Okay, so recently I got a Dewalt 735 Thickness Planer.

Dewalt 735

It's great at what it does. (It even LOOKS like it's ready to consume the tender flesh of lumber, like some twisted Victorian lumberjack... though, maybe that's just me).

Jack the Crosscutter

Anyway, the problem I've been experiencing with the tool is that, while planing, should a board exceed a certain length, such that the midpoint of the piece is no longer supported while some portion of it is still being cut, the board tips, resulting in snipe (a gouge in the board, resulting from the angle shifting mid-operation).

Illustrating snipe

Okay, no problem: I'll just build some supports. Now, I just so happen to have some Heavy Duty 22" full-extension 150-pound-rated drawer slides - about a dozen of em - from a different project. Cool. I'll run a roller bar between em and make the whole deal extendible at need. I can even add flip down leg supports if needed (though if the slides are strong enough, I'd rather not).

Extensive support

As I was doodling this out, my son walks in and asks what I'm looking at. I muse aloud that it's a shame I don't have longer slides. 22 inches isn't terribly huge for infeed/outfeed.

Then he asks me: "So why can't you just hook two together? Just stick another piece between em so you can screw into both sides of that." Heh. From the mouth... of... babes?


Now, intuitively, this seems like a bad call (certainly without the leg support). But I'll be damned if I can figure out WHY or what to tell him. Some long-forgotten, niggling memory of similar suggests this is a Static Equilibrium problem, which - despite it having been decades since I've looked at any of this - I have some vague recollection on how to calculate. But it's the addition of the strut that hooks the two slides together I cannot account for.

enter image description here

Now, although the finished version of either approach will have two slides with a bar spanning them, I'm reducing this to two dimensions for my own old brain to walk through.

  • HOW do I figure out how much load would be put on the roller if I were to take his approach?

  • COULD a 150 lb capacity (well, two combined, for the 2d example we're using, per side - four total in actual implementation) support a portion of the mass of a single, call it 25lb board?

  • WOULD it need legs?

Edit additional clarification: the dimensions of the mobile base are 38" wide (the axis perpendicular to the force) by 26" deep (the axis parallel) by 30" tall (including the locking steel casters it rests atop).

It is constructed of the better part of a full sheet of 3/4" Baltic Birch plywood (call it 75 lbs), and contains a small cabinet containing various bits and bobs (~30 lbs). Also seated alongside the planer is a cast iron bench top jointer (~65 lbs), which itself has 2 drawers beneath it containing various sleds, jigs, and blades (~20 lbs). This is all in addition to the four 4" steel casters (~2.5 lbs ea.) the fixture rests atop, and the 105 lbs. planer itself. This accounts for an approximate total weight of ~300 lbs and change. I've been simply treating it as the proverbial "immovable object" in the equation, Archimedes be damned, but that is, I admit, simply because it hadn't occurred to me it's stability might be compromised.

  • $\begingroup$ I hope someone will help you calc this - it's gone midnight here and I'm on my phone, so I'm going to say instead - if you have the slides already, just prototype and test 👌 $\endgroup$ Nov 30, 2020 at 0:11
  • 1
    $\begingroup$ The drawer slide rating will be assuming that load is spread over the length of the slide, and is downwards - adding another slide will mean the force on the first slide also has a rotation/bending element. Whether that's OK depends on the design details of the slides, really $\endgroup$ Nov 30, 2020 at 0:13
  • $\begingroup$ Yeah, I'm gonna have to. All my instincts say it's not gonna behave as I want it to, but this is compounded by an insidious voice suggesting that nor will it act as I EXPECT it to, either. $\endgroup$
    – NerdyDeeds
    Nov 30, 2020 at 5:40
  • $\begingroup$ Even if the slides support the load, they may deflect too much at full extension. $\endgroup$
    – Drew
    Dec 2, 2020 at 10:00

1 Answer 1


Your best bet is to rig the planer with a couple of carpenter's horses or something on the sides. However, if we want to go your way need to check the following.

  • Overturning moment on the fully extended setting with 44" cantilevered rollers.

  • Torque on the joint between the two drawer slides.

Let's say your machine and its base weigh 40# and the base is 24 inches wide. And half of 25#board weight is on the roller half on the blades, say 10lbs sum of the roller, and half of 4 sliders.

We compare the moment of the board's end with the moment due to the base and planer and the other end of the board.

$$\text{moment due to the base side}\ (40+12.5)12" = 630lbs.inch \\ \text{moment due to the roller side}\ (12.5 +10)(44+12)=1260lbs.inch>>630$$

Therefore the planer will tip over and we need legs.

As for the torque between the sliders, we have a torque of $\tau= (12.5+7.5)*22=177.5lbs.inch$

say 200, do you think you slider can take that much torque?

  • $\begingroup$ I updated the question with additional information, including the primary characteristics of the base itself. The issue I run into with factoring the weight of the board as a 50/50 split is it's a sliding load whose angle of repose shifts during its transition, though, again intuitively, I cannot see how I could avoid the legs myself, either. 1260lbs per inch, though... surely that's mitigated in SOME capacity, or the planer itself would suffer damage whenever it was used, with or without the stand... wouldn't it? I don't doubt your math, but that seems awfully high. Am I deluded? $\endgroup$
    – NerdyDeeds
    Nov 30, 2020 at 5:51
  • $\begingroup$ Also, to answer you question, the slides are rated for 150 lbs. +/- 20%, so we'll say max theoretical load of 180. It's the conjunction of two slides per side, four total, and the rigid central member connecting the two that makes me wonder if some of that load isn't distributed in some obtuse capacity. Likr $\endgroup$
    – NerdyDeeds
    Nov 30, 2020 at 5:56
  • $\begingroup$ it's not too much it is 1260/12 =~100 lbs times feet. which equal to your 22lbsx 6 feet. $\endgroup$
    – kamran
    Nov 30, 2020 at 5:59
  • $\begingroup$ Like, does having two pivot points (two slides affixed to the base) distribute the moment of deformation? Does having a rigid cross-member between the slides? Likewise, does the 300# mass of the base ameliorate the overturning moment? Or does it remain the same... regardless, will the slides themselves FAIL? $\endgroup$
    – NerdyDeeds
    Nov 30, 2020 at 6:05
  • $\begingroup$ we ignored your inadvertently leaning on the board or the vibration and possible jerking of the rig. yes by the new information the resisting moment of the base is is bigger but still not solid. $\endgroup$
    – kamran
    Nov 30, 2020 at 6:06

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