# Thread: Calculating forces on linear bearings ...

1. I've decided on my router that I'm going to use supported linear bearing rails on the X-axis. Two parallel rails, with two bearings on each fixed to the gantry.

Let:
• x = distace between two bearings
• F = horizontal cutting force parallel to X axis
• m = mass of gantry
• Z = distance from cutter to centre point of Z axis linear bearings.
• z = distance from X linear bearings to X axis leadscrew.
Using those I've arrived at the follwing formula (by taking moments and resolving the forces) for the force on a single bearing:

Max value = m*g/4 + F/(2x)*(Z+z)
Min value = m*g/4 - F/(2x)*(Z+z)

Which shows the force on the bearings could act upwards if the Z axis is long enough, or the bearing spacing is small.

My question is, have I mised anything here in the formulas and what would one normally chose for x in relation to Z:question:
If I put the numbers in for my router (x=0.3, z=0.5 ish), cutting steel, then I get -50N to 200N. Well within the 980N/-340N rating for 25mm bearings...and within rating even for 12mm bearings. I'm not sure which to go for. Two 25mm rails, or maybe get 4 rails and have 16mm...:question:

2. You've not allowed for the unequal force when cutting of centre.

Ideally, you should calculate the static load on each bearing.
Then calculate the force applied in the worst case scenario of the maximum cutting force being applied at one side of the gantry i.e. all the cutting force being applied through just 2 bearings.

In an ideal world, where the gantry is completely flex free, the cutting force would be transmitted through all 4 bearings, but in the real world it's not going to happen. Of course, some of the force will still be transmitted through the opposite side, but without lots of calculations or FEA modeling, it's simpler to use the worst case scenario.

3. Thanks for that. It did occur to me that the bearing forces on opposite sides of the gantry wouldn't necessarily be equal, but I wasn't sure ... so I neglected it

That makes it, for worst case, m*g/4 + F/x*(Z+z). That makes it 340N and -63N for my router which is still well within the ratings.

I'll try working out the equations for if you've got 8 bearings and 4 rails next. That's a bit more difficult though.

Another thing I've neglected is the cutting force parallel to Y (and Z if you're drilling). The former isn't going to be that great, and I think the drilling force is only significant if it's more than the weight of the gantry ...though for Z there is a moment on the bearings assuming the drill isn't centered between them.

Can anyone point me to how to work out cutting force when drilling?

4. For 4 rails, the forces are still the same, just the force is applied over more rails/bearings.

For simplicity, I'd assume each pair of rails/bearings will see an equal load, as the flex in the gantry/bearing mounts on each side is going to be pretty negligible. Of course, that depends on how you plan on mounting each rail in relation to the ballscrew.

5. Originally Posted by m_c
Of course, that depends on how you plan on mounting each rail in relation to the ballscrew.
I don't follow that last sentence?

I'm using supported round rail, with the supported surface mounted flat on the bed.

Just realised I should also include the additional load on the bearings from the gantry accelerating. Need to know the where center of mass of the gantry is for that...can get it from the CAD model I suppose. Might be taking it too far!

6. The pivot point for any cutting forces is going to be the ballscrew.

7. Originally Posted by m_c
The pivot point for any cutting forces is going to be the ballscrew.
Oh, yes i see now - if the screw isn't centered between the bearings the forces won't be equal. That's where I took moments about originally for convenience.

Bed time for me now - lecture at 9am!

8. Actually rephrase that, the pivot point for any cutting forces parallel with the x-axis is going to be the lead screw.

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