Sharing load between linear actuators
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 4 hours ago
RedHatter
1134
add a comment |
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 4 hours ago
RedHatter
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
add a comment |
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 4 hours ago
RedHatter
1134
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 4 hours ago
RedHatter
1134
asked 4 hours ago
RedHatter
1134
asked 4 hours ago
RedHatter
1134
asked 4 hours ago
RedHatter
1134
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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.
answered 2 hours ago
kamran
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
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered 2 hours ago
kamran
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
add a comment |
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.
answered 2 hours ago
kamran
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
add a comment |
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.
answered 2 hours ago
kamran
3,5291410
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.
answered 2 hours ago
kamran
3,5291410
edited 2 hours ago
answered 2 hours ago
kamran
3,5291410
answered 2 hours ago
kamran
3,5291410
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
add a comment |
@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
add a comment |
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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