Sharing load between linear actuators
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I'm building a motorized standing desk and am trying to calculate what linear actuators I need. If I have 4 linear actuators (one on each corner) each able to lift 25kg does that mean that the desk would be able to lift a full 100kg?
In other words, when linear actuators are sharing a load is the total drive force equal to the sum of the actuators individual drive force? Or is it more complicated than that?
linear-motion
asked Dec 28 '18 at 19:38
RedHatterRedHatter
1154
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add a comment |
$begingroup$
I'm building a motorized standing desk and am trying to calculate what linear actuators I need. If I have 4 linear actuators (one on each corner) each able to lift 25kg does that mean that the desk would be able to lift a full 100kg?
In other words, when linear actuators are sharing a load is the total drive force equal to the sum of the actuators individual drive force? Or is it more complicated than that?
linear-motion
asked Dec 28 '18 at 19:38
RedHatterRedHatter
1154
$endgroup$
$begingroup$
Is the weight on the table distributed equally? But if you simplify the problem then yes they sum up.
$endgroup$
– joojaa
Dec 28 '18 at 21:02
$begingroup$
The weight is mostly centered.
$endgroup$
– RedHatter
Dec 28 '18 at 21:08
add a comment |
$begingroup$
I'm building a motorized standing desk and am trying to calculate what linear actuators I need. If I have 4 linear actuators (one on each corner) each able to lift 25kg does that mean that the desk would be able to lift a full 100kg?
In other words, when linear actuators are sharing a load is the total drive force equal to the sum of the actuators individual drive force? Or is it more complicated than that?
linear-motion
asked Dec 28 '18 at 19:38
RedHatterRedHatter
1154
$endgroup$
I'm building a motorized standing desk and am trying to calculate what linear actuators I need. If I have 4 linear actuators (one on each corner) each able to lift 25kg does that mean that the desk would be able to lift a full 100kg?
In other words, when linear actuators are sharing a load is the total drive force equal to the sum of the actuators individual drive force? Or is it more complicated than that?
linear-motion
linear-motion
asked Dec 28 '18 at 19:38
RedHatterRedHatter
1154
asked Dec 28 '18 at 19:38
RedHatterRedHatter
1154
asked Dec 28 '18 at 19:38
RedHatterRedHatter
1154
asked Dec 28 '18 at 19:38
RedHatterRedHatter
1154
asked Dec 28 '18 at 19:38
RedHatterRedHatter
1154
1154
$begingroup$
Is the weight on the table distributed equally? But if you simplify the problem then yes they sum up.
$endgroup$
– joojaa
Dec 28 '18 at 21:02
$begingroup$
The weight is mostly centered.
$endgroup$
– RedHatter
Dec 28 '18 at 21:08
add a comment |
$begingroup$
Is the weight on the table distributed equally? But if you simplify the problem then yes they sum up.
$endgroup$
– joojaa
Dec 28 '18 at 21:02
$begingroup$
The weight is mostly centered.
$endgroup$
– RedHatter
Dec 28 '18 at 21:08
$begingroup$
Is the weight on the table distributed equally? But if you simplify the problem then yes they sum up.
$endgroup$
– joojaa
Dec 28 '18 at 21:02
$begingroup$
Is the weight on the table distributed equally? But if you simplify the problem then yes they sum up.
$endgroup$
– joojaa
Dec 28 '18 at 21:02
$begingroup$
The weight is mostly centered.
$endgroup$
– RedHatter
Dec 28 '18 at 21:08
$begingroup$
The weight is mostly centered.
$endgroup$
– RedHatter
Dec 28 '18 at 21:08
add a comment |
1 Answer
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oldest
votes
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Yes, the four will share the 100kg, each carrying 25kg.
Only thing is you want to connect them to same rocker switch and install the linear actuators symmetrically with respect to the center of your desk.
Edit
After some comments I modify my answer this way. Lets say there is a heavy vise you need to have on the desk and it throws the center of gravity of the desk from the center to x and y centimeters off the center. Lets call the distance between the base of the legs X on the long side and Y on the short side.
Then the loads distributed to the four legs are going to change as follows.
The load to the right hand side pair of legs $$ is = l00*x/X = R_{load}$$
$$ R_{load}*y/Y =load on the upper right leg. $$
We repeat this process and find the individual loads for each leg.
edited Dec 29 '18 at 1:05
RedHatter
1154
answered Dec 28 '18 at 21:27
kamrankamran
4,5452511
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, the four will share the 100kg, each carrying 25kg.
Only thing is you want to connect them to same rocker switch and install the linear actuators symmetrically with respect to the center of your desk.
Edit
After some comments I modify my answer this way. Lets say there is a heavy vise you need to have on the desk and it throws the center of gravity of the desk from the center to x and y centimeters off the center. Lets call the distance between the base of the legs X on the long side and Y on the short side.
Then the loads distributed to the four legs are going to change as follows.
The load to the right hand side pair of legs $$ is = l00*x/X = R_{load}$$
$$ R_{load}*y/Y =load on the upper right leg. $$
We repeat this process and find the individual loads for each leg.
edited Dec 29 '18 at 1:05
RedHatter
1154
answered Dec 28 '18 at 21:27
kamrankamran
4,5452511
$endgroup$
add a comment |
$begingroup$
Yes, the four will share the 100kg, each carrying 25kg.
Only thing is you want to connect them to same rocker switch and install the linear actuators symmetrically with respect to the center of your desk.
Edit
After some comments I modify my answer this way. Lets say there is a heavy vise you need to have on the desk and it throws the center of gravity of the desk from the center to x and y centimeters off the center. Lets call the distance between the base of the legs X on the long side and Y on the short side.
Then the loads distributed to the four legs are going to change as follows.
The load to the right hand side pair of legs $$ is = l00*x/X = R_{load}$$
$$ R_{load}*y/Y =load on the upper right leg. $$
We repeat this process and find the individual loads for each leg.
edited Dec 29 '18 at 1:05
RedHatter
1154
answered Dec 28 '18 at 21:27
kamrankamran
4,5452511
$endgroup$
add a comment |
$begingroup$
Yes, the four will share the 100kg, each carrying 25kg.
Only thing is you want to connect them to same rocker switch and install the linear actuators symmetrically with respect to the center of your desk.
Edit
After some comments I modify my answer this way. Lets say there is a heavy vise you need to have on the desk and it throws the center of gravity of the desk from the center to x and y centimeters off the center. Lets call the distance between the base of the legs X on the long side and Y on the short side.
Then the loads distributed to the four legs are going to change as follows.
The load to the right hand side pair of legs $$ is = l00*x/X = R_{load}$$
$$ R_{load}*y/Y =load on the upper right leg. $$
We repeat this process and find the individual loads for each leg.
edited Dec 29 '18 at 1:05
RedHatter
1154
answered Dec 28 '18 at 21:27
kamrankamran
4,5452511
$endgroup$
Yes, the four will share the 100kg, each carrying 25kg.
Only thing is you want to connect them to same rocker switch and install the linear actuators symmetrically with respect to the center of your desk.
Edit
After some comments I modify my answer this way. Lets say there is a heavy vise you need to have on the desk and it throws the center of gravity of the desk from the center to x and y centimeters off the center. Lets call the distance between the base of the legs X on the long side and Y on the short side.
Then the loads distributed to the four legs are going to change as follows.
The load to the right hand side pair of legs $$ is = l00*x/X = R_{load}$$
$$ R_{load}*y/Y =load on the upper right leg. $$
We repeat this process and find the individual loads for each leg.
edited Dec 29 '18 at 1:05
RedHatter
1154
answered Dec 28 '18 at 21:27
kamrankamran
4,5452511
edited Dec 29 '18 at 1:05
RedHatter
1154
edited Dec 29 '18 at 1:05
RedHatter
1154
edited Dec 29 '18 at 1:05
RedHatter
1154
1154
answered Dec 28 '18 at 21:27
kamrankamran
4,5452511
answered Dec 28 '18 at 21:27
kamrankamran
4,5452511
answered Dec 28 '18 at 21:27
kamrankamran
4,5452511
4,5452511
add a comment |
add a comment |
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$begingroup$
Is the weight on the table distributed equally? But if you simplify the problem then yes they sum up.
$endgroup$
– joojaa
Dec 28 '18 at 21:02
$begingroup$
The weight is mostly centered.
$endgroup$
– RedHatter
Dec 28 '18 at 21:08