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

In summary, the conversation is discussing a physics problem involving a model that simulates the phalanges of a human finger. Each phalanx is represented by a block with a specific mass, and they are linked by joints with cords and a rotary motor. The conversation addresses the need for data on torque, lengths, and masses to determine the tensions and forces applied by the fingertip. The parameters of r, a, b, and lf are also mentioned as known values in the equation that relates the moment of the motor to the force applied by the fingertip. The conversation also discusses the simulation of the finger being attached to the hand and the need for frictionless pass-throughs for the strings attached to the masses. Suggestions are made to
  • #36
ElectricVocaloid said:
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.
 
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  • #37
haruspex said:
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...
 
  • #38
ElectricVocaloid said:
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.
 
  • #39
ElectricVocaloid said:
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.
 
  • #40
Lnewqban said:
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?
 
  • #41
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.
 
  • #42
Lnewqban said:
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.
 
  • #43
haruspex said:
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 :(
 
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  • #44
ElectricVocaloid said:
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.
 
  • #45
haruspex said:
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
Sin títulossssss.png
Sin títulossssssssssssssssssssssssssssssss.png

Sin títulssssssso.png
Sin títulossssssddsssssss.png
 
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  • #46
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.

1658330500903.jpeg
 
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