Rotary to linear motion using a crank

[Pau Hernandez]

Hello guys,

since linear motors are too expensive I decided to buy a rotational motor and use some gears and a crank mechanism to get enough torque and translate rotation into linear motion.

I need to meet a certain force on the shaft though. The question is how do I calculate the resulting force on the shaft?

I do know the following formula: Torque = crossproduct( Radius, Force).

This is be the crank construction I am talking about:


I appreciate your help. God bless you all.

 

[Pau Hernandez]

https://en.wikipedia.org/wiki/Piston_motion_equations well, yes its that simple guys. I am so sorry!

 

[JBA]

Using F = ma with that information only accounts for the dynamic motion load on the system and does not account for any external forces applied to the slider.

 

[Pau Hernandez]

are you referring to my first post or second post?

 

[Baluncore]

For a low speed crank, the rotary torque of the motor is being converted through the gears and then the crank to a linear motion. Any external force that opposes that linear motion is passed back through the crank, then gears, to the motor shaft. Before you can calculate maximum torque on the motor shaft you must specify that external force and how it varies with crank position. What is that external load ?

 

[Pau Hernandez]

The external load is a small door for my dog.

 

[JBA]

I was referring to your second post, sorry for the confusion.

 

[Baluncore]

you must specify that external force and how it varies with crank position.

There are still too many unknowns and possibilities;
What is the geometry of the door ? How much does it weigh ? How is it connected to the crank ?

 

[OmCheeto]

The external load is a small door for my dog.

Make sure it has plenty of safety devices. Doors can be dangerous.

The Oregon Zoo's male lion, Zawadi Mungu, injured the tip of his tail Monday morning when it was caught beneath a closing hydraulic door
...
"He lost the tip of his tail just above the tuft,"​

[ref]

 

[ron taylor]

this is a simple torque question.
if you apply 1 pound of force to a crank that is located 12 inches away from the center of
the cranks shaft then you have 1 foot pound of torque or force applied to the center of the
cranks shaft.
by placing a gear on the other side of the cranks shaft that is say 2 inches in diameter
then you can apply a force of 1 foot pound x 12 inches to the linear gear or rack.
because the ratio between the two distances in inches from the center of the shaft is
12 inches to 1 inch
1 lbf x 12 (inches) = 12 lbf
one revolution of the crank will result in 1 revolution of the gear and the gear has a circumference
of 3.14 inches so the rack will move 3.14 inches with a force of 12 lbf.

 

[Baluncore]

this is a simple torque question.

In engineering, the simpler the question, the harder it is to guess which answer is wanted.

The most significant component of the torque on the crankshaft will be a function of angle between crankshaft and the link to the door.
Torque = force * Cos(crank angle). That will change as the door opens and closes. The motor torque can be minimised by optimising the linkage.

Is it a sliding door? Does it slide horizontally or vertically? Will door sliding friction be a significant problem? How is the crank coupled to the door?
Is the door hinged? Is the hinge axis horizontal or vertical? What is the distance from the hinge to the attachment on the door of the link from the crank? What is that link?

If the door slides horizontally, or swings about a vertical axis hinge then the motor and crank torque can be close to zero. A counterweight could balance the crank and link of the static system. If it is opened or closed quickly it will become a dynamics problem. The motor torque will then be a function of the speed and acceleration required. It will also depend on the weight of the crank and link.

The OP question is too simple to have a simple answer.