Homework2 Solution v4

7/24/2019 Homework2 Solution v4

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2.007 Design and Manufacturing 1

Homework #2

NAME:___School Solution_________________________________________________________

Date Issued:Thursday 14 FEB, 11AMDate Due:Tuesday 26 FEB, 11AM

Please answer the following 4 questions showing your work to the extent possible withinthe allotted time. Point allocations are listed for each question (all equal at 25 pointseach, all subsections within a question are equally important to that question). Thisshould not take more than 6 hours. If any problem takes much more than 1.5 hours, seekhelp from instructors, UAs, classmates, etc. As described in the course policiesdocument, this is one of four homeworks you will complete in this course. Eachhomework counts as 5% of your total grade. You will submit your work in hardcopy atthe beginning of lecture.

1. (25 points) You plan to use two brushed DC motors to drive the rear wheels of a robot (onemotor per wheel). The drive wheels are 100mm in diameter, and are 200mm apart (center-to-

center). The mass of the robot is 0.32 kg. The center of gravity of the robot is 80mm off theground and directly at the center of the four wheels. You have two brushed DC motors with a

planetary gear drive (Vigor BP-05) that meet the manufacturers specifications available on

the 2.007 web site. You plan to power the motor directly with a 7.4V battery.

a. Sketch the torque-speed curve of the motor at 7.4V. Label the stall torque, no-load speed, and maximum powerpoint. Give values for each of these in

appropriate units.

According to the spreadsheet from the manufacturer:

At 3VThe stall torque is 3.6 kgf*cm but I prefer the units 0.353 N*m

The no load speed is 131 rpm but I prefer the units 13.7 rad/s

At 7.4 V (which is what the 2 cell Li Po battery provides) both the torque and no-load speedscale linearly with voltage

The stall torque is 0.353 N*m * 7.4V/3V = 0.871 N*m

The no load speed is 13.7 rad/s *7.4/3V = 33.8 rad/sec

Torque speed curve

0 10 20 30 400

0.2

0.4

0.6

0.8

1

( )

New andImproved

Max power 1

4 0.871

N m

33.838

1

s

7.368W

rad/sec

Nm


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b. Estimate the initial acceleration of the robot if the motors are throttled to full froma dead stop. (Friction coefficient between the floor and the drive wheels is 0.7).

Draw a free body diagram of one of the drive wheels.

The initial acceleration of the robot could be limited by one of two phenomenaby friction or

by the motors properties.

The motor is initially in the stalled condition and can therefore provide 0.871 N*m. This would

allow a Force tangent to the bottom of the 50mm radius wheel of 0.871N*m/0.05mm=17.4N

There are two drive wheels, so the total force would be 2*17.4N=34.8N

But the friction with the floor may limit the force to a smaller value. The normal force on each

wheel is 0.32 kg*g/4=0.785N. The frictional force on each wheel is therefore at most

0.7*0.785N=0.549N. The friction allows far less force to be applied.

The initial acceleration can therefore be estimated using the static force on the two drive wheels /

mass of the robot. 2*0.549N/0.32 kg= 3.4 m/s^2 or 0.35 gs.

A more refined answer can be made by drawing a full FBD of the robot and solving a system of

3 dynamics equations in 3 variables. This solution increased the acceleration estimate to4.7m/s^2. This is due to the weight shift caused by the acceleration. The rear wheels have a

higher normal force and therefore get more traction. I dont feel sure that this answer is better as

a response to the question of what is the initial acceleration. On the real vehicle, it will take

time for the compliance in the system to allow the rear of the vehicle to sit down and actuallyapply the additional normal force that emerges from the model. I think the reality is somewhere

in the range between the two answers (3.4 m/s^2 and 4.7m/s^2).

The FBD of the drive wheel was requested. The reaction at the axle and torque at the axle do nothave to be determined in order to solve for the initial acceleration though. They are forces

internal to the overall robot. Only externally applied forces determine the acceleration.

N= 0.87kg*g/4

N=0.7* 0.87kg*g/4

N= 0.87kg*g/4

R


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c. Use forward Euler to estimate the time it take the robot to move forward by onemeter. Explain your process and give your estimate.

I used Matlab to implement a solution. The graphs suggest there is a period of constant

acceleration followed by a fairly abrupt transition to constant velocity. The specific velocity

attained is governed by the no-load speed of the motor.

I find that the time required to cover one meter is 0.78 sec. If I change my code to neglect the

rear shift of the weight on the wheels due to acceleration, the time is a bit longer -- 0.84 sec. I

think its OK if students neglect this modest and subtle effect this time around.

% This is a script to model a BP-5 motor driving the maxi-me robot% This is in 2.007 pset #2 problem #1c

clear all;delt = 0.01; % Time step for forward Euler (try changing it!)stop_time=1.1; % Total time to be simulatedJ = stop_time/delt; % Number of time steps

Kt=0.0052; % Motor torque constant (N*m/Amp)Rm=0.87; % Reistance of the motor windings (Ohms)Rw=0.05; % Radius of the wheel (meters)Vb=7.4; % Voltage of the battery (Volts)mu=0.7; % Friction coefficient wheel contacting floor

% Assuming kinentic and static friction are the same for these two% materials

Stall_torque=Kt*Vb/Rm;No_load_speed=Vb/Kt;Power_max=Stall_torque*No_load_speed/4;

m=0.32; % Mass of the robot (kg)G=40; % Gear ratiog=9.8; % Acc due to gravity

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8Velocity of the robot over time

Time (seconds)

Velocity(meters/second)

red line = st eady state velocity inferred

from no load speed

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6Horizontal position of the robot over time

Time (seconds)

Height(meters)

This is the target distance

This is the time 0.78

to reach the target distance


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% Initialization of state variables (this is a "dead start")omega(1) = 0; % angular vel of motorvel(1) = 0; % Velocity of persondist(1) = 0; % Distance the person has ascendedforce_shift(1)=0;

% This loop is the forward Euler method% Euler's method is a simple explicit scheme for solving ODEs% It is described at https://stellar.mit.edu/S/course/2/fa12/2.086%/courseMaterial/topics/topic5/lectureNotes/2.086_Unit_IV_V1.1/%2.086_Unit_IV.pdf% page 325fori=2:J

back_emf(i) = omega(i-1)*Kt;current(i) = (Vb-back_emf(i))/Rm;torque(i) = Kt*current(i);

% The force the motor can apply to the ground if NOT limited by frictionforce_motor(i) = torque(i)*G/Rw;% The force the wheel can apply to the ground if limited by frictionforce_friction(i) = ((m*g/2)+force_shift(i-1))*mu;

force_applied(i)=min([force_motor(i) force_friction(i)]);

acc(i) = force_applied(i)/m;vel(i) = vel(i-1)+ acc(i)*delt;dist(i) = dist(i-1)+ vel(i)*delt;omega(i) = vel(i)*G/Rw;dist(i)=dist(i-1)+omega(i-1)*(Rw/G)*delt;

force_shift(i+1)= acc(i)*m*80/(100*2);% I decided to model the aft shift of the weight on the wheels due% to forward acceleration. There was very little effect on the% time to reach the target distance

end

time=delt*min(find(dist>2.4))

figure(1)plot(delt*(1:J),vel)hold onvel_steady=(No_load_speed/G)*Rw;plot(delt*(1:J),vel_steady*ones(1,J)*1.001,'r')title('Velocity of the robot over time');xlabel('Time (seconds)');ylabel('Velocity (meters/second)');text(0.01,vel_steady-0.1,'red line = steady state velocity inferred',...

'HorizontalAlignment','left')

text(0.01,vel_steady-0.2,'from no load speed',...'HorizontalAlignment','left')

figure(2)plot(delt*(1:J),dist);target=1.0;

hold onplot(delt*(1:J),target,'r')title('Horizontal position of the robot over time');xlabel('Time (seconds)');ylabel('Height (meters)');


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text(0.01,target+0.1,'This is the target distance',...'HorizontalAlignment','left')

time_to_target=delt*min(find(dist>target))plot([time_to_target time_to_target], [0 target],'r')text(time_to_target,0.2,'This is the time',...

'HorizontalAlignment','left')text(time_to_target+0.3,0.2,num2str(time_to_target),...

'HorizontalAlignment','left')text(time_to_target,0.1,'to reach the target distance',...

'HorizontalAlignment','left')


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d. How steep a grade can this robot climb? What determines that limit? Power ofthe motors? Torque of the motors? Friction? Tipping aft?


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2. (25 points) Below is depicted a part that can be attached to a motor output shaft and usedas a link in a mechanism. Create this part geometry in Solidworks.

There are two options for completing this problem (check only one block below):

I have completed this and a 2.007 instructor or UA has recorded my result

I posted my models(not screen shots) on Stellar under HOMEWORK/HW2_2/

NOTE: All dimensions in millimeters.

2

Many people are mixing up

the 6 and the 7 in the cross

section. If they do that,there will be an overhang

where the drafted extrusion

meets the smaller cylinder.


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3. (25 points) Design a simple circuit with only switchesand LEDs and one 7.4 V battery. There will be sevenof each arranged in the pattern shown to the righthere (leave out the decimal). There will be a separatepattern of LEDs mounted to one board and anotherpattern of switched mounted to a different board. Thesegments are a both 1/6 meter tall and are 2 meters

apart. Design a circuit so that closing each switchresults in lighting each corresponding LED. Designthe arrangement so that the amount of wireconsumed is as low as you can make it. Estimate theamount of wire needed.

This is a good design. I mentioned in lecture the issue that if you turn

on 4 or more segments in series, then they will probably not light. But

these are in parallel. I mentioned that if you turn on only one LED with

7.4V source, I expect it will burn out if there is no resistance. But thatwas not something I really intended for the students to work out.

Maybe there will be enough resistance in this length of wire. Maybe

each segment should have 2 LEDs in series to make it brighter rather

than wasting power in a resitor. If the students do comment on it, thatwould be a nice bonus. But this is not really necessary for full credit.

There is some risk that students feel this question is not well linked to the lab. It

may be motivating for them to see this connection. Imagine you replace the

switches in this design with switches that are sensitive to a magnetic field. Then,

just when the switch is oriented correctly over Tims brain, the LEDs on the other

board will make a digit light up scoring the fMRI task.
http://www.google.com.sg/url?sa=i&rct=j&q=seven+segment+display&source=images&cd=&cad=rja&docid=QyDK_EN2ljJdmM&tbnid=aHz3l2KrWgymTM:&ved=0CAUQjRw&url=http://allaboutee.com/2011/07/04/arduino-proximity-sensor-circuit-tutorial-and-code/&ei=X8cLUc3wH6ey0AGM9YHwBQ&bvm=bv.41867550,d.dmQ&psig=AFQjCNErlUpJgGrLh8m_uLr2kqFc1t6UYg&ust=1359812759717788http://www.google.com.sg/url?sa=i&rct=j&q=seven+segment+display&source=images&cd=&cad=rja&docid=QyDK_EN2ljJdmM&tbnid=aHz3l2KrWgymTM:&ved=0CAUQjRw&url=http://allaboutee.com/2011/07/04/arduino-proximity-sensor-circuit-tutorial-and-code/&ei=X8cLUc3wH6ey0AGM9YHwBQ&bvm=bv.41867550,d.dmQ&psig=AFQjCNErlUpJgGrLh8m_uLr2kqFc1t6UYg&ust=1359812759717788http://www.google.com.sg/url?sa=i&rct=j&q=seven+segment+display&source=images&cd=&cad=rja&docid=QyDK_EN2ljJdmM&tbnid=aHz3l2KrWgymTM:&ved=0CAUQjRw&url=http://allaboutee.com/2011/07/04/arduino-proximity-sensor-circuit-tutorial-and-code/&ei=X8cLUc3wH6ey0AGM9YHwBQ&bvm=bv.41867550,d.dmQ&psig=AFQjCNErlUpJgGrLh8m_uLr2kqFc1t6UYg&ust=1359812759717788http://www.google.com.sg/url?sa=i&rct=j&q=seven+segment+display&source=images&cd=&cad=rja&docid=QyDK_EN2ljJdmM&tbnid=aHz3l2KrWgymTM:&ved=0CAUQjRw&url=http://allaboutee.com/2011/07/04/arduino-proximity-sensor-circuit-tutorial-and-code/&ei=X8cLUc3wH6ey0AGM9YHwBQ&bvm=bv.41867550,d.dmQ&psig=AFQjCNErlUpJgGrLh8m_uLr2kqFc1t6UYg&ust=1359812759717788


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4. (25 points total) This question concerns a mechanism for a gripper.

A) Design a gripper to the following specifications:

The gripper that can hold an object with 5N of grip force.

The object should be about 4cm in diameterlets imagine the object is a cylinder.

The gripper should be actuated by a single VS-2A servo. Upon opening, there should be at least 1 cm clearance with the object throughout the

approach motion which should be at right angles to the axis of symmetry of the cylinder.

If the VS-2A servo loses power, the grip force should not be released.

The design should be fully modular and bolt onto a flat surface with three #8 bolts in anequilateral triangle 4cm on each side.

A suggestion (at your discretion) -- make the gripper as simple as possible. Perhaps only onefinger should move and the other can be fixed with respect to a stationary wrist.

Sketch your design below or paste in a CAD screen shot.

I realize the grading on this will have to be fairly subjective.

I have worked up my own solution, just for fun.

Rubber bands attach here

between these two hooks andgenerate most of the grip force.

Servo actuates by push/pull rod


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B) Provide enough calculations and explanation to persuade a classmate that the arrangement in yoursketch does provide 5N of grip force and that it opens wide enough and that the VS-2A servo is

capable of performing the actuation.

I will try to get all of the

specified grip force from the

rubber. To provide 5N of gripat the tip of the fingers, we need

5N*79/45 or about 10N tension

in the rubber. Should be easy to

put enough rubber on thesehooks to accomplish that.

To overcome 10N of tension in the

rubber, we need 10N tension in the

rod (the rubber and the rod seem tobe about the same perpendicular

distance from the main pivot). To

overcome 10N of tension in the rod

we need 10N*17mm torque =0.17Nm. The VS2A is specified at

0.44Nm. That should be plenty ofmargin.


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C) Lets say we find that the gripper needs to be lighter, so we want to move the actuator (VS -2A)off of the arm and onto the base of the robot. Develop two options by which you can accomplish

this. Describe them with annotated sketches or other figures. Assume the arm must be able toarticulate vertically by 45 degrees. For both designs, estimate the percentage of actuation forcethat is diminished through the remote actuation device.

Cable housing is tied down rigidlyin two places, one on the arm andone on the base.

Push/ pull wire within thehousing emerges and is

attached to the lever on

the gripper. It would be

most effective for pulling,less effective for pushing

unless the wire is stout

and the free part is shortenough.

Push/ pull wire within the

housing emerges and is

attached to the horn on the

servo. There would begreat flexibility in

choosing the location of

the servo.

As the arm articulates vertically by 45degrees, the cable and housing would

have to bend. In the areas where the

cable went aroung a bend, it would act

like a capstan. If the may bend is 22.5deg (the 45 is split) then the losses

would be e^(-mu*22.5*pi/180). I

would say the losses in actuation forceare about 20% in the case of not verygood housing material (mu=0.5) and

maybe as low as 5% with a good

choice of housing and good surfacecondition.


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Pull/ pull strings applied to both the lever and the opposite side of the

moving finger and extending back to opposite sides of the servo horn.

The servo could be mounted on the arm to avoid coupling between gripactuation and arm elevation and still move the mass close to the base.

There would be essentially zero loss in actuation force. However, there

might be some dead band as string creeps over time. This arrangement

seems to need frequent adjustment.

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Homework2 Solution v4