Sharing load between linear actuators












1














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







share|improve this question






















  • Is the weight on the table distributed equally? But if you simplify the problem then yes they sum up.
    – joojaa
    3 hours ago












  • The weight is mostly centered.
    – RedHatter
    3 hours ago
















1














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







share|improve this question






















  • Is the weight on the table distributed equally? But if you simplify the problem then yes they sum up.
    – joojaa
    3 hours ago












  • The weight is mostly centered.
    – RedHatter
    3 hours ago














1












1








1







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







share|improve this question













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






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 4 hours ago









RedHatter

1134




1134












  • Is the weight on the table distributed equally? But if you simplify the problem then yes they sum up.
    – joojaa
    3 hours ago












  • The weight is mostly centered.
    – RedHatter
    3 hours ago


















  • Is the weight on the table distributed equally? But if you simplify the problem then yes they sum up.
    – joojaa
    3 hours ago












  • The weight is mostly centered.
    – RedHatter
    3 hours ago
















Is the weight on the table distributed equally? But if you simplify the problem then yes they sum up.
– joojaa
3 hours ago






Is the weight on the table distributed equally? But if you simplify the problem then yes they sum up.
– joojaa
3 hours ago














The weight is mostly centered.
– RedHatter
3 hours ago




The weight is mostly centered.
– RedHatter
3 hours ago










1 Answer
1






active

oldest

votes


















2














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 throughs 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 and.



then the loads distributed to the four legs are going to change as follows.



The load to the right hand side pair of leges $$ is = l00*x/X = R_{looad}$$
$$ R_{load}*y/Y =load on the upper right leg. $$
We repeat this process and find individual loads fore each leg.





share























  • @Wasabi, he says the weight is mostly centered in his comment. Yes of course it is assumed the wait is uniformly distributed, or balanced.
    – kamran
    2 hours ago











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1 Answer
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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 throughs 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 and.



then the loads distributed to the four legs are going to change as follows.



The load to the right hand side pair of leges $$ is = l00*x/X = R_{looad}$$
$$ R_{load}*y/Y =load on the upper right leg. $$
We repeat this process and find individual loads fore each leg.





share























  • @Wasabi, he says the weight is mostly centered in his comment. Yes of course it is assumed the wait is uniformly distributed, or balanced.
    – kamran
    2 hours ago
















2














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 throughs 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 and.



then the loads distributed to the four legs are going to change as follows.



The load to the right hand side pair of leges $$ is = l00*x/X = R_{looad}$$
$$ R_{load}*y/Y =load on the upper right leg. $$
We repeat this process and find individual loads fore each leg.





share























  • @Wasabi, he says the weight is mostly centered in his comment. Yes of course it is assumed the wait is uniformly distributed, or balanced.
    – kamran
    2 hours ago














2












2








2






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 throughs 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 and.



then the loads distributed to the four legs are going to change as follows.



The load to the right hand side pair of leges $$ is = l00*x/X = R_{looad}$$
$$ R_{load}*y/Y =load on the upper right leg. $$
We repeat this process and find individual loads fore each leg.





share














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 throughs 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 and.



then the loads distributed to the four legs are going to change as follows.



The load to the right hand side pair of leges $$ is = l00*x/X = R_{looad}$$
$$ R_{load}*y/Y =load on the upper right leg. $$
We repeat this process and find individual loads fore each leg.






share













share


share








edited 2 hours ago

























answered 2 hours ago









kamran

3,5291410




3,5291410












  • @Wasabi, he says the weight is mostly centered in his comment. Yes of course it is assumed the wait is uniformly distributed, or balanced.
    – kamran
    2 hours ago


















  • @Wasabi, he says the weight is mostly centered in his comment. Yes of course it is assumed the wait is uniformly distributed, or balanced.
    – kamran
    2 hours ago
















@Wasabi, he says the weight is mostly centered in his comment. Yes of course it is assumed the wait is uniformly distributed, or balanced.
– kamran
2 hours ago




@Wasabi, he says the weight is mostly centered in his comment. Yes of course it is assumed the wait is uniformly distributed, or balanced.
– kamran
2 hours ago


















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