旼컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴커 � � � 爐커 靈커 靈爐� 靈커 爐커 靈컴컴엿 � � � 픔 � � � � 픔쬔 � � � 獺 粕 懃켸 聃켸 � 聃켸 � � 聃컴컴엿 � � � Tutorial by Richard Nichols 납 � 聃컴� 聃컴컴冒 읕컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴켸 (c)1996 Richard Nichols 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴 Hi! You are reading a tutorial in vectors by Richard Nichols of Australia... This document is not to be reproduced in any other form than this original. ie. don't cut bits of it up etc. Also this MUST be distributed free of charge. The only charge from me for your use of it is that you come visit my website at 'http://bip.concept.se/user/rich/index.htm' and tell me what you think... Okay enough crap, let's get on to the main bit: 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴 Part One. In part one of this tutorial we will go over the very basic act of adding two vectors together in a variety of ways. I will give a few ways because, although methods used in programming are often faster and simpler, they usually don't give a good understanding of what you are doing. 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴 What are vectors? 컴컴컴컴컴컴컴컴� Vectors are units which have both magnitude and direction. For example 500 metres at 57� is a vector. The opposite of a vector is a scalar which is a unit with only a magnitude. eg. 500 m. Vectors are written in a few ways. If we assign a letter to a vector we could write (for example): � or � or i i ~ . This isn't very important though. For the purposes of this tute you can assume that any alphanumeretic I use is a vector. We will mainly be dealing with direction vectors. ie. A distance or force and a bearing eg. 134N @ 45�. I may also refer to position vectors which are the same as normal ones but originate from the origin. 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴 Vector Addition : Cartesian (rectangular coordinates) 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴� Vector Addition on our rectangular plane is easy. I will demonstrate: Let Point A = (5, 2) and B = (-4, 1).  � � � � 2� �A B� 1� <컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴> -4 � 5 Imagine a line coming from � the origin and going to each � A & B. These are our vectors.� � So A + B is somewhere in � between these.  All we have to do is add the respective x and y coords together: x coords: (+5) + (-4) = 1 y coords: (+2) + (+1) = 3 So A + B = (1, 3). Easy huh? Well on to polar coordinates then... 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴� Vector Addition : Method A - Least complex - Least Understanding 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴 For those of you not familiar with polar coordinates, imagine a plane with concentric circles as shown: - | - / | \ / | \ / | \ / | \ / | \ / More circles|here (sorry:Ascii) \ / | / | \ ---------------------------+------------------------------ \ | / \ | \ | / \ | \ | / \ | / \ | / \ | / Instead of giving an (x, y) coordinate we give a radius and degrees: (13, 45�). This says a line of 13 units 45� around from 0� (and of course originating from the origin). Where 0� is depends on what you are doing. In Maths usually we use to the right horizontally (-->) as 0 but in Physics it is more common to use bearings so that upwards is 0� (under this system we say something travelling at 0� is going straight ahead). The method I am about to show here simply uses what you have already learnt. We convert the Polar coordinates to rectangular and then add them. We then convert them back. We convert them like this: /| / | We know what r (radius) and � (angle) are r / | so we have to change them into x, y as shown. / |y /� | +-----+ x This is _very_ simple trig that everyone should know: x = r.cos.� y = r.sin.� If you don't get it do a few equations and you should see the significance. Convert the two vectors as above and then add them like in the first part. Converting them back is a little more difficult: 1st get �: � = arctan(y/x) now get r: r = x 컴컴컴� cos.� or alternatively use pythagoras: r = �(x� + y�) Again if you're not sure do a few examples. If that's as far into this as you want to go then you can skip past the next two methods. 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴� Vector Addition : Method B - Intermediate - Reasonable Understanding 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴 But real men like trig. :) This method uses what are known as the Sine and Cosine rules. This was the first method I was taught (in Physics). It would not be particularly useful in a programming situation. On a triangle we must name the sides in a certain way eg. + /C\ / \ Where A, B, C are angles (must use caps) a / \ b and where a, b, c are the sides. / \ / \ /B A\ +-------------+ c The point of this Ascii picture is that the angles have a corresponding side of the _same_ letter. This is essential while working out using this method. As long as you get this right it is plain sailing. Okay here are the rules. I won't bother with the proofs for them for two reasons a) I can't think of the top of my head b) It would take too long! Cosine Rule: a� = b� + c� - 2bc.cos.A Sine Rule: sin.A = sin.B = sin.C 컴컴� 컴컴� 컴컴� a b c This sine rule can also be inverted so that the top is the bottom and vice versa (it's a ratio after all). It just depends on what sort of numbers you want to work with... Anyway on to the main bit: The difference with this method is that we are using polar coords all the time. This means we must add the vectors correctly, that is end to end. The true representation of (5, 0�) + (3, 90�) for example would be like this: 3 컴컴컴�> 넨 / � / 5 � / R O = Origin � / �/ - - - - - - -O- - - - - - - - - Where the diagonal vector R is the resultant vector. It is now much easier to see what is going on isn't it? I will point out now that we will hardly ever have a right angle but I used one to make the diagram clearer okay? We must now name our sides like so: c 컴컴컴�> 넨A B/ � / b � / a O = Origin 쿎/ �/ - - - - - - -O- - - - - - - - - Yeah, I know it's a spas diagram but its the best I can do with ascii. So what do we know? We know c = 3 and b = 5 and that A = 90� ( (3, 90�) ). Ah! We can use the cosine rule to work it out: a� = b� + c� - 2bc.cos.A substitute: a� = 5� + 3� - 2x5x3xcos(90) = 25 + 9 - 2x5x3x0 = 34 therefore: a = �34 = 5.830951895....... So we now have the vector's magnitude but what about the angle? This is where the sine rule comes in: sin.A = sin.C 컴컴� 컴컴� a c sin(90)/5.83= sin(c)/3 0.172= sin(c)/3 (*3) 0.515= sin(c) therefore: c = arcsin(.515) = 31� Only one more thing: We want a bearing from 0�. This is easy we can see that all we have to do is subtract 31 from 90 which equals 59. That's it! Now we have our resultant: 5.83 @ 59�. 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴� Vector Addition : Method C - Expert - Excellent Understanding 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴 Personally I don't like this method; it is too much like hard work. :) I learnt this in Maths C. It works well but is pretty useless where computers are concerned but you should learn how to do it just to do it for the sake of better understanding. Suposing we had two vectors a and b. a=(10, 45�) and b=(5, 100�) 0� -\ a & b are the vectors | / - a / \ b | / - / \ | / / |/ 270�- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -90� | Basically this method entails drawing two triangles from the two vectors: 0� -\ a & b are the vectors | /|Z- a / x \ b I just chucked w,x,y,z | / |� - on there so you would we know a, b and �. / |-z- \ be able to reference | / | the next bit.... /� 쯻 |/------| 270�- - - - - - - - - - - - - - - - - - -w- - - - - - - - - - - - - - - - -90� | What we have to do is to resovle the vectors into their respective horizontal and vertical coords and then add them together... Horizontal= a.cos.� + b.sin.Z If you can see that then have a lok at the diagram. It's just basic trig. Now we have a problem: We don't know angle Z. We find it like so: | 0� |� - - - -\- - - - - - | /|Z\10� If b = 100� then a / | \ the angle shown must | / |� \ be 10� which means / |--- \ Z=80�. | / | /45� �| |/------| 270�- - - - - - - - - - - - - - - - - - -w- - - - - - - - - - - - - - - - -90� Great! now we can work out the horizontal distance. Horizontal= a.cos.� + b.sin.Z = 10.cos.45� + 5.sin.80� = 7.071 + 4.924 = 12 (near enuf) Now we do the vertical: Vertical = a.sin.� - b.cos.Z (we are finding side y on the 1st diagram) = 10.sin.45� - 5.cos.80� = 7.071 - 0.868 = 6.2 Okay now we have that lets draw our resultant diagram: 0� | + | -/| We can put out horiz & vert | -/ | nos on here now... | -/ |6.2 | -/ | |-/ | ------------------------------------12--------------------------------90� We get the length of the vector by just using pythagoras: a� = b� + c� = 12� + 6.2� = 144 + 38.44 = 182.44 a = 13.5 0� | + | -/| | 13.5-/ | | -/ |6.2 | -/ | |-/A | ------------------------------------12--------------------------------90� Now to get the angle is easy...We could cos, sin or tan..I'll use tan. tan.A = OP/AD = 6.2/12 = .51666... A = arctan(.5166..) = 27.324� 0� | + | -/| | 13.5-/ | | -/ |6.2 | -/ | |-/27.324� | ------------------------------------12--------------------------------90� Now just to finish off make it a bearing: 90-27.324 = 62.676� So the answer = 13.5 @ 62.7� --------------------------------------------------------------------------- Sooo..That concludes the tutorial. This should be simple stuff for most people but if you do have any trouble mail 'tiberius@mailhost.net' for help. Here are some questions for you. Use whichever method suits you: 1. Add (4, 10�) and (5, 60�) 2. Add (9, 100�) and (12, 170�) 3. A boat is travelling along a river at 3 knots with bearing of 39�. An easterly wind is blowing at 1 knot. What is the boat's true heading and speed? 4. A bit more thought: Say I have a traffic light spanning a road joined to two buildings by two ropes as shown: ----------------------------------- \60� 60�/ \ / \ / \ / \ / The light's weight is 150 newtons. +--+ |[]| |[]| +--+ Work out the tension (or pull) in the ropes.... ---------------------------------------------------------------------------- Check the answers with the vector addition program attached to this......... ---------------------------------------------------------------------------- This tutorial is (c)1996 Richard Nichols. I take no responsibility for the feasability or correctness of this document, or any damage it could do. Remeber to check out my cool website at: 'http://bip.concept.se/user/rich/index.htm/' Byee, byee May, 1996 - Richard Nichols