct shock absorber according to your application

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
²v_D	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

^{1 All mentioned values of W4 in the capacity charts are only valid for room temperature. There are reduced values at higher temperature ranges.}

^{2 v or vD is the final impact velocity of the mass. With accelerating motion the final impact velocity can be 1.5 to 2 times higher than the average. Please take this into account when calculating kinetic energy.}

^{3 ST =^ relation between starting torque and running torque of the motor (depending on the design)}

^{In all the following examples the choice of shock absorbers made from the capacity chart is based upon the values of (W3), (W4), (me) and the desired shock absorber stroke (s).}

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]
Für alle Beispiele gilt:

Q = (1,5 · W₃) / s

Stopping time t [s]
Für alle Beispiele gilt:

t = (2,6 · s) / v_D

Deceleration a [m/s2]
Für alle Beispiele gilt:

a = (0,75 · v_D²) / s

^{Approximate values assuming correct adjustment. Add safety margin if necessary. (Exact values will depend upon actual application data and can be provided on request.)}

Applications

Application	Formulae	Example
1. Mass without propelling force	W₁ = m · v² · 0,5 W₂ = 0 W₃ = W₁ + W₂ W₄ = W₃ · x v_D = v me = m	m = 100 kg v = 1,5 m/s x = 500 1/h s = 0,050 m (gewählt) W₁ = 100 · 1,5² · 0,5 = 113 Nm W₂ = 0 W₃ = 113 + 0 = 113 Nm W₄ = 113 · 500 = 56500 Nm/h me = m = 100 kg
2. Mass with propelling force	W₁ = m · v2 · 0,5 W₂ = F · s W₃ = W1 + W2 W₄ = W3 · x v_D = v me = (2 · W₃) / v_D² 2.1 bei senkrechter Bewegung nach oben W₂ = (F – m · g) · s 2.2 bei senkrechter Bewegung nach unten W₂ = (F + m · g) · s	m = 36 kg ¹v = 1,5 m/s F = 400 N x = 1000 1/h s = 0,025 m (gewählt) W₁ = 36 · 1,5² · 0,5 = 41 Nm W₂ = 400 · 0,025 = 10 Nm W₃ = 41 + 10 = 51 Nm W₄ = 51 · 1000 = 51000 Nm/h 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 ·HM · s) / v W₃ = W₁ + W₂ W₄ = W₃ · x v_D = v me = (2 · W₃) / v_D²	m = 800 kg v = 1,2 m/s HM = 2,5 P = 4 kW x = 100 1/h s = 0,100 m (gewählt) W₁ = 800 · 1,2² · 0,5 = 576 Nm W₂ = 1000 · 4 · 2,5 · 0,1 : 1,2 = 834 Nm W₃ = 576 + 834 = 1410 Nm W₄ = 1410 · 100 = 141000 Nm/h me = 2 · 1410 : 1,2² = 1958 kg ^{Note: Do not forget to include the rotational energy of} ^{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 · x v_D = v me = (2 · W₃) / v_D²	m = 250 kg v = 1,5 m/s x = 180 1/h (Stahl/Guss) μ = 0,2 s = 0,050 m (gewählt) W₁ = 250 · 1,5² · 0,5 = 281 Nm W₂ = 250 · 0,2 · 9,81 · 0,05 = 25 Nm W₃ = 281 + 25 = 306 Nm W₄ = 306 · 180 = 55080 Nm/h me = 2 · 306 : 1,5² = 272 kg
5. Swinging mass with propelling force	W₁ = m · v² · 0,5 = 0,5 · J · ω² W₂ = (M · s) / R W₃ = W₁ + W₂ W₄ = W₃ · x v_D = (v · R) / L = ω · R me = (2 · W₃) / v_D²	m = 20 kg v = 1 m/s M = 50 R = 0,5 m L = 0,8 m x = 1500 1/h s = 0,012 m (gewählt) W₁ = 20 · 1² · 0,5 = 10 Nm W₂ = 50 · 0,012 : 0,5 = 1,2 Nm W₃ = 10 + 1,2 = 11,2 Nm W₄ = 306 · 180 = 16800 Nm/h v_D = 1 · 0,5 : 0,8 = 0,63 m/s me = 2 · 11,2 : 0,63² = 56 kg
6. Free falling mass	W₁ = m · g · h W₂ = m · g · s W₃ = W₁ + W₂ W₄ = W₃ · x v_D = √2 · g · h me = (2 · W₃) / v_D²	m = 30 kg h = 0,5 m x = 400 1/h s = 0,050 m (gewählt) W₁ = 30 · 0,5 · 9,81 = 147 Nm W₂ = 30 · 9,81 · 0,05 = 15 Nm W₃ = 147 + 15 = 162 Nm W₄ = 162 · 400 = 64800 Nm/h v_D = √2 · 9,81 · 0,5 = 3,13 m/s me = 2 · 162 : 3,13² = 33 kg
6.1 Mass rolling/sliding down incline	W₁ = m · g · h = m · v_D² · 0,5 W₂ = m · g · sinβ · s W₃ = W₁ + W₂ W₄ = W₃ · x v_D = √2 · g · h me = (2 · W₃) / v_D² 6.1a bei senkrechter Bewegung nach oben W₂ = (F – m · g· sinβ) · s 6.1b bei senkrechter Bewegung nach unten W₂ = (F + m · g· sinβ) · s	m = 500 kg h = 0,1 m x = 200 1/h ß = 10 °C W₁ = 500 · 9,81 · 0,1 = 490,5 Nm W₂ = 50 · 9,81 · sin(10) · 0,075 = 63,9 Nm W₃ = 490,5 + 63,9 = 554,4 Nm W₄ = 554,4 · 200 = 11880,0 Nm/h
6.2 Mass free falling about a pivot point	W₁ = m · g · h W₂ = 0 W₃ = W₁ + W₂ W₄ = W₃ · x v_D = √2 · g · h · (R / L) me = (2 · W₃) / v_D² tan α = s / R	m = 50 kg h = 1 m x = 50 1/h R = 300 mm L = 500 mm W₁ = 50 · 9,81 · 1 = 490,5 Nm W₂ = 0 W₃ = 490,5 + 0 = 490,5 Nm W₄ = 490,5 · 50 = 24525,0 Nm/h
7. Rotary index table with propelling torque	W₁ = m · v² · 0,25 = 0,5 · J · ω² W₂ = (M · s) / R W₃ = W₁ + W₂ W₄ = W₃ · x v_D = (v · R) / L = ω · R me = (2 · W₃) / v_D²	m = 1000 kg v = 1,1 m/s M = 1000 Nm s = 0,050 m (gewählt) L = 1,25 m R = 0,8 m x = 100 1/h W₁ = 1000 · 1,1² · 0,25 = 303 Nm W₂ = 300 · 0,025 : 0,8 = 63 Nm W₃ = 28 + 9 = 366 Nm W₄ = 37 · 1200 = 36600 Nm/h v_D = 1,1 · 0,8 : 1,25 = 0,7 m/s me = · 366 : 0,7² = 1494 kg
8. Swinging arm with propelling torque (uniform weight distribution)	W₁ = m · v² · 0,17 = 0,5 · J · ω² W₂ = (M · s) / R W₃ = W₁ + W₂ W₄ = W₃ · x v_D = (v · R) / L = ω · R me = (2 · W₃) / v_D²	J = 56 kgm² ω = 1 1/s M = 300 Nm s = 0,025 m (gewählt) L = 1,5 m R = 0,8 m x = 1200 1/h W₁ = 0,5 · 56 · 1² = 28 Nm W₂ = 300 · 0,025 : 0,8 = 9 Nm W₃ = 28 + 9 = 37 Nm W₄ = 37 · 1200 = 44400 Nm/h v_D = 1 · 0,8 = 0,8 m/s me = 2 · 37 : 0,8² = 116 kg
9. Swinging arm with propelling force (uniform weight distribution)	W₁ = m · v² · 0,17 = 0,5 · J · ω² W₂ = (F · r · s) / R = (M · s) / R W₃ = W₁ + W₂ W₄ = W₃ · x v_D = (v · R) / L = ω · R me = (2 · W₃) / v_D²	m = 1000 kg v = 2 m/s F = 7000 N M = 4200 Nm s = 0,050 m (gewählt) r = 0,6 m R = 0,8 m L = 1,2 m x = 900 1/h W₁ = 1000 · 2² · 0,17 = 680 Nm W₂ = 7000 · 0,6 · 0,05 : 0,8 = 263 Nm W₃ = 680 + 263 = 943 Nm W₄ = 943 · 900 = 848700 Nm/h v_D = 2 · 0,8 : 1,2 = 1,33 m/s me = 2 · 943 : 1,33² = 1066 kg
10. Mass lowered at controlled speed	W₁ =m · v2 · 0,5 W₂ = m · g · s W₃ = W₁ + W₂ W₄ = W₃ · x v_D = v me = (2 · W₃) / v_D²	m = 6000 kg v = 1,5 m/s s = 0,305 m (gewählt) x = 60 1/h W₁ = 6000 · 1,5² · 0,5 = 6750 Nm W₂ = 6000 · 9,81 · 0,305 = 17952 Nm W₃ = 6750 + 17 952 = 24702 Nm W₄ = 24702 · 60 = 1482120 Nm/h me = 2 · 24702 : 1,5² = 21957 kg

Application

Formulae

Example

1. Mass without propelling force

W₁ = m · v² · 0,5

W₂ = 0

W₃ = W₁ + W₂

W₄ = W₃ · x

v_D = v

me = m

m = 100 kg
v = 1,5 m/s
x = 500 1/h
s = 0,050 m (gewählt)

W₁ = 100 · 1,5² · 0,5 = 113 Nm
W₂ = 0
W₃ = 113 + 0 = 113 Nm
W₄ = 113 · 500 = 56500 Nm/h
me = m = 100 kg

2. Mass with propelling force

W₁ = m · v2 · 0,5

W₂ = F · s

W₃ = W1 + W2

W₄ = W3 · x

v_D = v

me = (2 · W₃) / v_D²

2.1 bei senkrechter Bewegung nach oben

W₂ = (F – m · g) · s

2.2 bei senkrechter Bewegung nach unten

W₂ = (F + m · g) · s

m = 36 kg

¹v = 1,5 m/s

F = 400 N

x = 1000 1/h

s = 0,025 m (gewählt)

W₁ = 36 · 1,5² · 0,5 = 41 Nm

W₂ = 400 · 0,025 = 10 Nm

W₃ = 41 + 10 = 51 Nm

W₄ = 51 · 1000 = 51000 Nm/h

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 ·HM · s) / v

W₃ = W₁ + W₂

W₄ = W₃ · x

v_D = v

me = (2 · W₃) / v_D²

m = 800 kg
v = 1,2 m/s
HM = 2,5
P = 4 kW
x = 100 1/h
s = 0,100 m (gewählt)

W₁ = 800 · 1,2² · 0,5 = 576 Nm
W₂ = 1000 · 4 · 2,5 · 0,1 : 1,2 = 834 Nm
W₃ = 576 + 834 = 1410 Nm
W₄ = 1410 · 100 = 141000 Nm/h
me = 2 · 1410 : 1,2² = 1958 kg

^{Note: Do not forget to include the rotational energy of}
^{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 · x

v_D = v

me = (2 · W₃) / v_D²

m = 250 kg
v = 1,5 m/s
x = 180 1/h
(Stahl/Guss) μ = 0,2
s = 0,050 m (gewählt)

W₁ = 250 · 1,5² · 0,5 = 281 Nm
W₂ = 250 · 0,2 · 9,81 · 0,05 = 25 Nm
W₃ = 281 + 25 = 306 Nm
W₄ = 306 · 180 = 55080 Nm/h
me = 2 · 306 : 1,5² = 272 kg

5. Swinging mass with propelling force

W₁ = m · v² · 0,5 = 0,5 · J · ω²

W₂ = (M · s) / R

W₃ = W₁ + W₂

W₄ = W₃ · x

v_D = (v · R) / L = ω · R

me = (2 · W₃) / v_D²

m = 20 kg
v = 1 m/s
M = 50
R = 0,5 m
L = 0,8 m
x = 1500 1/h
s = 0,012 m (gewählt)

W₁ = 20 · 1² · 0,5 = 10 Nm
W₂ = 50 · 0,012 : 0,5 = 1,2 Nm
W₃ = 10 + 1,2 = 11,2 Nm
W₄ = 306 · 180 = 16800 Nm/h
v_D = 1 · 0,5 : 0,8 = 0,63 m/s
me = 2 · 11,2 : 0,63² = 56 kg

6. Free falling mass

W₁ = m · g · h

W₂ = m · g · s

W₃ = W₁ + W₂

W₄ = W₃ · x

v_D = √2 · g · h

me = (2 · W₃) / v_D²

m = 30 kg
h = 0,5 m
x = 400 1/h
s = 0,050 m (gewählt)

W₁ = 30 · 0,5 · 9,81 = 147 Nm
W₂ = 30 · 9,81 · 0,05 = 15 Nm
W₃ = 147 + 15 = 162 Nm
W₄ = 162 · 400 = 64800 Nm/h
v_D = √2 · 9,81 · 0,5 = 3,13 m/s
me = 2 · 162 : 3,13² = 33 kg

6.1 Mass rolling/sliding down incline

W₁ = m · g · h = m · v_D² · 0,5

W₂ = m · g · sinβ · s

W₃ = W₁ + W₂

W₄ = W₃ · x

v_D = √2 · g · h

me = (2 · W₃) / v_D²

6.1a bei senkrechter Bewegung nach oben

W₂ = (F – m · g· sinβ) · s

6.1b bei senkrechter Bewegung nach unten

W₂ = (F + m · g· sinβ) · s

m = 500 kg
h = 0,1 m
x = 200 1/h
ß = 10 °C

W₁ = 500 · 9,81 · 0,1 = 490,5 Nm
W₂ = 50 · 9,81 · sin(10) · 0,075 = 63,9 Nm
W₃ = 490,5 + 63,9 = 554,4 Nm
W₄ = 554,4 · 200 = 11880,0 Nm/h

6.2 Mass free falling about a pivot point

W₁ = m · g · h

W₂ = 0

W₃ = W₁ + W₂

W₄ = W₃ · x

v_D = √2 · g · h · (R / L)

me = (2 · W₃) / v_D²

tan α = s / R

m = 50 kg
h = 1 m
x = 50 1/h
R = 300 mm
L = 500 mm

W₁ = 50 · 9,81 · 1 = 490,5 Nm
W₂ = 0
W₃ = 490,5 + 0 = 490,5 Nm
W₄ = 490,5 · 50 = 24525,0 Nm/h

7. Rotary index table with propelling torque

W₁ = m · v² · 0,25 = 0,5 · J · ω²

W₂ = (M · s) / R

W₃ = W₁ + W₂

W₄ = W₃ · x

v_D = (v · R) / L = ω · R

me = (2 · W₃) / v_D²

m = 1000 kg
v = 1,1 m/s
M = 1000 Nm
s = 0,050 m (gewählt)
L = 1,25 m
R = 0,8 m
x = 100 1/h

W₁ = 1000 · 1,1² · 0,25 = 303 Nm
W₂ = 300 · 0,025 : 0,8 = 63 Nm
W₃ = 28 + 9 = 366 Nm
W₄ = 37 · 1200 = 36600 Nm/h
v_D = 1,1 · 0,8 : 1,25 = 0,7 m/s
me = · 366 : 0,7² = 1494 kg

8. Swinging arm with propelling torque (uniform weight distribution)

W₁ = m · v² · 0,17 = 0,5 · J · ω²

W₂ = (M · s) / R

W₃ = W₁ + W₂

W₄ = W₃ · x

v_D = (v · R) / L = ω · R

me = (2 · W₃) / v_D²

J = 56 kgm²
ω = 1 1/s
M = 300 Nm
s = 0,025 m (gewählt)
L = 1,5 m
R = 0,8 m
x = 1200 1/h

W₁ = 0,5 · 56 · 1² = 28 Nm
W₂ = 300 · 0,025 : 0,8 = 9 Nm
W₃ = 28 + 9 = 37 Nm
W₄ = 37 · 1200 = 44400 Nm/h
v_D = 1 · 0,8 = 0,8 m/s
me = 2 · 37 : 0,8² = 116 kg

9. Swinging arm with propelling force (uniform weight distribution)

W₁ = m · v² · 0,17 = 0,5 · J · ω²

W₂ = (F · r · s) / R = (M · s) / R

W₃ = W₁ + W₂

W₄ = W₃ · x

v_D = (v · R) / L = ω · R

me = (2 · W₃) / v_D²

m = 1000 kg
v = 2 m/s
F = 7000 N
M = 4200 Nm
s = 0,050 m (gewählt)
r = 0,6 m
R = 0,8 m
L = 1,2 m
x = 900 1/h

W₁ = 1000 · 2² · 0,17 = 680 Nm
W₂ = 7000 · 0,6 · 0,05 : 0,8 = 263 Nm
W₃ = 680 + 263 = 943 Nm
W₄ = 943 · 900 = 848700 Nm/h
v_D = 2 · 0,8 : 1,2 = 1,33 m/s
me = 2 · 943 : 1,33² = 1066 kg

10. Mass lowered at controlled speed

W₁ =m · v2 · 0,5

W₂ = m · g · s

W₃ = W₁ + W₂

W₄ = W₃ · x

v_D = v

me = (2 · W₃) / v_D²

m = 6000 kg
v = 1,5 m/s
s = 0,305 m (gewählt)
x = 60 1/h

W₁ = 6000 · 1,5² · 0,5 = 6750 Nm
W₂ = 6000 · 9,81 · 0,305 = 17952 Nm
W₃ = 6750 + 17 952 = 24702 Nm
W₄ = 24702 · 60 = 1482120 Nm/h
me = 2 · 24702 : 1,5² = 21957 kg

Use calculation tool and order online

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 v_D = 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 v_D = v = 2 m/s s = 0,1 m W₁ = 200 Nm W₂ = 200 Nm W₃ = 400 Nm me = (2 · 400) / 4 = 200 kg ^{FORMEL: ME = (2 · W3) / VD2}
C Mass without propelling force direct against shock absorber	m = 20 kg v_D = v = 2 m/s W₁ = W₃ = 40 Nm me = (2 · 40) / 2² = 20 kg ^{FORMEL: ME = M}
D Mass without propelling force with mechanical advantage	m = 20 kg v = 2 m/s v_D = 0,5 m/s s = 0,1 m W₁ = W₃ = 40 Nm me = (2 · 40) / 0,5² = 320 kg ^{FORMEL: ME = (2 · W3) / VD2}