# Tag Info

## Hot answers tagged linear-motion

6

Does anyone know if this would be a good idea? The proposed design has poor rigidity towards torsional loads. It doesn't matter whether the actuator is off-center or if the load is off-center: if load's center of gravity and actuator are not at the same position, the torsional load will likely damage the bearings. In your example with two guide rails, ...

6

Your mechanism is going to lock up like the old car jacks immediately, due to differential raising of the platform. I am sorry but it looks like a textbook example of what not to do. in most real-life cases where the lifting actuator has to be off-center, they use a linked pair of scissors mechanism, like what they do in trucks' gate lifter, or the hydraulic ...

5

In a scissors jack design, items one and three, you'll have the least mechanical advantage when the assembly is retracted. Using drawer slides or the equivalent is a good idea, but will not provide for a balanced lift if there are loads applied that are not uniform over the cabinet's bottom surface. Your selected motor, with a thirty to one reduction should ...

3

This is a diagram for the mechanism, . And here is Wikipedia link, click.

3

Looking at the demo video, this machine looks fake to me. The person who assembled and exhibited it in the video used some trick to move the machine. I believe there is a platform that can tilt and contains a background attached to it, a table on top of it, a demonstrator standing on the platform and a camera (And lighting) kept on this platform. This ...

2

Constrain the driven shaft and use a coupling to the motor.

2

it may be necessary to construct more than one design (known as iteration) in order to refine your final objective. "Very rigid and with no slop" is a rather general description. "A tolerance of 0.10 mm" would be more precise. My understanding of pad printing is that one requires the impression pad to be re-inked from a source located near the destination. ...

2

Obviously cranks and pistons have been used for hundreds of years for this. Your other option would be fast-acting motors or hydraulic rams. But a 500N (or is it 600?) force isn't what happens when something hits an object. Hitting an object imparts kinetic energy, and a 1 HP motor just doesn't have a lot to give because you have to speed something up so ...

2

Initial thoughts? I'm not a mechanical engineer but I think you'd be disappointed. Figure 1. A quick pictorial analysis shows that there is almost no reason for the lift arm to move up. Also, most of the effort in the linear actuator would be taken up by a force vector from C to A with very little of the effort going into rotation. Figure 2. An online ...

2

The idea is mostly sound. I have a load F that is too high for my linear rails....so instead of beefing up my rails, I increase the load sharing across 3 or more rails, thus dividing it. In reality, I have actually seen such things....but only on poorly built machines as a half solution. In general one can do anything they want....but that doesn't mean ...

2

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 throws the center of gravity ...

2

IMHO (and I guess you already know that), from a structural point of view putting 100 kg at 1/4[m] will create a huge amount of bending moment on a actuator that it is rated for 120[kg]. An although you don't clarify whether the actuator electric/hydraulic, if bearings are involved you are bound to get failure sooner rather than later. Normally this type of ...

2

Either one would probably work as long as you gear it properly; but like Eric Shain, I'd advise you to go with the scissor jack because it's simpler. However, because the force varies with the angle of the scissors, you'll need to gear it down somewhat more than my suggestion. Lifting the setup is around 201 joules (41 kg * 0.5 m * 9.8 m/s (g)), and it would ...

2

Take a look at the illustration below, which is how I understand your question. You're asking, what would be the minimum value for the blue arrows in order to make each object move? First, let's look at the simpler case - the red box. I don't think this needs another diagram - you've rightly stated in your question that if there is a coefficient of friction ...

2

As a quick basic estimation approach we annotate the following: The length of bar 3 = L Top and bottom bearing force on each box $F_b$ The height and width of boxes 4 and its counterpart on channel 2 H and W. $$\Sigma M=0 ,\quad 220*L/4 = H/2*4F_b \rightarrow \quad F_b=220*L/8H$$ Now you have to figure if the bearings and their connections are okay to ...

1

What happens in the arrangement is that you are applying a force away from the center of gravity. Imagine you have a pencil on a flat surface and you try to push. If you don't push it around the center of gravity you will cause it to rotate. Although using linear bearings (especially longer ones) will mitigate the problem, you will be creating more normal ...

1

I might look into medical equipment, I was looking at actuators the other day, and they weren't too expensive.

1

This is one of Newton's Axioms. I.e., strictly speaking the rule does not need to be able to be derived from some other physical law, but an empirical fact supported by a huge number of actual and possible experiments. This axiom (together with the second of Newton's Axioms), leads to the inertial principle - the total momentum of a closed system is ...

1

Rather than think of it as 'resisting', think of it as force. In a vacuum, you are correct that there is nothing to push against, nothing resisting. The resisting, as you put it, comes from pushing out something that was taken into space with the rocket, the gas from the rocket motor. In force terms, it takes force to push that gas out. The force is in ...

1

Let's call the point where the hinge is welded to the boom D. Scaling CD from your diagram we pick 30mm. That means the mechanical leverage of this gadget is approximately $$F_{boom}= F_{actuator}\frac{30}{400}= 0.075F{actuator}$$ This gets even worse because we need to consider the reduction of leverage due to the location of D, a nearly ratio of 1/20 ...

1

Use a screw press mechanism and reducing gearbox. Then it becomes an issue of calculating the force multiplication of the screw and gearbox mechanism to get 500 newtons. It won't be fast, but it will provide the force you require.

1

If budget was not an issue you could design a crank / piston assembly from scratch... As it is an issue then consider re-purposing a small single cylinder engine - lawnmower or chainsaw perhaps. Consider lubrication as well...

1

In my opinion, the device motion itself is not faked , but there may be camera trickery involved to hide an accomplice, acting as an additional energy source. Look at 0:49 seconds - he gives the blue wheel a spin - he is putting energy into the device. It is certainly believable that the inertia stored here would be sufficient to propel the device to the ...

1

check this graph, Now if you attach a spring loaded vertical ramrod released upwards when the little roller cam, bearing inside this rotating gear falls into the step, you have your hammer. It is based on Leonardo De Vinci's hammer, except like a negative or reverse mechanism. the cam is inside the gear not outside.

1

A consideration of an upward-striking hammer is that you have to have some method of preventing damage to the mechanism as the hammer reaches the surface being struck. This could be the absorption of energy or the displacement of the item/surface being struck. Recoil is the method used in the gravity-fed DaVinci design. A couple of methods come to mind, but ...

1

Basically it works like a lever with the force or weight arm length as AC and resisting arm which is connected to the actuator as BC. The ratio of mechanical advantage remains constant as the jack lifts the load because the ratio of the horizontal projection of the arms which is proportional to their respective lengths multiplied by the cosine of angle ...

1

I doubt that the calculator you referenced is relevant for headphones. A fundamental difference between headphones and loudspeakers is that headphones transmit energy to the small, (almost) sealed, volume of air trapped between the headphone and the structure of your ear. That is completely different from a loudspeaker which is designed to transmit energy ...

1

BK, BF refer to the mount format - in this case a rectangular face mounted block with one (BF) or two (BK) bearings and mount holes. 1st Letter == mount format; 2nd letter == bearing set ; numeric == ID of the bearings in mm; A quick internet search will identify all of these bearing types and specifications.

1

As you see, you use one fixed and one floating bearing to compensate for stress due to thermal expansion of the rod, manufacturing or assembly tolerances. Floating bearing can’t take axial loads, where fixed bearings take both axial and radial loads. I guess the letters e, b, f specify the mounting type, for example But to be sure you have to provide a ...

Only top voted, non community-wiki answers of a minimum length are eligible