 So now let's consider another example of vector addition in two dimensions applied to aerospace. In this particular problem we have a micro drone, a small drone somewhat sketched here, that is traveling with an airspeed of 8 meters per second and a heading directed 30 degrees south of west. There's a steady northerly breeze with a wind speed of 5 meters per second and a bearing of 150 degrees. If no correction is made for the wind, what is the bearing and ground speed for the track of the drone and what is the drift angle between the drone heading and the track bearing? So this problem is going to require us to define some vectors, to draw some vectors and to put those vectors together to add the airspeed and the windspeed to determine the ground speed. And so again, what we're looking for here, we'll recognize the equation that we're going to use here. Ground speed equals airspeed plus windspeed. Or to write that a little more simply, the velocity over the ground is equal to the velocity through the air plus the velocity lent to the flying drone in this case from the wind itself. So let's start by going ahead and drawing what we actually see here. So we see there's an airspeed of 8 meters per second with a heading directed 30 degrees south of west. Now notice that's a little bit different than if we just set a heading, typically that's considered to be from the north and measured in a clockwise direction. In this case, that's not what's happening, we're actually measuring 30 degrees south of west. So I'm going to go ahead and sketch, we'll go ahead and determine west as being to our left here and we'll estimate something here that's about 30 degrees south of west. 30 degrees and we'll give that a magnitude of 8 meters per second. Typically, I like to go ahead and draw vectors, velocity vectors coming out from whatever's actually moving. Continuing to read, we say there's a steady northerly breeze with a wind speed of 5 meters per second and a bearing of 150 degrees. Now northerly means coming out of the north, so, and here we have a bearing of 150 degrees. Now that's where we're going to use our typical description of what bearing is, which is measured from a northerly direction. Here's north and that was our west and we went south of west, but now we're going to go typically clockwise. Okay, there's our clockwise 150 degree bearing and it's going to be at 5 meters per second is the magnitude that we're considering here. So now that we've sort of drawn our picture here, let's actually take these two vectors, you can sort of see that the wind is adding here, but what we're going to ultimately end up doing is adding our vectors and to add the vectors, we probably want to consider them a little bit more head to tail in our vector sketch here. So let me go ahead and take my 8 meters per second and my 5 meters per second and add those two things together. Actually, I just didn't add them head to tail. Let's go ahead and add them head to tail. There's our 5 meters per second and so what we're looking for is the addition of those two. There's our ground speed as a combination of our air speed and our wind speed. So now that we've sketched our problem a little bit here and we've recorded the information from the word problem over here in our sketch, let's go ahead and work through our process. The first step in considering this is to define a basis. In other words, to consider an orientation that we want to think about the problem so that we can break our problem down into component parts. And again, we're going to take advantage of our sort of standard method of measuring things where we're going to go ahead and define north as being positive for our y-axis and east or to the right in this case as being positive for our x-axis. So there's our x-axis and our y-axis. So when we label a component with x, we mean positive to the east, negative to the west and when we label it with a y positive to the north and negative to the south. So now that we've defined our basis, our second step is to go ahead and determine components for each vector that we have already described in its magnitude and direction form. So I think our first vector that we want to look at for that is our airspeed vector. So I'm going to go ahead and sketch a triangle associated with their airspeed. Notice it's pretty handy that we use that description of 30 degrees south of west because now we have an acute angle that we can use in our right angle. And here is our value, the airspeed x component and here is our airspeed y component. So now we revisit our trigonometry remembering our letter mnemonic saucatoa and we start to identify relationships between vax and vay and the hypotenuse which has a value of 8 meters per second and the angle that we know. So now if I look at the relationship, I recognize the sign of the angle. In this case the sign is my y component over my hypotenuse. So I can recognize that the sign of 30 degrees is equal to vay divided by 8 meters per second. Well if I re-solve that we know that vay is equal to my 8 meters per second times the sign of the 30 degrees. And you should recognize that the sign of 30 degrees is actually a number that you should know. Alright that's one half so we end up with a value here of 4 meters per second. Notice however there's an important piece that we haven't looked at carefully yet. Look at the direction that I'm pointing my vay, my velocity of my y component, or the y component of my velocity, which direction it's pointing. It's pointing down or south, down on our page, south in our description here and we defined positive as being up or north. So in this case we're going to go ahead and give it a negative value. So it's pointing in that particular direction. Notice we didn't need to think about that positive or negative sign when we were building our triangle but now that we've established what the magnitude of that is we'll go ahead and give it a direction in terms of a sign. Let's do a similar thing for vax recognizing from this equation that we're going to have the cosine of our angle and we're going to end up with 8 meters per second times the cosine of 30 degrees. The cosine of 30 degrees, 0.866, and we multiply that out we're going to get a value of 6.93 meters per second. Can you see the sign of this particular value? Let's take a look. It looks like our x component is going to the left or to the west and we defined east or to the right as being positive so we'll also put a negative value there. And it will be important to remember that negative value when we add our components later. So now we've established our components for our airspeed let's also follow the process for establishing the components for our windspeed. If we look at our windspeed here we notice we have a reference axis here of north and we can bring the windspeed down and it's 150 degrees. Now in this case we would like to go ahead and create again for ourselves let's make this a little bit longer so we have some more space we'd like to create a right triangle again so we can use our Socketoa trigonometric relationships. Well I know that this is 150 degrees so I can draw my triangle like this and recognize that this angle here is 150 degrees the entire angle minus 90 degrees. So this is a 60 degree angle. And then this would be our velocity of our windspeed in the y direction and the velocity of our windspeed in the x direction. Notice I put the arrows so they end up going head to tail and end up with the heads in the same position. It's not the only way I could have done this one I could also have done a triangle that looked like this. If we look at that we can see that this angle here is 180 degrees 180 minus 150 is 30 degrees. So the other way I could have built this is to build this triangle recognizing a 30 degree angle and defining the velocities the velocity components like this. So now let's go ahead and see if we can find the components for the windspeed. If I look first here I can recognize again using my trigonometric relationship that I have, let's go ahead and see if we can find VWY the windspeed, the y component of the windspeed. And I can recognize here that here's my angle and I'm going to look across here so I know that the sine of my angle is the opposite VWY over the hypotenuse my 5 meters per second. So I can translate that there multiplying both sides by the 5 meters per second 5 meters per second times the sine of my angle which in this case is 60 degrees. Now I'm going to do the same thing down here on this triangle but notice on this triangle I now have the adjacent angle that the cosine of my angle is equal to VWY over 5 meters per second. Well up here it's the sine but down here it's the cosine but you'll notice that there are different angles in this case it's 30 degrees. The other way I could have recorded this is 5 meters per second times the cosine of 30 degrees. And remember these complementary angles here are going to have the sine of one angle is going to be equal to the cosine of the complement so they're actually the same value it doesn't matter how you set it up as long as you're careful with which one is the sine and which one is the cosine you should end up with similar values. And in this particular case the similar values I'm looking for let's see here 5 meters per second is going to be times 0.866 so there's the value we're looking for is going to be 4.33 4.33 meters per second. Let's also consider which direction this 4.33 is moving in notice our Y component is down or south so that's going to have a negative value. Now let's go ahead and go with the other piece VWX and now I can choose either one to look at I'm going to go ahead and use this top one recognizing it's the cosine 5 meters per second times the cosine of 60 degrees or down here the sine of the 30 degrees and when we calculate that's a value we should know of 1.5 we get 2.5 meters per second again let's check which direction we're going in in this case we're going to the right or to the east and notice that's positive and positive value here and that will be important for our next step so now our third step now that we've established the components of the vectors that we know is to take those components and add them or should I say combine them because addition and or subtraction in this particular case we recognize our relationship ground speed equals air speed plus wind speed we're going to break it down to the two component parts we're going to combine to find my velocity my ground speed velocity we'll find the x component and the y component by adding the appropriate components from here so let's see here vgx is going to be equal to vax here's my x component here negative 6.93 meters per second and I'm going to add to it my x component of plus 2.5 meters per second and when I add those two things together I get negative 4.43 meters per second similarly with vgy there's our y component of the air speed negative 4.0 meters per second and then our component our y component of the wind speed negative 4.33 meters per second and when we combine those two things together we end up with negative 8.33 meters per second it can often be helpful in a case like this to go and look at your graphical addition where you drew it before and see if that makes sense if we look at our components here let's consider first of all our x components in our case here our x component of our air speed had a relatively large value to the left or to the west and then our x component of our wind speed had a positive value in the other direction so you can see how we went negative and then positive and our resulting vector ended up having a negative value and that's what we see if we look for the y component you'll see we had a negative component for the air speed and we had an additional negative component for the wind speed and those combined to give us a large negative component for the ground speed so it seems like our directions are lining up appropriately in this particular case alright now we've done step 3 which is to add our components our next step is to translate components we have all the information we need to define the ground speed vector however we would like to be able to express this ground speed vector in terms of magnitude and direction so we're going to go ahead and represent our ground speed velocity now in terms of our two components here's our first component of negative 8.33 meters per second notice the negative is only important in the case of me pointing the direction here but now I'm going to turn it into a triangle so really it's only the magnitude that we care about and similarly I'm going to go ahead and get my ground speed of negative 4.33 meters per second it's kind of important which direction we're going in here but that's the direction is defined by that negative and now we care about that particular length so I'm trying to find the magnitude and direction of the vector vg so what do we do to do that we go back to our triangle trigonometry once again pick an angle I'm going to go ahead and choose this angle to do my calculation of the angle we'll define that again as angle theta alright and we recognize here that the tangent of our angle the tangent is typically going to be the one that we use when we have the components and we want to work backwards the tangent of the angle is going to be equal to the opposite side over the adjacent side in this case our opposite side is 4.33 meters per second and our adjacent side is 8.33 meters per second noting that the meters per second cancels out we get a ratio of .532 when we take the inverse tangent making sure we're set for degrees we find an angle value of .28 degrees now it's a little too early to call that the actual bearing of my drone because notice it's relative to this particular triangle we could say that it's .28 degrees west of south that's one of our possible descriptions or we need to figure out is there another way that we can sort of describe it let's think about this in terms of a northern bearing and we recognize that we have something that is 28 degrees right here there are actually two descriptions that we can use two bearing descriptions we can either go in our standard clockwise direction which is going to be 180 degrees plus the additional 28 degrees to keep going or a total of 208 degrees or we can go the other direction in that case if we're going in the counterclockwise direction that's going to be a negative value that's going to be the equivalent of 180 degrees if we went all the way 180 degrees but then came back 20 180 degrees minus the 28 degrees gives us a value of 152 degrees but because we went in the counterclockwise direction it's a negative value either of these two values are legitimate values for describing the actual bearing of our drone either of those two values is legitimate as they define the same angle okay so we've established the bearing of the drone the last piece that we need to actually establish here is the magnitude or the speed the ground speed of the drone itself so to complete the calculation of the ground speed we'll recognize again that we have a right triangle and use the Pythagorean theorem a squared plus b squared equals c squared where my hypotenuse is the unknown ground speed that we're trying to find and we'll square both of the sides here notice again the signs are not important especially in this case because the squared will remove the negative times the negative is the positive so I take 8.33 meters per second quantity squared and I add it to 4.33 meters per second quantity squared when I do that I end up with the value vg squared is equal to roughly 89.0 meters squared per second squared and then when I take the square root of both sides I get a value of 9.4 meters per second and in this case I could say 9.43 but if I look carefully although I kept a number of significant digits throughout the problem when I look at the end I realize that the best I could say this was 8.0 and perhaps 5.0 since it's a word problem I can assume maybe one more significant digit but that's probably the extent of what I consider comfortable to include so there's my ground speed velocity of 9.4 meters per second now there's one more piece that the problem asked us for it asked us for the drift angle in other words the difference between the angle that we're actually heading and the angle that we're actually bearing the track bearing basically the track being the path of the drone itself so if we want to consider that we're going to need to look at each of these pieces here so I'm going to look at the air speed I recognize that the air speed is 30 degrees south of west and if I look carefully here I know that my final track of my drone was 28 degrees from south west of south and so the angle I'm looking for here is this angle that's in between the difference between those two things there's a couple ways to calculate this but if you recognize that in this quadrant this entire thing needs to add up to a total of 90 degrees then we subtract the sum of these two things and we establish that we have a 32 degree difference so in this particular problem you'll notice that the wind is actually if you look carefully directly perpendicular to my air speed I'm traveling in what's called a crosswind this is a crosswind actually almost all winds have what's called a crosswind component you can break the wind down into the component that's parallel and the component is perpendicular well in this case the entire wind is a crosswind and that crosswind changes the direction we're going and establishes what we call a drift angle in this case of 32 degrees as it turns out there are some ways to solve this problem to simplify the problem by thinking about it a little differently we'll see that in the next video