Vigneswaran Ramamoorthy's Blog

A.ஒளியை பின்தொடரும் இரட்டை உருளைப் பொறியன்.

A.Differential Wheeled Light Follower Robot.

இந்த பொறியன், முன்புறம் எதிரெதிர் மூலைகளிலும் வைக்கப்பட்டுள்ள இரு ஒளி உணர்கருவிகள் மேல் படும் ஒளிச்செறிவின் மூலம் இயக்கப்படுகிறது.

This robot's movement is controlled by the light intensity it receives using two photo sensors placed at front opposite corners of the robot.

தேவையான பாகங்கள்:

Required Components:

1. நுண்கட்டுப்படுத்தி : நிரவுகோண் 8 துணுக்கு நுண்கட்டுப்படுத்தி (பிஐசி18ஃ452).

1. Microcontroller : Augumented advanced 8 bit microcontroller (Pic18f452).

• உள்ளமை மின்ஓடியக்கி,படிநிலை மின்ஓடியக்கி,எல்இடிக்கள் மற்றும்பல ஒப்புமை உள்ளிடு/வெளியிடு முகப்புகளை சரிபார்க்கவும்.

• check for built in motor driver,stepper motor driver,Leds and many analog input/output ports.

• திறன் சீராக்கம் மின்சுற்றை சரிபார்க்கவும்(0-5வி).

• check for power regulation circuit(0-5v).

2. பிஐசி நிரலர்.

2. Pic Programmer

• இந்த பொறியனை நிரலாக்க ஒரு தொடர்நிலைத் துறைசார் ஐசிஎச்பி நிரலர் வேண்டும்.

• a serial port based ICSP Programmer for PIC microcontroller is must for programming this robot.

• ஐசிஎச்பி நிரலர் ஐசி-ப்ரோக் மென்பொருளுக்கு ஏற்றதாக இருக்க வேண்டும்.

• programmer should be compatible with IC-Prog software.

3. ஒளி உணரி.

3. Light Sensor

• ஒளிசார் மின்தடை 7எம்எம் (இரண்டு).

• Light Dependent Resistor 7mm (2 nos).

4. மறுமின்னூட்டும் மின்கலத்தின் குணங்கள்.

4. Rechargeable Battery Characteristics:

• அதிக எம்ஏஅச் (ஆற்றல் கொள்திறன்).

• high mAh(energy capacity).

• எப்பொழுதும் 1ஆம்ப் ஆவது வெளியீடு வேண்டும்.

• can output atleast 1amp at any time.

• அதிகமாக மறுமின்னூட்ட இயல வேண்டும்.

• can recharge as much as we want.

• மறுமின்னூட்ட 10 மணி நேரம் பிடிக்கும்.

• 10 hours to recharge.

• உயர் தன்வய மின்னிறக்கம்.

• self discharge rate is high.

• நிகாடு மின்கலத்தைவிட அதிக மறுமின்னூட்ட சுழற்ச்சி.

• more number of recharge cycles than Nicad battery.

• பரிந்துரைக்கப்படடும் மறுமின்னூட்டும் மின்கலம்.

Recommended Rechargeable Battery:

• இரு என்ஐஎம்அச் மின்கலக்கட்டு.

• two Nimh 6v battery packs.

• விசைப்பொறித்திறனை சுற்றுத் திறனிலிருந்து தனிமைப்படுத்த இரு மின்கலகங்களை பயன்படுத்தவும் (ஒரே மின்கலத்தை பயன்படுத்தினால் சுற்றமைப்பை உருக்க நேரும்).

• two batteries to isolate motor power from circuitory power(single battery for both motor and circuit may melt sensitive circuitory).

5. விசைப்பொறி.

5. Motor

• இரு 5-8வி டிசி விசைப்பொறி.

• two 5-8v DC motor.

• முறுக்கு விசை தேவையைவிட இருமடங்கு இருக்க வேண்டும்.

• torque should be double the requirement.

6. உருளை.

6. Wheel

• 2-3 இன்ச் குறுக்களவு உருளை.

• 2-3 inch diameter wheel.

 

பிஐசி தொகுப்பி:

Pic Compiler:

பிஐசி-18 நுண்கட்டுப்படுத்தியை நிரலாக்க, ஐ-டெக் சி தொகுப்பி வேண்டும்.

• Hi-Tech C compiler for Pic 18 microcontrollers.

• இந்த லைட் பதிப்பில்(இலவசப்பதிப்பு), உகப்பாக்கம் தளர்வுறச்செய்யப்பட்டுருக்கிறது (அதாவது தொகுத்த கோப்பின் அளவு பெரியதாக இருக்கும்).

• In this lite version(free version), optimization is disabled(i.e size of the compiled file is big).

Download C Code

Note: PhotovoreRobot_code\dist\default\production\Photovore.X.production.hex (HEX file to be loaded in PIC MCU)

B.Phenomenon observed by Vignesh Chandraguru.
Explanation by Vigneswaran Ramamoorthy.

When the fingers are open, the length between the reference y-axis and tip of fingers noticeably varies. But if they are folded, the tip of all the fingers seems to touch a horizontal line, somewhere below y-axis. Why?

 

The reference y-axis is drawn at the bottom tip of the baby finger. The lengths y1 is from the y-axis to bottom tip of middle finger, y2 is from the y-axis to bottom tip of index finger, and y3 is from the y-axis to bottom tip of ring finger are such that y1>y2>y3.


Let us assume the distal, middle, and proximal phalanx are equal in length for all the fingers, that is
b1=b2=b3
r1=r2=r3
m1=m2=m3
i1=i2=i3


If the fingers are closed, the angle between the folded phalanges should be 90 degrees and consequently the middle phalanx of the fingers doesn’t contribute to y-distances any more. The length of the up-going Proximal Phalanx is negated by the down-coming distal phalanx of the fingers.


Thus when the fingers are folded, the distance between y-axis and the tip of fingers are y1, y2, y3, and 0 for middle finger, index finger, ring finger, and baby finger respectively.


Contradicting the second line of the observation, the tip of all fingers doesn’t touch a horizontal line.See the image below.


Uncertainties:
1)The length of the three phalanges are not always the same for everyone, and in most cases the order is Proximal>Middle>Distal.
2)The order y1>y2>y3 may not hold for some hands.
3)If y1, y2, and y3 are equal in length, the tip of all folded fingers touch the y-axis.

C.Phenomenon observed by Vigneswaran Ramamoorthy
Explanation by Vigneswaran Ramamoorthy

When viewed from a moving train, the objects which are closer seems to move faster than those which are far away in the direction opposite to the motion of train.Why?

Consider two objects, in our case trees, which are seen by a person at rest, from a train moving at a speed of S m/s.

P1 and P2 are planes which lies in the middle of the near and far planes of objects O1 and O2 respectively. See fig a below.

Let h1 and v1 be the horizontal width and vertical height of plane P1, and h2 and v2 be the horizontal width and vertical height of plane P2. See fig b below.

The distance between the planes is such that h2=2h1 and v2=2v1 and the line of sight touches the centre point of both the object's bounding box. See fig b below.

Consider that the train is moving with a constant velocity V such that the eye, inturn the planes P1 and P2 moves h1/2(half the horizontal width of plane P1) distance in t1 seconds.

From the fig c, we can note that at time t=t1, the object O1 is about to go out of scope of view in plane P1, but the object O2 is still within the plane P2.

D.Phenomenon observed by Vigneswaran Ramamoorthy.
Explanation by Vigneswaran Ramamoorthy.

The explanation is not updated completely. Please ignore the below explanation.

In a circus, a rescuer saves a person who falls down during a rehearsal by dashing him the horizontal direction, from a few feets above the ground. The impact, the person might have experienced if none had saved him is considerably reduced by the rescuer's action.
Analyse the phenomenon.

 

<img src="images/file000998133122.jpg" alt="Alternative Media" >

In case of problem in playing the above video, the clip can also be seen in this youtube link. See the time period 4:41 to 4:50 in this youtube video.Sorry for the inconvenience.

Overview:

The position, velocity, and acceleration components along with energy and momentum for this problem are analysed below.

For the explanation, we can consider two cases
case1: If the person falls to the ground without any rescue.
Case2: If he is rescued as shown in the above video.

In case2, we can separate the trajectory in which the casualty falls into two independent trajectories. i.e, the trajectory from A to C in fig b is separated into two independent trajectories, from A to B and from B to C.

At the end of the analysis, for a range of vo's (rescuer's horizontal velocity), the corresponding loss of total mechanical energy to other systems after collision are calculated.

Position, Velocity, and Acceleration Components:

Equations of 1D motion with constant acceleration

Position function, xt=xo+voxt+1/2 axt2 (ax is constant)
Velocity function, vxt=vox+axt.
Acceleration function, axt=ax

Replace x with y&z for y&z dimensions.

Lets consider this to be a 2D problem, so we wont be mentioning Z-direction components in rest of the explanation.

Case1: If the person falls to the ground without any rescue.

The time when he starts to fall at A be 0 and the time when he touches the ground at D be t1.

Case1: For time 0 and t1, the x and y direction components of position, velocity, and acceleration are decomposed in the table below.

x direction components for fig a
at time 0 at time t1
xo=0 (my free choice) xt1=0+0+0
vxo=0 vxt1=0+0
ax0=0 axt1=0
y direction components for fig a
at time 0 at time t1
yo=h1 (my free choice) yt1=h1+0-1/2 gt12
vyo=0 (initial velocity is 0) vyt1=0-gt1
ayo=-g ayt1=-g

Case2: If he is rescued as shown in the above video.

Here for case2, we can separate the motion showed in fig b to two independent motions as shown in fig b(i) and fig b(ii).

 

In fig b(i), the time at point A is 0 and time at point B be t2. In fig b(ii), the time at point B is 0 and the time at point C (where the casualty touches the ground) be t3.

In case2, for both the motions, the position, velocity, and acceleration components in x and y directions are decomposed in the below tables.

x direction components for fig b(i)
at time 0 (at pt A) at time t2(at pt B)
xo=0 (my free choice) xt2=0+0+0
vxo=0 vxt2=0+0
ax0=0 axt2=0
y direction components for fig b(i)
at time 0 (at pt A) at time t2(at pt B)
yo=h1 (my free choice) yt2=h1+0-1/2 gt22
vyo=0 (intial velocity is 0) vyt2=0-gt2
ayo=-g ayt2=-g
 
x direction components for fig b(ii)
at time 0 (at pt B) at time t3(at pt C)
xo=0 xt3=0+vxt3+0=w
vxo=vx (+ve value) vxt3=vx+0
ax0=0 (vx is const) axt3=0
y direction components for fig b(ii)
at time 0 (at pt B) at time t3(at pt C)
yo=h3

figb2_displacement_y_t3

vyo=vy1 (-ve value)

figb2_vel_y_t3

figb2_acc_y_t0

figb2_acc_y_t3

 

 

 

 

 

 

 

 

 

 

 

 

 

Velocity components at point B

Velocity components of casualty just before the collision: vxt1=0 to vyt1=vBy

Velocity components of rescuer just before the collision: vxt1=vo to vyt1=0

Velocity components of combined mass(casualty+rescuer) just after the collision: vxt1=vx to vyt1=vy1

Energy equations:

1. Work done by gravity to move the mass (casualty) from A to D:








2. Work done by gravity to move the mass (casualty) from A to B:








3. Work done by gravity, when the mass (rescuer+casualty) moves from B to C:

     




















Momentum equations:

















































 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



Imagine the rescuer and the casualty to be in outer space. Let us assume that the force experienced by them due to gravity of any celestial body is zero. Now let us give the casualty an initial velocity  and consider the rescuer to be stationary.
 It is an inelastic collision (the colliding bodies stick together after collision). In this case, after the casualty hits the rescuer, their masses stick together and continue to move with a velocity 








Conclusion:

 

.

 

என் வலைப்பதிவைப் பார்த்தமைக்கு நன்றி.

Thanks for visiting my blog.