Dynamics problem (force and torque analysis) of a human finger

[ElectricVocaloid]

Did you try virtual work as I suggested?

I've tried but I really can't find a way to fix it.I'm pretty lost with this problem.

And assuming that the torque of the servomotor is the same in all the joints, it seems to me that it is not correct because the number of phalanges on which it must act is an important factor to consider if I want to find the force in the tip of the last phalanx.

 

[haruspex]

I've tried but I really can't find a way to fix it.

Do what I said, use the geometry to figure out what a small advance dx of the finger tip corresponds to as a change in length of the contracting cord. Assume the angles are equal.

 

[ElectricVocaloid]

FINGER PAD CENTER (3 phalanges) = (dx, dy)

= ( ( 2b + lf ) . sen(β) + ( 2b + lf ) . sen(2β) + ( b + lf/2 ) . sen(3β) + a . sen(3β - 90°) ; ( b + ( 2b + lf ) . cos(β) + ( 2b + lf ) . cos(2β) - (b - l/2 ) . cos(3β) + a . cos(3β - 90°) )Here I have obtained the parameterized equation of the point where the fingertip would be...

and I need to conveniently transform the axes so that the positive y is up and the positive x is forward...

I was trying vector algebra to calculate at least the position of the point of the fingertip by linking it with the angle β of each joint.

 

[Lnewqban]

The cylinder on which the chain is wound or released is connected from its center of rotation to a servomotor which, according to the data, performs a torque M = 1.6 kgf. cm

Since this system models a contracting finger linked to a hand, it can be seen that the first of the phalanges is linked to a "static wall" that would represent the hand.

This torque is quite weak for which it is valid to consider that the binding chain will not suffer deformations due to the tensions, for which it is considered an ideal chain. This chain is a way of modeling the tensor muscles that make the necessary torque so that our fingers can flex or extend depending on whether they are released from the inside and extended from the outside, or vice versa.

I have thought of not considering the first 2 phalanges, but the problem is that if I did not consider them, I would be stating that for an engine that generates 1.6 kgf. cm of torque it is the same to rotate 1 phalanx or 3 phalanges, or even infinite phalanges. I suppose that it is an inconsistency with the laws of energy conservation, because with a finite torque it would be generating infinite forces in all the buds (inner faces) of infinite phalanges linked by that ideal chain.

We understand how the mechanism works.
Are you familiar with work and mechanical advantage concepts?
The number of falanges is irrelevant, higher number of those only makes the curve smoother.

Being an ideal mechanism, work in (tension tendon x tangential displacement) must be equal to work out (F apl x dispacement of finger's contact point).
There are axial forces on each falange and joint as reactions to the tension in the tendon; falanges here work as columns.

 

[ElectricVocaloid]

We understand how the mechanism works.
Are you familiar with work and mechanical advantage concepts?
The number of falanges is irrelevant, higher number of those only makes the curve smoother.

Being an ideal mechanism, work in (tension tendon x tangential displacement) must be equal to work out (F apl x dispacement of finger's contact point).
There are axial forces on each falange and joint as reactions to the tension in the tendon; falanges here work as columns.

taking into account that the first phalanx is attached to a static point that prevents it from falling (that is, the hand)... does the fact that there are more phalanges not imply that there is more weight that the chains must support when stretching or contracting?

all the phalanges in this problem we assume that they have the same length, which is lf, and we also assume that they all have the same mass, that is, their weights will be equal.

I am not sure that the torque M that the servomotor performs to rotate the cylinder is exactly the same as that experienced in the last of the joints...

 

[haruspex]

FINGER PAD CENTER (3 phalanges) = (dx, dy)

= ( ( 2b + lf ) . sen(β) + ( 2b + lf ) . sen(2β) + ( b + lf/2 ) . sen(3β) + a . sen(3β - 90°) ; ( b + ( 2b + lf ) . cos(β) + ( 2b + lf ) . cos(2β) - (b - l/2 ) . cos(3β) + a . cos(3β - 90°) )Here I have obtained the parameterized equation of the point where the fingertip would be...

and I need to conveniently transform the axes so that the positive y is up and the positive x is forward...

View attachment 304423

I was trying vector algebra to calculate at least the position of the point of the fingertip by linking it with the angle β of each joint.

Not sure how you are defining β. I'll assume it's the angle of turn at each joint.
The three phalanges, the base support and the red arrow form a pentagon. To return to the start, two different angles must be navigated. In your diagrams, the base support and the last phalanx are about equal length. If we take that to be near enough then those last two turning angles are equal, α say, then 3β+2α=2π.
Taking the red arrow as the x axis, you can now right down its length as a function of β.

 

[ElectricVocaloid]

Not sure how you are defining β. I'll assume it's the angle of turn at each joint.
The three phalanges, the base support and the red arrow form a pentagon. To return to the start, two different angles must be navigated. In your diagrams, the base support and the last phalanx are about equal length. If we take that to be near enough then those last two turning angles are equal, α say, then 3β+2α=2π.
Taking the red arrow as the x axis, you can now right down its length as a function of β.

the beta angle is the inclination of each phalanx, this formula is used for any inclination they have as long as they are all equal, in this way I obtain the distance to the fingertip, although I do not know if it is really a useful data...

 

[haruspex]

the beta angle is the inclination of each phalanx, this formula is used for any inclination they have as long as they are all equal, in this way I obtain the distance to the fingertip, although I do not know if it is really a useful data...

The next step is to figure out the relationship between β and the length of the contracting cord.

 

[Lnewqban]

taking into account that the first phalanx is attached to a static point that prevents it from falling (that is, the hand)... does the fact that there are more phalanges not imply that there is more weight that the chains must support when stretching or contracting?
...
I am not sure that the torque M that the servomotor performs to rotate the cylinder is exactly the same as that experienced in the last of the joints...

That additional weight does not affect F apl as it is supported by the top tendon via the internal mechanism of the servo and the anchorage point, always.
Work-in, which is angle times moment, remains constant all the way to the last falange, which is where work-out happens.
The other falanges do not produce any useful work or waste any, as everything is ideal; they only transfer energy or work from the servo to the last falange.
It is a geometrical problem, nothing else.

 

[ElectricVocaloid]

That additional weight does not affect F apl as it is supported by the top tendon via the internal mechanism of the servo and the anchorage point, always.
Work-in, which is angle times moment, remains constant all the way to the last falange, which is where work-out happens.
The other falanges do not produce any useful work or waste any, as everything is ideal; they only transfer energy or work from the servo to the last falange.
It is a geometrical problem, nothing else.

there you are right, the only thing I don't know is if the weight provided by the phalanges makes the system have more effort. How do you suggest setting up the equations?

 

[Lnewqban]

As represented (finger horizontally extended), the weights will help the motor when curling down, and will make it work a little harder when curling the finger up.
I don’t know about setting up the equations.
I would recommend following all the directions provided above by @berkeman and @haruspex.

 

[haruspex]

That additional weight does not affect F apl as it is supported by the top tendon

I disagree. The servo could idle, exerting no torque, and there would still be a force at the tip. That said, I would have thought the torques from the phalanx weights were small compared to that available from the motor.

I note that all the diagrams show the load (the force opposing the fingertip) as normal to the distal phalanx. If that is accurate, as the finger curls the load changes direction. Seems unlikely.

 

[ElectricVocaloid]

I disagree. The servo could idle, exerting no torque, and there would still be a force at the tip. That said, I would have thought the torques from the phalanx weights were small compared to that available from the motor.

I note that all the diagrams show the load (the force opposing the fingertip) as normal to the distal phalanx. If that is accurate, as the finger curls the load changes direction. Seems unlikely.

I agree that the number of phalanges increases the load that the servomotor must support. In this case, the force that the yolk would exert if it were in contact with a surface is requested. This force should be related to the torque of the motor. I am really lost in the formulation of those equations, I have already done everything possible including the free body diagrams and the geometric analysis of distances :(

 

[haruspex]

the force that the yolk would exert if it were in contact with a surface is requested

Yes, but that will depend on the way the surface moves when pushed.
If the surface only moves straight down, its lever arm about the base joint is the horizontal length of the finger. That will reduce a bit as the finger curls, leading to a slightly larger force exerted.
But if it swings around so that its resistance is always normal to the fingertip then the lever arm will get a lot shorter, allowing a substantially larger force.
If you only care about the minimum force then just take the finger as straight.

 

[ElectricVocaloid]

Yes, but that will depend on the way the surface moves when pushed.
If the surface only moves straight down, its lever arm about the base joint is the horizontal length of the finger. That will reduce a bit as the finger curls, leading to a slightly larger force exerted.
But if it swings around so that its resistance is always normal to the fingertip then the lever arm will get a lot shorter, allowing a substantially larger force.
If you only care about the minimum force then just take the finger as straight.

the original problem has a part a) and a part b)
Part a) is with the finger extended, that is, the angle between phalanges is β = 0°
And in item b) you have to calculate the force but considering a β ≠ 0° for example an angle between phalanges of β = 50°
The problem is that I was still able to formulate the equations for part a) which consists of calculating the force exerted by the fingertip on an object at the instant where the finger begins to flex (it is supposed to be the simplest part of the problem ).
Part b) of the problem is with the finger already flexed, at an angle between phalanges of β ≠ 0°, but here it is the same problem but applying trigonometry and linear algebra, using that formula that appears in the image above to calculate the coordinates of the point of the fingertip and from there estimate the force.

Keep in mind that both for part a) and for part b), the force exerted by the finger on a surface will always be normal to the surface of the fingertip that exerts said force, and will be opposite to the reaction force generated by the object receiving the force of the finger

 

[Lnewqban]

If you leave the last falange to be horizontal, Faplic will have horizontal and vertical components, as the pivot is located at a higher level.
The wrist could be located low for the fully curled hand, keeping a shape that makes the value of Faplic maximum, just like climbers shape their hands for grabing the rocks.

For a rough equation, you could see that [Faplic]y should be directly proportional to dimension a, and inversely proportional to dimensions r and lf.
Dimension b is variable and only determines the maximum angle between two consecutive falanges.
Moment of the servo is constant and given.