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Standard Grade Technological Studies Summary Notes Compiled By Mr. A. Cunningham May ‘04 An acknowledgement must go to the authors of the LT Scotland Support Notes for the Technological Studies course, as some of the diagrams and text has been used from these notes in this document.
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Standard Grade Technological Studies Summary Notes Compiled By Mr. A. Cunningham May ‘04

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Page 1: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Standard Grade Technological

Studies

Summary Notes

Compiled By

Mr. A. Cunningham

May ‘04

An acknowledgement must go to the authors of the LT Scotland Support Notes for the Technological Studies course, as some of the diagrams and text has been used from these notes in this document.

Page 2: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Contents 

 

Systems 

  Page 3

Pneumatics 

  Page 5

Modular Electronics 

  Page 9

Component Electronics 

  Page 10

Logic Electronics 

  Page 14

Mechanical Systems 

  Page 18

Energy 

  Page 30

Programmable control 

  Page 32

Page 3: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Standard Grade Technological StudiesSystems Summary Notes

The Universal SystemAll systems can be analysed in terms of input, process and output. A diagram called the universal system diagram consists of these three basic elements.

process

input output

Universal system diagram

input sub-system

process (control) sub-system

output sub-system

OutputInput

System boundary

Sub-SystemsThe sub-system diagram shows the internal detail of the system. Each box, called a sub-system, can be thought of as a system within a system and has its own input and output. The dashed line around the sub-system is called a system boundary and this marks the area of interest to us. The ‘real world’ input and output are shown as arrows entering and leaving the sub-system diagram.

Open & Closed Loop Control Systems

Open loop systems are the simplest of systems. They take in an input, process it and produce an output. They do not use sensors or feedback to try and alter what the system is doing.

Closed-loop control is a more accurate system of control and at the same time more expensive. It employs self-monitoring, where a sensor is used to read the condition being controlled and adjust the output if necessary. This monitoring takes place through a feedback loop. Here an input sensor checks the output and adjusts it when it does not meet the requirements.

Types of Control

•Manual control is performed by the actions of humans.

•Automatic control is performed by technological devices (often electronic).

•Closed Loop Manual, e.g. A person filling a bath. The feedback is provided by the persons eyes watching the level of the bath.

•Closed Loop Automatic. System controlled by a machine usually automatic. Electronic transducers are used as sensors and provide feedback for electronics/computer to make decisions.

Positional control – used to control the exact position of an object/machine. Used in robotic arms to ensure arm rotates to the correct place.

Sequential Control Systems

 Sequential control is used where the outputs are required to follow a fixed cycle of events; that is to switch on or off in a particular sequence

Page 4: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

More On Open & Closed Loop Control Systems

In closed-loop control the value of the output is constantly monitored as the system operates and this value is compared with the set (or reference) value. If there is any difference between the actual value and the set value (an error), then the input to the system is varied in order to reduce the output error to zero.

A closed-loop system can always be identified by the presence of a feedback loop. An open-loop system never has a feedback loop.

Darkness sensor

controlsub-system

Room heated automatically

output driver

temperature sensor

heater+

Switch on

cold

Set levelElectrical energy

The diagram shows a control diagram for a typical closed loop system. The error detector takes in two signals – one form the temperature sensor (feedback sensor) and one from the set level. It subtracts one from the other. If there is any difference and error signal is produced. This is passed to the control sub-system which will decide how much to turn the heater on. This is called negative feedback and is typical in closed loop systems. The graph below shows how the Set Level and Actual level compare in the system above. AC TU AL TEM PER ATU R E

SET TEM PER ATU R E

TEM

PERA

TURE

T IM E

Feedback signal

Set level signal

Error detector symbol

Page 5: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Standard Grade Technological Studies

Pneumatic Systems – Summary Sheets

Safety

Learn all the safety rules, e.g.

• Wear safety goggles

• Don’t blow air at anyone, not ever yourself

• Don’t let compressed air come in contact with your skin

• Check all connections are secure before turning on the air

• Don’t leave pipes trailing along the floor

Advantages of Pneumatic System

Clean

• Pneumatic systems are clean because they use compressed air. If a pneumatic system develops a leak, it will be air that escapes and not oil.  

Safe

• Pneumatic systems are very safe compared to other systems. We cannot, for example, use electronics for paint spraying because many electronic components produce sparks.  

Reliable

• Pneumatic systems are very reliable and can keep working for a long time.  

Economical

• If we compare pneumatic systems to other systems, we find that they are cheaper to run. This is because the components last for a long time.

Flexible

• Once you have bought the basic components, you can set them up to carry out different tasks.

Page 6: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Describing How A Circuit Works

You will be asked to name components in circuits and describe how the circuits operate. In the General paper, you will only be given either AND control or OR control style circuits. At Credit level you will usually be given a sequential circuit (one which follows a particular sequence). A few examples are shown below:

Valve A Valve B

Valve A Valve BShuttle valve

AND CONTROL

OR CONTROL

In order to get the single acting cylinder to outstroke, you need to actuate valve A AND valve B.

The cylinder will outstroke if valve A OR valve B is actuated.

Could you describe how this circuit works?

ANSWER

When the push button is pressed, the 5/2 valve changes state and the cylinder outstrokes. As it outstrokes, it pushes the former together and the hot plastic sheet is pressed into shape. As this happens it also actuates the roller. Air now flows through the restrictor and starts to fill up the reservoir. Once the reservoir is full, the 5/2 valve changes state and the cylinder instrokes, ready for the process to begin again.

Page 7: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Completing Circuits

You can be given a pneumatic circuit and be asked to finish the piping to a given specification. If you want to do well in these questions, start by learning where the pipes go to basic valves. Try adding the piping to the circuits below.

Answers

Exam Questions

You must practise answering lots of Pneumatics questions with circuits to get a feel for the level of difficulty and the types of question you could be asked. There is no substitute for hard work I’m afraid!

Air Bleed Circuits

They use a diaphragm valve. When the air tube is blocked the air can no longer escape and is forced into the diaphragm valve which changes state and causes the cylinder to outstroke.

Page 8: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Calculations

All the formulas you need for pneumatics are given in the data booklet. (An extract is shown below)

Force in a single acting cylinder on Outstroke (Easiest calculation)

You will get the air pressure, P and the piston diameter, d.

Use d to get the area, a. You can either use or use a=r2 and ½ d to get r. Then you just use F=P x a to find the force.

Try these questions:

Find the force for the pressures and diameters given

1.P=0.3N/mm2, d=12mm

2.P=0.5N/mm2, d=23mm

Force in a double acting cylinder on Outstroke & Instroke

The outstroke calculation is the same as for the single acting cylinder.

To find the instroke force you need to work out effective area of the cylinder. ( remember there is less surface area on the instroking side of the piston because of the space taken up by the piston rod & so the force is always less than the outstroke force.)

Effective area = piston area – piston rod area

Once you work this out simply use it with the F=P x a formula to find the force.

Try these questions:

Find the instroke force for the pressures and diameters given

1.P=0.4N/mm2, piston diameter = 15mm, piston rod diameter = 4mm

2.P=0.8N/mm2, piston diameter = 20mm, piston rod diameter = 3mmAnswers: Single acting 1) 33.9N 2) 208N Double acting 1) 65.8N 2) 246N

Page 9: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Analogue and digital signals

All components in electrical and electronic circuits are either receiving or transmitting electrical signals. These signals can be either analogue or digital.

Analogue devices

An analogue signal varies according to the physical surroundings. For example, the E&L light-sensing unit will send out a voltage that is proportional to the amount of light falling on the LDR.

Typical analogue input transducers are:

•input voltage units

•light-sensing units

•temperate-sensing units

•moisture/rain sensor units

•sound-sensing units.

Digital devices

A digital signal is one which has only two settings, on or off. In electronic terms it has only two levels, high or low.

 

The push switch unit is a typical simple digital transducer.

Output transducers

 Output transducers take an electrical signal and change it into a physical output. They include the output boards in modular systems or output components in any electronic system.

Examples

Bulb Unit, Motor Unit, Solenoid, Relay & Buzzer

Relays

You must be able to complete a diagram showing a system with a relay, a motor and a separate power supply.

POS

SIG

0V

NEG

R ANG E+5V D C TO +8V DC

+

S0V-

+

S0V

-

E & L INSTRU M ENTS Ltd

TP

0v

O /PIND

AC TIVEHIG H

+

S0V-

+

S0V

-

E & L INSTRU M ENTS Ltd

TP

0v

0v

+V

0V

+

S

-0V

+

S

-

E & L INSTRU M ENTS Ltd

+ -6V

Sub-Systems Boards

You need to know what the following boards can be used for and how they can be linked together to produce a system.

Transducer Driver, Switch unit, light sensor, temperature sensor, moisture sensor, latch, comparator, transducer driver, buzzer, lamp, d.c. motor and solenoid (including actuating 3/2 valve). AND, OR, INVERTER(NOT), NAND & NOR

Remember all systems will start with an input sensor board (light, temperature,etc) and end with a transducer driver followed by an output transducer, e.g. a Motor, Buzzer, etc.

Remember relays are used to switch on higher powered circuits using low power control circuits.

Standard Grade Technological StudiesModular Electronics Summary Notes

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Logic Gates & Truth Tables

You must learn the symbols, truth tables and Boolean expressions for the logic gates shown

OR

AND

NAND

NOR

NOTZ = A

Z = A.B

Z = A+B

Z = A.B

Z = A+B

A Z

0 1

1 0

A B Z0 0 00 1 01 0 01 1 1

A B Z0 0 00 1 11 0 11 1 1

A B Z0 0 10 1 11 0 11 1 0

A B Z

0 0 1

0 1 0

1 0 0

1 1 0

Z A

BBAZ A B Z

0 0 0

0 1 1

1 0 1

1 1 0

XOR

Standard Grade Technological StudiesLogic Electronics Summary Notes

Page 15: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Boolean Expressions from truth tables

You must be able to take a truth table and produce Boolean expressions from it.Steps to follow:

•Find the 1s in the Z column

•Write the Boolean expression for each 1, e.g. Z = A.B.C

•Write the expressions out in words, e.g. Z = ( A AND NOT B AND C) OR (A AND B AND C)

•Write out the inputs, e.g. A , B C

•Draw in any NOT gates

•Draw in the AND gates

•Finally draw in the OR gates if required

A B C Z

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 0

1 0 0 0

1 0 1 1

1 1 0 0

1 1 1 1 Z=A . B . C

Z=A . B . C

Z = (A AND NOT B AND C) OR (A AND B AND C)

A

B

C Z

NAND Equivalents

NOR

XOR

NOT

AND

OR

Page 16: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

21 3 4 5 6 7

14 13 12 11 10 9 8

+Vcc

Gnd(0V)

21 3 4 5 6 7

14 13 12 11 10 9 8

+Vcc

Gnd(0V)

OUTPUT

7408 7432

+Vcc

0v

Pin-out Diagrams & Drawing Circuits

You must be able to select suitable logic ICs (chips) and draw in the connections for a given logic system. An example is given below. Don’t forget to draw in the connections for +Vcc ( the positive supply voltage) and 0v.

Input AInput B

Remember you don’t need to use all the logic gates in a chip – if you only need one, you only use one!

Drawing Logic Diagrams form Written Descriptions

You have two choices here to solve this type of problem:

1. Try to draw a diagram out straight away from what you have been told, or

2. Draw a truth table out for the problem and then go through the process outlined on the previous page for turning a truth table into a logic diagram.

The choice you make depends on the difficulty of the system. See over for an example.

Page 17: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Logic Diagram from a Written Description – Example

A machine (M) is to start under the following conditions

•The guard is down ( Guard Down: G=1, Guard Up: G=0)

•The operator is sitting on the seat ( On seat: S=1, Not on seat: S=0)

•Either the fast button or the slow button is pressed ( FST – Fast, SL – Slow)

•The machine temperature is low ( Temp Low: T=0, Temp High: T=1)

The logic diagram can be produced by simply reading the specification. We know that all the 4 conditions must be true before the machine will start – so we will need a 4 input AND gate. There are 5 inputs in total – G, S, FST, SL & T. The FST & SL inputs need to go through an OR gate because we are told that only one needs to be pressed and the temperature input must go through a NOT gate so that we will get a 1 out from it when it is low.

Diagram

Draw the inputs on the LHS first Next add in any OR or NOT gates And finally add the AND gate

G

S

FST

SL

T

M

Page 18: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Standard Grade Technological Studies

Mechanical Systems – Summary Sheets

Types of Motion:

Rotary

Turning in a circle. This is the most common type of movement, for example wheels, clock hands, compact discs, CD-ROMs.

Linear

Movement in a straight line, for example movement of a paper trimmer cutting a straight edge on paper or a lift moving between floors.

Reciprocating

Backwards and forwards movement in a straight line, for example the needle in a sewing machine or the piston in a car engine.

Oscillating

Swinging backwards and forwards in an arc, for example the pendulum of a clock, a playground swing or a rocking horse.

Levers

Levers can be used as force multipliers or distance multipliers. A force multiplier allows you to get a large force out for a small force in. A distance multiplier allows you to get a large distance out for a small distance in.

Mechanical Advantage (MA) = Load

Effort

Velocity Ratio (VR)=Distance Effort is moved

Distance Load is moved

Efficiency = = MA X 100%

VR

Page 19: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Moments

A moment is a turning effect.

Moment = Force x Distance

If a body is in equilibrium the sum of the clockwise moments must equal the sum of the anticlockwise moments. Or

CWM = ACWM

F1 d1 = F2 d2

This can apply to straight lever, angled levers and beams.Free-body diagrams

This is a diagram showing all the forces acting on a body.

Example: draw a free-body diagram representing the forces acting on the fork-lift truck.

R1 & R2 are the reaction forces. They push up to balance the forces pushing down due to the weight of the truck and its load.

Another condition of equilibrium is:

upwards forces = downwards forces

This is used with the principle of moments to calculate reaction forces in structures.

R1 R2

2kN

Weight

2m 0.7m

0.8m

R2W

Page 20: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

 

Free-body example

• Draw a free-body diagram for the car

• Calculate the reaction forces R1 and R2.

R1 R2

1.5m

1m

9.5kN

Take the moments about R1 (just think about it as being like a pivot)

CWM = ACWM

F1 d1 = F2 d2

9.5k x 1.5 = R2 x 2.5

R2 = 14250 2.5

R2 = 5700N

Now use

upwards forces = downwards forces

R1 + R2 = 9.5kN

R1 + 5700 = 9500

R1 = 9500 – 5700

R1 = 3800N

Page 21: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Gears

Gears are toothed wheels designed to transmit rotary motion and power from one part of a mechanism to another. Gears are used to increase or decrease the output speed of a mechanism and can also be used to change the direction of motion of the output.

Gear Ratios

Gear Ratio = Number of teeth on driven gearNumber of teeth on driver gear

Gear Ratio = DrivenDriver

or

In the simple gear train above the gear ratio would be:

If gear A is still rotating at 100 rpm in a clockwise direction then gear B will now rotate at 50 rpm in an anticlockwise direction.

Gear ratio=2412

21 or 2:1

Output speed = Input speedGear ratio

Idler gears

To get the driven gear to rotate in the same direction as the driver, a third gear is inserted in the system. This idler gear has no effect on the gear ratio of the system. The size of the idler is not important and is normally a small gear.

Page 22: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Ratchet and pawl

A wheel with saw-shaped teeth round its rim is called a ratchet. The ratchet wheel usually engages with a tooth-shaped lever called a pawl. The purpose of the pawl is to allow rotation in one direction only and prevent rotation in the opposite direction. Ratchet and pawl gears are used in winches to prevent slippage.

CABLE

RATCHET

PAWL BAR

CRANK HANDLE

WINCH DRUM

Gear Trains

When 2 or more gears are meshed together they form a simple gear train. The overall gear ratio of the train can be worked out using the first driver gear and the last driven gear.

A B C

D

+ + + +

The gear train shown includes 4 gears meshed together. The number of teeth each has is:

A = 50 teeth

B = 10 teeth

C = 25 teeth

D = 120 teeth

If A is the input driver the overall gear ratio is:Gear ratio=

Teeth on DTeeth on A

12050

or 12:5=2.4:1

It doesn’t matter if you have 2 or 52 gears in a simple train, the overall gear ratio is still calculated using the first and last gear in the train.

500rpm

Output Speed = Input Speed

Ratio

= 5002.4

208rpm

Page 23: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Compound gears

 If gears are required to produce a very large change in speed, for example if the multiplier ratio is 100:1, then problems can arise with the size of gear wheels if a simple gear train is used.

This problem can be overcome by mounting pairs of gears on the same shaft, this arrangement is called a compound gear system.

Compound Gear Ratio

Calculate the gear ratio of each pair of gears and then multiply the ratios together.

Gear ratio AB=DrivenDriver

8020

41

Gear ratio CD=DrivenDriver

=6010

61

Overall Ratio=41

61

241

If the input speed was 100rpm:

Output Speed=Input SpeedGear Ratio

=10024

=4.17rpm

Worm and wheel

Another way of making large speed reductions is to use a worm gear and wormwheel. The worm, which looks rather like a screw thread, is fixed to the driver shaft. It meshes with a wormwheel, which is fixed to the driven shaft. The driven shaft runs at 90 degrees to the driver shaft. When considering the speed changes in most worm gear systems, you can think of the worm as if it were a spur gear with one tooth. It is a single tooth wrapped around a cylinder.

Page 24: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Torque

Torque is the amount of turning produced by a force.

Example 1

How much torque is required to tighten the nut if the force required is 45 N and the radius of the tool is 200 mm.

radius

Torque Force radius

Nm N m

Torque Force radius

=45 0.2

=9Nm

Belt-and-chain drives

These are used when rotary motion has to be transmitted over a relatively long distance and gear trains would be too prone to losses due to friction.

Belt & Pulley Drive

A belt is wrapped around two or more pulleys. The belt is tightened or tensioned by pulling one of the pulleys out and locking it in place. Pulleys are thin metal discs with a groove cut into the circumference of the disc. 40 mm

160 mm

DRIVENPULLEY

DRIVERPULLEY

12

The tensioned belt transmits the rotary motion from pulley 2 to pulley 1. The belt is angled as shown in figure 2 to give better grip to prevent the belt from slipping. A change in speed can be accomplished by varying the diameter of the driver pulley and driven pulley. The driver pulley will turn in the same direction as the driven pulley. 

V-belt

Page 25: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Belt & Pulley Drive (continued)

A change in direction can be achieved by crossing the belt over. See diagram below.

DRIVEN

DRIVER

Multiplier Ratio

Like gears, you can vary the speed of a pulley system by using different sizes of pulley. The multiplier ratio is very similar to the gear ratio and is calculated as follows:

Multiplier Ratio=Diameter of Driven PulleyDiameter of Driver Pulley

Output Speed =Input Speed

Multiplier Ratio

Speed & Torque

When a pulley, belt or chain system produces an increase in speed, you get a corresponding decrease in Torque. Likewise, when a system produces a speed decrease, you get a corresponding increase in Torque.Belt tensioning – Jockey Wheels

One advantage of a belt system is that it can absorb shock loads by slipping. Too much slippage is undesirable however and the inclusion of a small pulley called a Jockey wheel can ensure that a belt remains in tension.

JO C KEYPU LLEY

D R IVERD R IVEN

Page 26: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Toothed belts

Belt drives tend to use their ability to slip to their advantage. However, where slippage would damage a mechanism, toothed belts have been developed that retain the advantages of normal belts but do not slip.

Chain drives

Where large forces have to be transmitted, and there can be no slippage allowed, chain drives are used. Instead of a pulley, a toothed wheel known as a sprocket is used to drive a chain. The chain in turn drives another toothed wheel. Once again, the speed can be varied by making the sprockets different sizes.

D R IVEN

D R IVER

Chain tension

Chain-drive systems must also have a means to tension the chain. If the chain is over-tensioned there will be excessive wear on the chain, sprockets and bearings in the system. In some bicycles and even motorcycles, the chain is tensioned by gently pulling the wheel back until the chain is tight and then tightening the locking wheel nuts. However, to give better control, a spring-loaded jockey wheel such as that used in Derailleur gears on racing bikes and mountain bikes is used

Spring loaded jockey wheels to maintain the tension as different sized gear wheels are chosen.

Page 27: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Converting Motion

R O TARYC AM

R O TARYM O TIO N

FO LLO W ER R EC IPR O C ATIN GM O TIO N

Cams

Changes rotary motion into reciprocating motion.

C R AN K

SLID ER

Rack & Pinion

Changes rotary motion into linear motion.

PIN IO N

R AC K

Crank & Slider

Changes rotary motion into reciprocating motion.

Worm

Worm & Nut

Changes rotary motion into linear motion.

Every full rotation of the worm results in the nut moving a distance equal to the pitch of the worm.

Example: A worm has a pitch of 2.5mm. If the worm is rotating at 100rpm how long will it take for the nut to move 1m?

Number of revolutions = 1000 2.5

= 400

The time taken will be 400 100 = 4 minutes

Page 28: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Efficiency & Friction

No machine is ever 100% efficient. This is because there are always losses in a system, e.g. energy is lost due to friction causing heat, or sound or light.

Uses of friction

Friction can be useful in some circumstance, e.g. in car brakes, car tyres or in belt drive systems where the belt needs to grip the pulley in order to make it turn.

OUTER RACE

INNER RACE

BALL BEARING

CAGE

Disadvantages of Friction and ways to reduce it

Friction causes unwanted wear on components and results in energy losses in machines as energy is converted into heat & sound. There are a number of ways of reducing friction, examples of which are given below.

We can reduce friction by oiling ("lubricating") the surfaces. This means that the surfaces no longer rub directly on each other, but slide past on a layer of oil. It's now much easier to move them. Other methods include:

•using "ball bearings" or "roller bearings", where balls or rollers allow the surface to move easily without actually touching each other

•using special materials, for example, Teflon, which have a very low coefficient of friction and thus slide easily (Teflon is used in "non-stick" frying pans for this reason)

Page 29: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Compound Levers

When you require a large force multiplication or mechanical advantage, levers can be linked together (A bit like compound gears) to produce a compound mechanical advantage. The example below illustrates this and shows how to calculate the output force.

0.6m

0.25m

0.15m

0.08m

The compound lever system shown is used to control a manual brake on a trolley. If an input force of 12N is applied what is the output force at the wheel?

Fulcrum

40.8NF0.25

0.8512F

0.25F0.8512

ACWMCWM

link. the togoes which

out force the find tomoments of principle the

use and handle) (the lever first the Consider

LINK

Easy!

76.5NF0.08

0.1540.8F

0.08F0.1540.8

ACWMCWM

again.moments

of principle the usingcrank bell the from force output the

outis work now do to need you Allis 40.8N.crank bell

the on down pushing force the thatknow now weSo

out

out

out

Fout

Fin

6.38M.A.12

76.5M.A.

In ForceOut Force

AdvantageMechanical

Page 30: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Standard Grade Technological Studies Standard Grade Technological Studies

Energy Summary NotesEnergy Summary Notes

Forms of energy

Non-renewable - finiteOil, Coal, Natural Gas & Nuclear

Renewable – infiniteSolar, Wind, Wave, Tidal & Hydro electric

Advantages of renewable sourcesWind – no pollution caused, can be built in remote locations.Solar – again no pollution, allow the trapping and concentration of the Sun’s energyWave - Wave power systems use kinetic energy in the waves to turn turbines. No pollution generated

Disadvantages of renewable sourcesWind – turbines are noisy and unsightly. Winds vary and so continual generation of electricity is not guaranteed.Solar – relies on long periods of sun light to be most effective. Not so good in Scotland in the winterWave – expensive to install and prone to a lot of wear as they have to be out at sea.

Formulae

Ep = mgh potential energyEk = ½mv2 kinetic energyEe = ItV electrical energyEh = CmTheat energyP = E/t PowerWork Done = F x d Work done

Note: you don’t have to learn these as they will be given in the data booklet.

The Law of Conservation of Energy

The law of conservation of energy asserts that for a closed system, where no energy goes in or out, the total energy within the system must always be the same, although its form may change.

Forms of energy

Kinetic – energy a body has when moving

Potential – energy stored in a body when it is raised up a height, or energy stored when, for example, a spring is compressed.

Electrical - the most versatile form of energy, it can be converted into many other forms of energy easily.

Heat - the energy transferred to a body which causes an increase in its temperature.

Conserving (Saving energy)

Most of the energy we use comes from non-renewable sources like fossil fuels. These energy sources will run out eventually and so it is important that we make them last as long as possible by limiting their use. You can save energy by: insulating homes and buildings, using energy saving light bulbs, driving fuel efficient cars, walking/cycling rather than driving,etc.

Page 31: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Energy transformations

You must be able to examine a diagram of a system (usually a power generating systems like hydro electric or wind power) and write down the energy transformations which take place, e.g. Kinetic Potential Kinetic Electrical

The diagram shown is a typical example of the type of question you could be asked.

A

B

C

D

E

F

Identify the forms of energy at points A (wind vane), B (generator), C (pump), D (water tank), E (water wheel) and F (generator).

Calculating efficiency

The efficiency of an energy transformation is a measure of how much of the input energy appears as useful output energy.

The efficiency of any system can be calculated using the equation:

Note: is the ratio of output to input energy. This can never be greater than one. In order to convert to a percentage, the efficiency, , is multiplied by 100.

Efficiency = =Useful Energy OutUseful Energy In

Energy OutEnergy In

Energy Audits

An energy audit is a list of all the energies coming IN and going OUT of a system. The total for the energies IN must be the same as the totals for the energies OUT.

Once you have calculated all of the energies in and out you should construct a systems diagram like the one shown below.

No machine is 100% Efficient

No machine is ever 100% efficient, there are always losses due to friction, heat loss and a fundamental principle of physics: some energy is always "lost" or wasted when one form of energy is converted to another. The "lost" energy is usually in the form of heat.)

Kinetic

Electrical

Kinetic

Potential

Kinetic

Electrical

Page 32: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Standard Grade Technological Studies

Programmable Control – Summary Sheets

Uses of Microcontrollers

Microcontrollers are single-chip ‘computers’ designed to control specific processes or products. They are found in a variety of products, e.g. household appliances (for example a microwave), alarm systems (for example a fire alarm), medical equipment (for example an incubator for premature babies) and electronic equipment (for example a computer mouse).

Advantages of Microcontrollers

•One microcontroller can often replace a number of separate parts, or even a complete electronic circuit.

•increased reliability and reduced quantity of stock (as one microcontroller replaces several parts)

•simplified product assembly and smaller end products

•greater product flexibility and adaptability since features are programmed into the microcontroller and not built into the electronic hardware

•rapid product changes or development by changing the program and not the electronic hardware.

Disadvantages of Microcontrollers

To program a microcontroller you need a computer. This can make it more expensive than building an electronic circuit.

Parts of the Microcontroller

Microcontrollers contain both ROM (permanent memory) and RAM (temporary memory).

The ROM (Read Only Memory) contains the operating instructions (that is, the ‘program’) for the microcontroller. The ROM is ‘programmed’ before the microcontroller is installed in the target system, and the memory retains the information even when the power is removed.

The RAM (Random Access Memory) is ‘temporary’ memory used for storing information whilst the program is running.

The ALU (Arithmetic and Logic unit) is used to perform calculation and to make logical decisions within the microcontroller.

The clock circuit within the microcontroller ‘synchronises’ all the internal blocks (ALU, ROM, RAM, etc.) so that the whole system works correctly.

Buses: Information is carried between the various blocks of the microcontroller along ‘groups’ of wires called buses. The ‘data bus’ carries data between the ALU and RAM, and the ‘program bus’ carries the program instructions from the ROM to the ALU.

I/O Port: This is the INPUT/OUTPUT port which connects the microcontroller to ‘real world’ inputs and outputs, e.g. from switches and sensors, motors and lights. The Basic Stamp has eight I/O ports which can be configured as either inputs or outputs.

EEPROM: electrically erasable programmable read-only memory. This is where Pbasic programs are stored for use by the microcontroller. Like ROM this memory is not lost when the power is cut and like RAM this memory can also be erased and a new program stored on it.

RAM

ROM

ALU I/O Port

Clock

BUS

Page 33: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Decimal to Binary Conversion

Example:

158 =

Method: Draw up the binary place values (see the table) and put a 1 in the largest place value which is less than the number, in this case 128. Subtract this place value from the number, 158-128=30, and then put a 1 under the largest place value which is less than this number, I.e. 16, subtract again to get 30-16=14, and so on.

128 64 32 16 8 4 2 1

1 0 0 1 1 1 1 0

Binary to Decimal Conversion

This is easy! Just write the binary number under the place value table and add up the place values with a 1 under them.

128+64+32+4+2= 230 Easy!

128 64 32 16 8 4 2 1

1 1 1 0 0 1 1 0

Flowcharts: these are used to plan out how a control sequence will work. Different symbols are used depending on whether the sequence is testing an input, switching on an output or pausing. The symbols you have to use are shown below:

Used for outputs, e.g. Switch pin 5 high

Line showing the flow of data. Arrows added to show directionProcess symbol, used to show a pause or a calculation, e.g. Wait 1sec

Decision, used to test inputs or if a loop has been completed, e.g. Is switch 1 on?, or have we looped 5 times?

Entry to or exit from a subprocedure

Terminator, either START or STOP

Page 34: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Programs

It isn’t easy to summarise programming as it is something you really have to “do”. Instead I have included example of some of the main concepts you need to know.

Switching on pins

High 5 ‘ switches on pin 5

Let pins=%01100000 ‘ switches on pins 6 and 5 and all the rest off

Let pins=0 ‘ switches all the pins off

Time delays

Pause 4000 ‘waits fro 4 seconds before moving on

If, then – testing inputs

Loop1: if pin1=0 then loop1 ‘if pin1 isn’t on then go back to loop1

Loop2: if pin3=1 then turn ‘if pin3 is on then go to turn

Loops

If you want to repeat a section of program a set number of times you use a FOR, NEXT loop. You must use a variable, ie b0,b1,b2 etc in the loop or use the symbol command to set up a variable name.

Symbol counter=b0 ‘sets counter to represent the variable b0

For counter = 1 to 10 ‘ start of loop

high 5 ‘ switch pin 5 on

pause 1000 ‘ wait 1 second

low 5 ‘ switch pin 5 off

pause 1000 ‘ wait for 1 second

Next counter ‘ end of loop (10 times)

Continuous Loops

If you want a program to repeat over and over forever, you should include a goto main at the end of the program, this will make the program jump back to the main label which is usually at the start of a program.

Setting up Input and Output pins

Let dirs=%11110000 ‘ sets pins 4-7 as outputs and 0-3 as inputs

Page 35: Standard Grade Technological Studies Summary Notes Compiled By  Mr. A. Cunningham May ‘04

Controlling the speed of motors

Pulse width modulation (PWM) is used to control the speed of d.c. motors. It works by switching the power to the motor on and off rapidly. The ratio between the time on and off is called the mark-space ratio and variation of this allows control over the speed. The main advantage of this type of speed control is that it is possible to maintain a reasonably high torque at low speeds compared to slowing motors by simply reducing the voltage. The diagrams below show how it works.

mark mark

on

off

spac

e spac

e spac

e

Slow speed Faster speedsp

ace

space

space

on

off

Sample program using PWM

main: high 7 ' output high (mark)

pause 5 ' pause for 5 ms

low 7 ' output low (space)

pause 10 ' pause for 10 ms

goto main ' loop

More on programs

GOSUB is used if you want to jump to a sub-procedure in a program. Sub-procedures start with a label, e.g. left: and end with a RETURN. This sends the program back to the line just after the sub-procedure was called using GOSUB.

Example

Gosub lights ‘jump to sub procedure lights

Gosub flash ‘jump to sub procedure flash

Flash: let pins=%11110000 ‘switch all lights on

pause 2000 ‘wait 2 seconds

let pins=0 ‘switch all pins off

return ‘ go back