 Have you ever wondered how spacecraft maneuver while they are in orbit? They have small groups of thrusters known as reaction control systems or RCS that can control the orientation and in some cases translate the spacecraft. They work by ejecting mass to create a reaction force that can move the spacecraft around its center of mass. Can we examine such a design problem using what we've learned in STATICS thus far? Examining a complete spacecraft in three dimensions is perhaps a bit too complex for us at this moment. But we can look at an overly simplified case to understand the analysis. What I will call the wonky space capsule for reasons that will hopefully become apparent in a minute. Our hypothetical spacecraft is illustrated here. It has a main thruster that is aligned with the central axis of the spacecraft and two smaller thrusters that are oriented at an angle of 70 degrees and 55 degrees with respect to the central axis of the spacecraft. Our problem statement for this spacecraft is what is the equivalent force couple at the center of mass, or point O, as a function of the magnitudes of the three thruster forces. For this we will define an axis system aligned with the central axis of the spacecraft. We then need to determine the location of these thrusters in our defined coordinate system which is provided here as coordinates from the center of mass in the units of meters. We now have all the information we need to determine what the resultant force R and resultant couple M sub R acting at the center of mass due to the thrusters is. Now that we understand the problem statement we now have to choose a solution methodology. We have seen that we can analyze what a resultant force moment system is by using vector formulation or vector decomposition. We've already looked at an example that uses vector decomposition so we will use the vector formulation for this example. We now examine our equations to find our resultant force and moment and the resultant force comes from summing all of the force vectors. So our resultant force R becomes F1 plus F2 plus F3. We will call this equation I. We also can look at some of the moments that tells us that our overall moment couple M R will be the summation of all the cross products of the position vector R crossed with the force of the thruster F where the position vectors will be defined as the distance between our center of mass and our thruster. Now we can already see a simplification in this problem in the sense that force F1 passes through the center of mass. So it will not actually create a moment about that point. So we see that R1 cross F1 will actually be zero. So we can already simplify the equation. We will call this equation equation two or sub-II. Now that we have our two summation equations we know what vectors we need. We need to determine what these vectors are. We can start with the force vectors and very easily see that force vector F1 will have a magnitude of F1 and it is acting completely in the y direction. So it will have a direction cosine of negative one. If we look at force two it will have a magnitude F2 and it will have an x component of the sine of 70 degrees but it's in the negative direction, so negative sine 70 and a y component in the positive y of cosine 70. We can then evaluate those expressions and get our numerical values. Again, similarly for F3 it will be in the positive x and positive y and we will get the x component will be sine 55, the y component will be cosine 55 and there is no component in the z direction because it's acting within the plane. We can then calculate those numerical values. Now we also for our moment equation here we need two position vectors R2 and R3. So we'll start by defining R2 as a distance between the center of mass and our thruster F2 and here we can see that we can actually just use our coordinates for the location of the thruster because that actually defines the position vector because we conveniently located our coordinate system at the center of mass. Similarly for R3 our position vector will be given by the coordinates. So x is one meter, y is four meters and z is zero meters in this case. Now we can go on to calculate things because we've determined all our vectors. So we'll start with calculating the reaction force which will be F1 plus F2 plus F3 subbing in our results that we calculated for our vectors here. We will get the following equation and we can see here that you're going to get a complex summation of all these magnitudes and we see that the z values are zero. So to clean up the notation when I add them I will break it up into the x and y components. So the i components becomes zero F1 minus 0.940 F2 plus 0.819 F3. So that's this component here. We can then add the y component from this line and our z component is zero in this case. So don't be afraid by the alternative formulation of it. It just saves a little bit of space here. We can now look at our resultant moment from our equation two where the force one component we said was zero or the cross product so we only have two cross F2 plus R3 cross F3. So we can now evaluate each of these cross products. So R2 cross F2 we can determine using the Amsterdam method where we stack our position vector on top of itself twice. So that's just R2 here stacked on itself twice and our force vector F2 right here stacked on itself twice. And in this method if you recall we cross out the top and bottom row and then we evaluate the cross products of the various rows. So we'll get minus two times zero minus zero times 0.342 F2 which ends up just being zero minus zero. And similarly we will get zero minus zero for the next evaluation next line of the evaluation and the last one will be non-zero. We will get minus one times 0.342 F2 this value here minus negative two times negative 0.94 F2 and this is in the K direction. So this gives us our three components and we will get that it's 2.22 F2 in the K direction. And note here that I've put units of meters not because the units of this whole thing is meters but the units of the number F2 we can still put the units in which would have newtons so forth. So our result so far is just from this cross product we need to evaluate the next cross product. So to do that again we can use the Amsterdam method and I'll go through it a little bit more quickly but we get those first two components are zero because of the multiplication of zero on each line. And then our last one we will get one times 0.574 F3 minus four times 0.819 F3 which results in negative 2.70 F3 in the K direction and then we can add this in and this becomes our final expression for our resultant moment. Again the units meters reflects the numbers we would still have to substitute in the units of our force when we place those values into the equation. So here we have our final results we should take a moment to think whether they make any sense and we can actually look at a few things since our model is two dimensional we would expect our resultant force to be contained within the XY plane and indeed it is to have an X and Y component to our resultant force. Similarly if you think about the resultant moment it should be causing a rotation or would tend to cause a rotation within the XY plane so about the Z axis coming in and out of the page and indeed our moment only is acting in the Z direction so at least from the directions and all of that, that makes sense. Now it's hard to evaluate the numbers because we did not actually put thrust values in here we kept it in parametric form and you may ask well why keep it in parametric form you'll see a lot of examples within the textbook have numbers and you're evaluating an actual fixed value we do that in a lot of textbook problems to practice but I wanted to show you this parametric form because this is very useful in an engineering analysis and I will show you that in the next few slides but I will warn you this kind of goes a little bit beyond what you would do in most problems but it shows you how applicable such a parametric equation is so let's take a look we can first look at our moment equation this was the moment equation for the function of our two thrusters f2 and f3 remember f1 was aligned with the center of mass so it would not cause a moment I can plot this as an actual contour surface so here we see a contour plot where the color is showing you the value of the moment and then I have on the x-axis f2 so I treated f2 as my x variable and f3 as my y variable starting this function now why is this useful well you can imagine you might want to create a certain resultant moment to cause a rotation of the spacecraft but as you can see here these contour lines tell me that I have multiple thrusts that can give me that value so I can look here for instance and at this point of the contour line it's approximately 275 newtons for this thruster or sorry for the for f2 and approximately 600 newtons for f3 alternatively I could produce the same rotational moment but with a thruster force of 650 newtons and 900 newtons now obviously hopefully it doesn't come as much of a surprise to you that this would be way less efficient you have to apply two larger thrusts so you would have to eject a lot more mass or use a lot more fuel to get that rotation so such a plot could help you evaluate the efficiency of certain maneuvers it also allows you to observe something else and that is what this plot is missing this doesn't show us the resultant force so if we did certain maneuvers you might get a different translation out of it but what we can observe is a particular line where our resultant moment is zero and in this case that is where f2 is equal to 1.22f3 you can actually solve for this line with this equation equal to zero and solving for f2 as a function of f3 this is an interesting line because along this line this is the combination of thruster forces that would not cause a rotation so the spacecraft would only translate so what we could do is examine all the possible maneuvers where you would avoid the rotation we could take our resultant we already came up with parametric form if we apply the condition that f2 is equal to 1.22f3 which was the condition where our resultant moment was zero then this equation simplifies a little bit and we get that our reaction force is negative 0.324f3 in the i direction plus 0.99f3 minus f1 in the j direction I can now see that this isn't a simple magnitude value this is a vector equation still so what I can now do is plot a vector field so if I look here on one axis I have f3 and on the other axis I have f1 and I've plotted for different combinations what the resultant force would be of a vector field so obviously if we're only applying thruster f1 and 0 on the other thrusters then I'm only going to get a downwards thrust because our thruster is down here it would be pointing downwards but you can get different combinations of translations without rotating the aircraft you could then evaluate the aspect of fuel efficiency but you have to be a little bit careful here this axis is only showing f3 but we also have to ensure that f2 is equal to 1.22f3 so there's actually a lot more thrust than what is shown on this axis so we could draw an alternative axis which would be f2 plus f3 and here you see the total thrust from the two smaller thrusters corresponding to this vector field and you could see if I wanted to translate the aircraft or apply a downwards force the thruster 1 would probably be more efficient but if I wanted to go in the other direction I could go here but now I'm using up a lot of thrust from my other thrusters so that's not so fuel efficient you might actually want to rotate the aircraft first to get into orientation where you can use this more efficient direction so hopefully that gave you an idea of some of the things that you can do in terms of further analysis I admit it goes a bit beyond what we would expect for you to be doing within a lot of statics problems but I think it's really important for you to see where the skills and tools you are learning now are going towards so that you will be able to see that you can apply it to much more interesting problems, design problems and decision making situations which is really the crux of engineering