John2020

But the reactionless drive has nothing to do with perpetual motion.

Well, it is a little bit difficult to agree because the References I use in my paper claim orherwise, however not in classical mechanics. If in the discipline of statistical mechanics and optic is possible then, it should be also possible in classical mechanics. I personally do not see other way than the fictitious forces utilized to transfer mass in a reactionless manner.

I think so. I just thought an induced magnetic field upon a magnetic material that since the translation mechanism is massless then momentum transfer in the opposite direction wouldn't be possible even through the field. What do you think about the above? Could a design that utilizes either Euler (the initial idea of my design) or centrifugal force work as a reactionless drive?

This is what I have envisioned with the construction in Fig.1-Upper that mimics (wrongly) an Euler force. The problem as you all show to me is, there is always a force component on x-axis that implies a reaction force upon the screw on x-axis. It will require a different design that has to demonstrate the creation of an Euler force exerted upon a mass without translation mechanism.. Would you agree with the above? If yes what that could be? Would it work for a massless (using magneric fields) bolt at steepness angle e.g. 30°?

Because I addressed the rotational energies coming from the torque (nut) and counter torque (upon the screw), where both have opposing direction of rotation (conservation of angular momentum), thus they cancel each other. You are right then, the only possibility to work then would be if the bolt was massless and steepness e.g. 30°. It practically means the translation mechanism should be conducted entirely through e.g. magnetic fields. As I think it further maybe it won't. Then probably, the only way to work such a device (not using a translation screw) is to have a rotating body that will induce an Euler force upon a small mass being both an intrinsic part of a non-rotating body (that consists of the system as a whole). In principle it should work based on reactionless mass displacement inside the system. Reactionless due to the Euler force.

Based on the fact there is no conversion of rotational energy to kinetic through the same mechanism, thus the reaction part will have just rotational energy that does not violate the Energy conservation. In general this can be expressed as follow: \[U_{rotation_{action-on-nut}} + U_{rotation_{reaction-on-screw}} = 0, \\ U_{rotation_{action-on-nut}} = U_{kinetic_{nut}}, \\ U_{kinetic_{screw}} = 0 \Rightarrow U_{rotation_{reaction-on-screw}} = -U_{kinetic_{nut}}.\] I understand the reasoning of swansont but something does not feel right (I give much attention to intuition and observation than the maths (maths are by nature non-intuitive)). With your comment (a big thanks) I just realized my idea will work when the steepness (decreasing the cross section of the screw and increasing the lead) goes to a very large finite value (not infinity), meaning the inclination of the slope will be almost 90°. Note: Swansont and maybe others have mentioned something similar (as the massless bolt) but back then I didn't realize what was the purpose of such suggestion. Now I see. That way the Fn*cos(θ) (action on x-axis) will be almost zero that would imply an almost zero reaction upon the screw. Theoretically, for an infinite steepness the device becomes an ideal reactionless drive. As you see my intuition was right from the beginning, the problem was to justify the idea with maths (where I admit I struggle a lot since it is not my daily business (testing software, test scripts etc). The derivation I provided in my first post (not so nice actually), it actually depicts the ideal case. A big thanks to all @Ghideon, @swansont, @joigus, @pzkpfw , @Phi for All and all all readers who like or dislike my comments. Consequently, when we could have a carbon fiber screw with a relative high steepness and a small mass relative to the nut (screw 1 unit and nut 10) then it would be possible to practically demonstrate the principle of the reactionless drive in classical mechanics.

Thank you for your practical view on this subject. No worries we just make a theoretical discussion by removing all those technical matters that make the entire undertaking more complex and more challenging. The main focus is the action reaction principle and momentum conservation on a pure ideal device on a theoretical level. I would like to add my view on the above: If the normal force on x-axis presupposes the presence of nut then,from the reaction side (upon the screw) we shouldn't have a reaction on x-axis. An additional argument that may support the above is, in absence of nut there cannot be a normal force component on x-axis. Consequently, in absence of nut we have just the rotation of screw as result and nothing else. Note: I understand the above breaks the action reaction symmetry, however if the above arguments are true then, we have to see what is wrong in our analysis.

It is all those you mention without the torque from the guiding lines. I don't understand where you are pointing with the thought experiment. When the mass of the bolt is very small compared to the nut and the system as a whole, what will the system do while nut is being displaced?

Because the normal forces that are exerted on the bolt result just in its rotation. Furthermore, there is no mass coupled (as we have with the nut) in the screw that will result in an opposite direction to nut displacement.

Ok we have already clarified this with the x and y components of Normal forces. 1 So we confirm there is a translation mechanism where a torque (over the x,y normal forces components) is converted to nut displacement. 2.On the other hand, the reaction is a counter torque (again over the analysis with the x,y components of the normal force) exerted upon the screw, that does not result in a mass being displaced (as we have with the nut) since nothing is being displaced in that direction (opposite to nut displacement). Is there any objection on the above? We speak now about the reaction of the screw while the nut is being displaced.

I admit I didn't communicate the entire idea as most would expect as also I made several mistakes while defining some things. However, I would be interested to have your view on the above three points a,b,c, too.

Then from the action-reaction principle: a) an opposing normal force component on x-axis is exerted on the screw b) an opposing normal force component on y-axis is exerted on the screw due to the applied torque upon the nut c) Since (a) and (b) are manifested because of the torque on the nut then, the reaction should be of the same type, namely a counter torque upon the screw Are the above statements correct? Note: The counter torque (as above has nothing to do with the torque of the system but that of the screw. We assume there is no system rotation (this is a mistake I have to correct in my paper. So we assume ideally the system will not rotate and the torque will be entirely converted to nut displacement as I declared this in my first posts) while the torque applies due to the conservation of angular momentum. We have gone all through these. You understand very well what I mean. This is all perfectly clear but you still miss the point. The mistake from my side (that someone from you pointed out) is when on applies a torque, the entire system will rotate. I have to fix this on my paper (I will explain in another occasion). Then, I said, let us for a moment assume the system will not rotate when we apply the torque on the nut and this torque will be totally converted to nut displacement. All other you mention are perfectly clear and I understand how the mechanism works.

Correction on the previous post: I am not saying this. We have a translation mechanism that implies the perpendicular component (in our case the counter torque and not that applies on the bolt) is converted to an induced force along the x-axis that is equals to the Normal force along the x-axis (the concept of the simple machine). If not then, what is the cause behind the Normal force on x-axis? Isn't the Normal force on the x-axis pushing the nut or not? Note: We know the nut does not rotate, however the counter torque is responsible for ascribing the thread contact of the nut a counter clockwise evolution in contrast with the applied torque on the bolt (clockwise). I think it is backwards, otherwise it does not make sense. In absence of torque there is no Normal y-component. So, on applied torque implies Normal y-component.

Yes, this is all clear. We don't have to repeat ourselves. The torque on the bolt induces a counter torque on on the nut although this does not rotate, however as we have already said the contact of the thread ascribes a helix in opposite direction of that of the bolt. Here the cause is the induced counter torque and due to the translation screw mechanism, it makes sense the perpendicular component a.k.a counter torque to be converted to the Normal force on the x-axis. I asked swansont the same (see above). If the converted counter torque to Normal force on the x-axis is not the cause of the nut motion then what is?

I am not saying this. We have a translation mechanism that implies the perpendicular component is converted to an induced force along the x-axis that is equals to the Normal force along the x-axis (the concept of the simple machine). If not then, what is the cause behind the Normal force on x-axis? Isn't the Normal force on the x-axis pushing the nut or not?

Thank you for clarifying that. I thought since two mechanical parts contact each other then it should be there besides friction (if we had), a mechanical force. Let's continue. The force that creates the torque (therefore I called it torque force), equals to the normal force along the x-axis that pushes the nut to the right, correct? Now acording to action-reaction principle a reaction should appear upon the screw, right?

It is wrong what I wrote about balancing. I just wanted to stress the fact a torque force is required to overcome the normal force on x-axis (thw same applies from the side of the screw). I assume you still speak about the case in absence of a torque force, like being idle (no motion is taking place). I just would like to clear out something for my own understanding: Are there contact forces (normal mechanical forces) in the construction (screw is contacting the nut) in absence of external forces e.g. gravity and torque forces?

Addressing just the normal forces along the x-axis without a torque force, there is no net force according to Newton's laws. A net force due to the torque force is required to overcome the normal force along the x-axis (since there is no additionally a friction force to overcome). In this case we have a non zero net force that accelerates the nut. This is how Newton's laws work. Because there is a normal and a counter normal force along the x-axis, otherwise you break Newton's 3rd law between the screw and the nut. One has to introduce a torque force to overcome this balancing of forces on two bodies. If you may confirm this then we may contnue.

If the above is the original drawing you posted with a single contact point of the nut thread, then yes it is expected as I claim. As I also mentioned to swansont, there is not net force along the x-axis if we don't t introduce the torque force. What swansont claims is flawed becaused it is based on static analysis (no torque force) that implies no net force on x-axis.

The analysis you made in absence of a torque force does not imply acceleration, thus no net force on x-axis.

But all those forces you are speaking about address a stationary situation (in absence of a torque force) as also apply to a dynamic (with a torque force) one. Consequently, they cannot affect the motion of the system. What is relevant for the motion of the nut, is just the net torque force (as I did in my analysis in the first post).

Of course, I understand a vector is the sum of its components. OK, regarding what I said about having no normal force on the x-axis (the x-component) was when one does not make the analysis at the slope level. Let's continue, there is an x-component of the normal force. What do you claim as next? edit: removed "negligible due to the helix angle". Wrong assumption. Somewhere has the torque force to be introduced, otherwise we speak nonsense. The torque force is perpendicular to the x-axis.

I don't understand this point. What force are you addressing now? Isn't that the action-reaction principle applies to contact forces and in our case those contact forces are the normal forces being perpendicular to the slope?

Your diagram with the N1 is the same as that of Ghideon with N. This means the contact of the thread of the nut moves along the slope (actually the contact is motionless and slope of the screw is the one that moves). Is it correct or not?

Again, the problem is not the friction but that motion occurs across the slope that presupposes a torque force is at play. In a previous post of yours if I remember correctly you claimed the nut is pushed to the right and the reaction pushes the rest of the system over the screw and along the axis of rotation. There are two misconceptions in this approach: a) Checking the diagram of Ghideon, motion occurs along the thread slope where the thread contact is relevant to forces analysis. b) I used as a counter argument a thin threaded cylinder having no inner volume (no inner mass). Your assertion of having a reaction force along the axis of rotation, obviously does not hold because assuming there would be a force, it would have no mass to push (in the opposite direction). Consequently, the momentum of the nut is not counteracted by an opposing momentum (since there is no mass to push along the axis of rotation).