A couple more diagrams on axe balance.

[FortyTwoBlades]

I maintain that this is incorrect. The middle axe handle "experiences" a given (unchanging) torque resulting from the head imbalance being "countered" by the two hands supporting the handle, no matter where the two hands are supporting the handle along the green axis. As long as there are two hands holding the handle, the axis of support is the green line, and it does not get "less off-axis" with the natural "axle" as the hands move. The green line is at a consistent geometric angle with the red line, even at the point where the two lines converge.

An easy example:

The weight of the axe head plus handle is 6 pounds.
The Center of Gravity is 2 inches (measured perpendicular) from the green line.
The axe is held horizontally by two hands.
The resulting torque "experienced" by the handle (as well as total torque "experienced" by the hands) is equal to
6 pounds times 2/12 feet = 1 foot pound.

This 1 foot pound of torque is what the two hands (in combination) are providing, no matter where on the handle they are located. This 1 foot pound of torque does not diminish or go away, as long as the axe head is being held horizontally by two hands.

It's measured perpendicular to the force (red line) and so, as far as I can see, as the point of the rotation converges with the axis of the force the torque decreases. Torque is measured about a point rather than about an axis, and as the point changes so can the torque. Since I used a vector program I upscaled the image to a 36" handle and the points 1, 2, and 3 were serendipitously 2.2", 1.2", and 0.2" perpendicular from the red line. If we made things super simple and use just 1lb of force, you have 2.2 foot pounds of torque at point 1, 1.2 foot pounds of torque at point 2, and 0.2 footpounds of torque at point 3. The closer your hand (the point being used to take the measurement) is to the red natural axle from which the force is deriving its strength, the lower the torque experienced.

 

[Square_peg]

I believe 42 is correct in that the torque changes as the location of the grip on the handle changes.

Imagine this, look at the center image in his first diagram. Imagine the handle is a round shaft. Suppose that round shaft is supported solely by a bearing at location 1. It's easy to see that the axe will roll down. Now imaging the shaft is supported solely by a bearing at location 2. The axe will still rotate down but not with as much force as it would have from location 1. Now slide the bearing back to the blue dot at the bottom of the handle. Do you see that it now coincides with the center of gravity of the axe? The bearing at the blue dot experiences no off-center torque because it lays on the axe's rotational center.

 

[Steve Tall]

It's measured perpendicular to the force (red line) and so, as far as I can see, as the point of the rotation converges with the axis of the force the torque decreases. Torque is measured about a point rather than about an axis, and as the point changes so can the torque. Since I used a vector program I upscaled the image to a 36" handle and the points 1, 2, and 3 were serendipitously 2.2", 1.2", and 0.2" perpendicular from the red line. If we made things super simple and use just 1lb of force, you have 2.2 foot pounds of torque at point 1, 1.2 foot pounds of torque at point 2, and 0.2 footpounds of torque at point 3. The closer your hand (the point being used to take the measurement) is to the red natural axle from which the force is deriving its strength, the lower the torque experienced.

Sorry, but that is wrong. The forces on the handle and head are transmitted via the green line, not the red line.

(While the green line is the actual physical axis of the handle, the red line in this case is imaginary, resulting from dangling the axe upside down from a pivoting support at point 3. The red line signifies that there is an equal mass on either side of it, but it doesn't dictate how the forces are applied to/from the head.)

Re: Torque is measured about a point rather than about an axis, and as the point changes so can the torque.
Torque actually can be transmitted (and measured) about an axis. The driveshaft of a car is an example. So is a torque wrench with an extension and socket (at 90 degrees to the wrench handle axis). The torque doesn't change along the length of the shaft (assuming no deformation of the shaft).

The actual axis of rotation for an axe depends on where the handle is being held/supported. For the middle axe in the top diagram, this is simple because it's a straight handle, and the axis of rotation between the supporting hands is always the green line. (Keep in mind that we are only discussing the axis of rotation due to an imbalanced head held horizontally.)

The lever arm is measured perpendicular to the axis of rotation (green line).

The problem can be simplified down to just a 3'-long weightless shaft, with a 6-pound weight attached to one end having a 2" offset. The shaft is held in two hands, keeping the weight (with offset) horizontal. The resulting "torque" applied to the shaft by the offset weight is 1 foot pound (as calculated in a previous post), and the torque transmitted along the shaft toward the supports is this same 1 foot pound, and the torque countered by the combined resistance ("counter torque" it was called) of the two hands, wherever they are located along the shaft, holding it and keeping it from rotating, is the same 1 foot pound.

The green line is the axis of rotation at the center of the shaft (axe handle), and the red line doesn't even come into play in this example.

 

[Steve Tall]

I believe 42 is correct in that the torque changes as the location of the grip on the handle changes.

Imagine this, look at the center image in his first diagram. Imagine the handle is a round shaft. Suppose that round shaft is supported solely by a bearing at location 1. It's easy to see that the axe will roll down. Now imaging the shaft is supported solely by a bearing at location 2. The axe will still rotate down but not with as much force as it would have from location 1. Now slide the bearing back to the blue dot at the bottom of the handle. Do you see that it now coincides with the center of gravity of the axe? The bearing at the blue dot experiences no off-center torque because it lays on the axe's rotational center.

I wrote and posted my recent comment before seeing this. I think that's a good example, however I disagree with your conclusion. One important feature is that the shaft is straight (like the handle) and the bearings at locations 1, 2, and 3 (blue dot) are thus all aligned in the same direction, to spin along the axis of the shaft (handle). The bearing at the blue dot might overlap or coincide with the axe's rotational center (or "axle" as 42 calls it), but the bearing is constrained from spinning along that "axle", since it can only spin around the axis of the shaft (handle).

Another important feature is that the handle is supported by two hands (or two bearings). This constrains the rotation (due to an imbalanced head) to keep it only rotating along the axis between the two supports (bearings or hands).

And even if the shaft were supported solely by a bearing at location 2, I'm not seeing how the axe wouldn't rotate down as much, unless we consider the nature of loaded bearings having some braking effect on rotation (resulting from the cantilevered weight of the axe head), but this is not part of the picture here when we are looking only at the effects of an unbalanced head.

 

[Square_peg]

The bearing at the blue dot might overlap or coincide with the axe's rotational center (or "axle" as 42 calls it), but the bearing is constrained from spinning along that "axle", since it can only spin around the axis of the shaft (handle).

That's a good argument. Hmmm.

 

[Square_peg]

And even if the shaft were supported solely by a bearing at location 2, I'm not seeing how the axe wouldn't rotate down as much, unless we consider the nature of loaded bearings having some braking effect on rotation

I didn't suggest a braking effect but just less force. Suppose the bearing needed greasing and required some minimum force to to begin rotating?

 

[Woodcraft]

https://van.physics.illinois.edu/qa/listing.php?id=294

"Why does it seems like you can develop more torque with a screwdriver with a longer shaft than with a shorter one (with the same diameter tip and same diameter handle)?
- Roger Kaiser, University of Illinois Alum
West Grove, PA
A:
This may seam odd, but the different length of screwdriver really doesn't affect the torque directly. Torque is the force applied times the distance from the axis you are rotating about. So the torque of the screwdriver only really depends on the thinkness of the handle.

But as you said, there seems to be a difference. I would suspect that you feel more torque because the longer handle gives you more of a grip with your hand. It will be easier for you to apply the force with your hand and that will add more torque.

It is important to note that I wrote this as if you were turning a screw. If insted, you were using the screwdriver to open a can of paint, there is a difference. The long handle will move your hand further from the rotation and give you more torque."

What is happening with a bit heavy ax is like a screw, not a paint can.
This "discussion" seems like someone is using an old fashioned hand drill (brace) and doing 95% of the drilling by cranking it with the hand twisting the knob, then claiming how much of a difference the offset handle makes.
The bottom hand position is where most of the stabilizing/correcting for "wobble" bit wise in a horizontal swing with an ax.

 

[Woodcraft]

https://van.physics.illinois.edu/qa/listing.php?id=298

"how could you you explain the reqson that a mechanic chooses to undo a tight nut using a long spanner and a door handle is placed well way from its hinges
- ilhan (age 12)
john taylor high school, england
A:
The idea is that of increasing the torque by applying
an external force, like the push of a hand on a door or
the pull of the mechanic's hand on the spanner ("wrench" to readers in the United States), at a farther distance from the axis of rotation.

It's not enough to push or pull far away, but the separation from the axis of rotation to the place where the force is applied needs to be perpendicular to the force (or have a substantial component perpendicular to the force). Pushing on the rim of a merry-go-round along a line pointing at the center pivot will not make it spin. Pushing along the rim will get it spinning. Similarly, pulling on a spanner in a direction perpendicular to the spanner will give the nut a torque proportional to the applied force and to the length of the spanner. This allows a tight nut to be turned with less force (of course the mechanic's hand travels farther if his spanner is longer).

Pushing a door on the side where the hinges are is a good exercise in futility -- the hinges just push back and the door doesn't move. Pushing along a line that goes through the axis of rotation doesn't cause rotation, but pushing on lines that miss the axis of rotation do the job."






The easiest explanation of this applied to a curved ax handle in a horizontal swing is like having a long socket with two handles. A long handle in the middle and a short handle at the end. If you were to use both handles, or put both hands on the long handle, it would be easier to get the bolt off. (Hold the bit from falling, pulling down) But the curves in the ax handle are the socket handles. And you can not swing an ax that way. You are constantly applying pressure to the short handle. (The bottom curve/hand position) At best you very lightly and briefly in passing get to apply a little pressure with one hand on the long handle (inside the large curve in the handle) as it makes its way to the short handle.
So the absolute minimum amount of force one would have to apply to keep the bit straight, (and maximum amount of hand travel for adjustment I might add) would be to swing the highly curved ax with hands separately in the curves. And as I said that is not going to work out to well for you. (It also happens to be the secret to how you are holding the ax flat and pretending it is balanced when it is not, one finger for a balance test, not two)
So the maximum amount of corrective force needed is on the bottom hand, the minimum amount of hand/wrist twist travel to correct the bit would also be the bottom hand position.

 

[Park Swan]

Real nice work on that offset handle!

 

[Steve Tall]

I didn't suggest a braking effect but just less force. Suppose the bearing needed greasing and required some minimum force to to begin rotating?

In that case, the force from the head imbalance (and the resulting torque) would not be diminished, it would just be opposed to some extent by friction in the bearings. If the bearings were frozen/seized, then it would be analogous to somebody gripping the handle and not letting the imbalanced head cause the axe to rotate.

A key point to this whole discussion is this: Around what axis does the unbalanced head (on a straight handle) want to rotate?
Doing the balance test while holding the axe horizontally with two hands (left hand palm down near the end, and right hand palm up near the head) will show that the unbalanced axe wants to rotate along the axis of the handle (the green line). The green line, not the red line, should therefore be used for considerations of torque (etc.) resulting from this imbalanced axe being held (and constrained) by two hands.

 

[FortyTwoBlades]

Sorry, but that is wrong. The forces on the handle and head are transmitted via the green line, not the red line.

(While the green line is the actual physical axis of the handle, the red line in this case is imaginary, resulting from dangling the axe upside down from a pivoting support at point 3. The red line signifies that there is an equal mass on either side of it, but it doesn't dictate how the forces are applied to/from the head.)

Re: Torque is measured about a point rather than about an axis, and as the point changes so can the torque.
Torque actually can be transmitted (and measured) about an axis. The driveshaft of a car is an example. So is a torque wrench with an extension and socket (at 90 degrees to the wrench handle axis). The torque doesn't change along the length of the shaft (assuming no deformation of the shaft).

The actual axis of rotation for an axe depends on where the handle is being held/supported. For the middle axe in the top diagram, this is simple because it's a straight handle, and the axis of rotation between the supporting hands is always the green line. (Keep in mind that we are only discussing the axis of rotation due to an imbalanced head held horizontally.)

The lever arm is measured perpendicular to the axis of rotation (green line).

The problem can be simplified down to just a 3'-long weightless shaft, with a 6-pound weight attached to one end having a 2" offset. The shaft is held in two hands, keeping the weight (with offset) horizontal. The resulting "torque" applied to the shaft by the offset weight is 1 foot pound (as calculated in a previous post), and the torque transmitted along the shaft toward the supports is this same 1 foot pound, and the torque countered by the combined resistance ("counter torque" it was called) of the two hands, wherever they are located along the shaft, holding it and keeping it from rotating, is the same 1 foot pound.

The green line is the axis of rotation at the center of the shaft (axe handle), and the red line doesn't even come into play in this example.

I wrote and posted my recent comment before seeing this. I think that's a good example, however I disagree with your conclusion. One important feature is that the shaft is straight (like the handle) and the bearings at locations 1, 2, and 3 (blue dot) are thus all aligned in the same direction, to spin along the axis of the shaft (handle). The bearing at the blue dot might overlap or coincide with the axe's rotational center (or "axle" as 42 calls it), but the bearing is constrained from spinning along that "axle", since it can only spin around the axis of the shaft (handle).

Another important feature is that the handle is supported by two hands (or two bearings). This constrains the rotation (due to an imbalanced head) to keep it only rotating along the axis between the two supports (bearings or hands).

And even if the shaft were supported solely by a bearing at location 2, I'm not seeing how the axe wouldn't rotate down as much, unless we consider the nature of loaded bearings having some braking effect on rotation (resulting from the cantilevered weight of the axe head), but this is not part of the picture here when we are looking only at the effects of an unbalanced head.

I think I see the issue here and it's that I was conflating torque of the head itself on the handle with wobble induced by any lateral forces of the upper hand during the stroke. The weight of the head would provide a constant torque on the green line, however, because the upper hand is free to slide and induce lateral forces, you do have force during the stroke (as opposed to simply holding the tool in the hands) where you're causing forces to act on the red line instead. The hand converging reduces your lever arm for those forces and so counter-torque for the lower hand to oppose them and keep the axe on track are reduced. However, the lower hand really has to be the one providing counter-torque for the head anyhow, and so long as the second hand is present then there's a constant amount of torque for the lower hand to be resisting. That also serves to explain why I find that one has to "think from the bottom hand" when using such axes, because you can't use your upper hand for resisting the torque other than at the resting state, so if you don't make the switch before release that induces the dreaded wobble effect, while "thinking from the bottom hand" will give a steady trajectory.

 

[FortyTwoBlades]

Managed to get those pictures snapped.

 

[Woodcraft]

As long as you keep that hand on the apex of bottom curve of the handle the other hand can be anywhere and you still are minimizing the effort needed to hold the bit level. It is exactly how the offset handle on a hand drill works. The ax isn't " balanced. If it were you could do that display in any position on the handle. (As in away from the curves.)As well as the all important one balance point display.

 

[FortyTwoBlades]

Note that the lower hand could also be moved up on the handle and both the offset neck handle and the Council would remain level. The hand positions along the length are arbitrary, but moving the upper hand makes the most sense photographically since that's the hand that's actually going to be moving in use. However, because the center of gravity on the offset handle example still lies external to the body of the object it's not possible to do a one-point balancing act with it. I can perhaps take a video with it showing the lower hand sliding while the axe remains level. Doing so with the upper hand is a bit trickier since it has the weight of the axe bearing down on it as the vertical fulcrum.

 

[Woodcraft]

Again, I can do this same thing with my Wetterlings hudson bay. Including holding moving the holding points to the rear of the handle. I can actually do it with one finger and a chair. Are you saying the Wetterlings is a balanced ax?

You have not "balanced the axe". (The pictures are at different angles but it looks from the plumb line pictures you actually made the ax MORE bit heavy with the new handle) You have however moved more of the handle into the rotation bit to poll of the ax. It also looks like with a blade that thin, and a handle that offset, you are having issues within the haft being struck behind the heel. Directly in the offset.

The council does balance inside the handle and can be tested with one hand. It is bit heavy. Not perfect balance as you claimed

 

[Woodcraft]

This test proves nothing. It is just a trick.

From above to show it is a single point of contact by the hand.

 

[Woodcraft]

One last point. In a balance test you find the point of balance along the handle and while balancing the head will show if the ax as a whole is bit heavy or balanced. If the point of balance along the haft is outside the handle, then the points of contact to show balance of the head and haft would be 1. Above the head. That you can do on a stationary object. And 2, somewhere along the handle. Just a thought.

 

[nzedge]

No, several of your previous assertions run counter to facts. Like, Most American axes a plumb line falls in front of the handle/eye." Or "the Council tool classic Jersey has perfect balance." That one is funny because it runs counter to the first claim,, and this one "there is no such thing as perfect balance."
But carry on..........

I agree. My american axes generally balance about inline with the front of the eye

 

[Steve Tall]

...because the upper hand is free to slide and induce lateral forces, you do have force during the stroke (as opposed to simply holding the tool in the hands) where you're causing forces to act on the red line instead. The hand converging reduces your lever arm for those forces and so counter-torque for the lower hand to oppose them and keep the axe on track are reduced...

FWIW, I think that simple leverage would provide a better explanation (instead of torques around the red line). The further away the "upper hand" is from the lower hand, the more leverage it has (to inadvertently overcome the control of the lower hand, "inducing lateral forces"). Like having the left hand holding the middle of a see-saw (at the fulcrum), and the right hand has an easier time moving the see-saw when it's further away from the left hand.

This would explain why there's less potential for "wobble" when the hands are closer together on the handle.

 

[DeadboxHero]

Managed to get those pictures snapped.

Cool stuff man, the string helps me see what's going on better.
Thanks