# HW 8 – Problem 1

Wednesday, April 20th, 2016

1. You are given the position of the shoulder, elbow, wrist, and 3rd knuckle with respect to time in the attached file (pitching_data.xls). The subject in was a Division 1 baseball pitcher with a mass of 78 Kg and a total height of 1.85 m. Segmented mass information is given in the following table:

 Plagenhoef, 1983 Body Segment Mass Length Location of COM Left Foot 1.38 1.88 50 Left Leg 5.05 11.17 42.65 Left Thigh 11.125 10.66 43.05 Right Foot 1.38 1.88 50 Right Leg 5.05 11.17 42.65 Right Thigh 11.125 10.66 43.05 Trunk 46.02 13.35 41.75 Left Hand 0.65 5.75 46.8 Left Forearm 1.87 15.85 43.2 Left Upper Arm 3.25 17.25 44.7 Right Hand 0.65 5.75 46.8 Right Forearm 1.87 15.85 43.2 Right Upper Arm 3.25 17.25 44.7 Head 8.23 4.77 55 Location of COM is the % segment length from proximal end of segment Length is given as the % of total body height Mass is given as the % of total body mass

1a) Draw the following FBD’s (10 points).

• The Upper Arm
• The Lower Arm
• The Hand
• The Ball

1b) Numerically differentiate the position data in order to calculate the velocity of the center of mass for the humerus, forearm, and hand.  Display your results in separate plots of velocity versus time (10 points).

1c) Numerically differentiate the velocity data calculated in (1b) in order to calculate the acceleration of the center of mass for the humerus, forearm, and hand. Display your results in separate plots of acceleration versus time (10 points).

1d) Calculate the magnitude of the distraction force (normalized by body weight) at the shoulder vs. time (10 points).

1e) Calculate the magnitude of the total force at the shoulder, elbow, and wrist (normalized by body weight) vs. time (10 points).

## 704 thoughts on “HW 8 – Problem 1”

1. Joshua J Liddy says:

Does it matter what derivation method we use? (i.e., two-point, three-point, five-point, or interpolation-based).

1. Alison Ling says:

I was planning to get a best-fit equation and then take derivative. Probably not the most accurate but should get a somewhat decent approximation

1. Herberto Dutra says:

I believe that by “numerical differentiation of the data” the problem is asking to calculate the Delta X/ Time (1/frequency). Keep in mind that the first step is to determine the position of the CM and do the vector corrected from the proximal extremity . I like the idea of developing a best fit equation. Are you thinking regression analysis ?

2. Ryan Asher says:

I was planning to do a best fit in excel as well. I was thinking polynomial. Anybody have any thoughts on what order polynomial fit would make sense?

1. Alison Ling says:

interesting, I did best-fit polynomial equations but they look pretty different than the way shown in class

2. Katie Roth says:

I would say play around with the order in Excel until the best curve approximation is reached? For me, it always seemed more trial-and-error than anything

3. Andrew Wilson Harmon says:

I think the polynomials might clip off the peaks in the measured data- I think the data varies too much to fit a polynomial to it unless the order is high.

2. Alison Ling says:

Do we analyze x, y, z directions separately, which means we will have three plots for velocity in the x, y, and z direction vs time and three plots for acceleration in the x, y, z direction vs time?

1. Desarae M Diedrich says:

I believe you should have three different velocity vs time graphs for each segment you are analyzing!

2. Nathan Shaw says:

It makes sense to me to plot the magnitude of the velocity and acceleration vectors, not the individual x,y,and z coordinates. I will plot one graph each for the COM of the 3 segments, so a total of 6 plots (3 vel., 3 accel.) in parts b-c..

1. Erik Hansen says:

The problem does not specifically ask for magnitude, like those that follow. I made one for each direction and segment.

2. Mariana Abreu Caffaro Lyle says:

FYI. I used the velocity2=x2+y2+z2 equation to get a velocity, taking the x,y,z components into account.

3. Eryn Horton says:

I plotted X, Y, Z, and Magnitude for each of the 3 segments and both Vel and Accel; so 6 plots total with 4 series per plot. It looks a tad messy but also cool to see how the Vel and Accel change per coordinate.

3. grabers says:

I’m confused by the ‘normalized by body weight’ requirement for parts D and E. I’ve been calculating the arm segment masses using the percentages from the table and the overall body weight – is there something more I should be doing for normalizing?

1. Nathan Shaw says:

I am doing it the same way, but I would also like some clarification on the wording “normalized by body weight”.

1. Chris Valentino says:

The distraction force magnitude that you calculate in part 1d can be divided by the total weight of the person. This gives you a non-dimensional force magnitude.

4. Erik Hansen says:

Should we be assuming the mass of the baseball is in the center of the hand, or somewhere else?

1. Chris Valentino says:

I assumed it was at the center of mass of the hand.

1. Katie Roth says:

I did as well

5. Erik Hansen says:

Does anyone know if we are supposed to graph 1d and 1e, or just show a table of results?

1. Herberto Dutra says:

I’m doing both. Once the sheet is created it’s easy to do the graph. I will turn a PDF with the homework body and two EXCEL attachments

2. Myra Fabro says:

I think a graph would make sense because it’s provides a visual profile of a large data set.

6. Alison Ling says:

For 1E, I am getting a little confused by the term “total magnitude”, if I take the usual approach and break up the “Sum of Forces=ma” into an equation for each direction, it seems like we can just solve one of them to get magnitude of the force on the hand/elbow/shoulder. That doesn’t seem right, am I missing something?

1. MacKenzie J Tweardy says:

I think they mean to find the magnitude of the elbow, of the hand, and of the wrist. So you should end up with three magnitudes. There isn’t a way to combine all three of them if we are using three separate FBD of them to solve.

1. Katie Roth says:

To find these magnitudes, do you take the TJR at each joint and then find total R from R1 & R2 (for example)?

7. Herberto Dutra says:

In order to calc the distraction force over time , should we simply multiply the mass of the segment by the acceleration in each Cartesian component (xyz) obtained in 1c) , except we should use the shoulder joint instead of the upper body CM?
Any thoughts?

8. Chris Valentino says:

Herberto, I was able to calculate a force vector with three components at the shoulder. “Doting this vector with a unit vector in the direction of the longitudinal axis of the upper arm should give you the distraction force based on the professor’s definition in class.

1. Herberto Dutra says:

Thank you Chris . Which value did you choose for the acceleration? Did you get the peak value from the chart drawn at part 1 c)?

9. Chris Valentino says:

I used all three upper arm acceleration components for each time step. I didn’t just choose the peak from 1c but did the distraction force calculations for the entire time slice.

10. Jacob C McGough says:

Common Problems so far:

1) Make sure you are in the correct units at each step when you do your calculations.

2) You should be normalizing the force data with respect to WEIGHT! not mass.

3) Make sure you include gravity in your sum of forces

1. Maru Cabrera says:

To include gravity in the sum of forces, should we be considering a particular assignment of x,y,z axis for the data that we have (e.g. going from feet to head is the positive z axis)?

11. Daniel Joseph Conway says:

Can someone explain the difference between part d and e…Is the distraction force the force from the deltoid on the shoulder?

1. Austin James McDonald says:

The distraction force will be the component of the force on the shoulder that is pulling it directly out of its socket toward the elbow.

2. Herberto Dutra says:

One comment that helped me was that parts d) and e) are in switched order . You should solve e) first and use the data from the shoulder force to get the distraction force. The DF will be a scalar , which will be obtained by dot multiplying the Unit vector by the force in the shoulder . This will need to be done for each coordinate and plot against time

1. Daniel Joseph Conway says:

That’s what I was thinking first. So to get distraction force, I’m guessing we just use the Total joint reaction at the shoulder and then take the dot product with the unit vector for the humorous?

1. Herberto Dutra says:

That’s what I did

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