 Welcome to Caltrans LSIT LS exam preparation course. One aid in your preparation for California licensure examinations. A word of caution. Don't use this course as your only preparation. Devise and follow a regular schedule of study which begins months before the test. Work many problems in each area, not just those in this course's workbook, but problems from other sources as well. This course is funded by Caltrans, but you and I owe a profound thanks to others, the courses instructors from the academic community, the private sector, other public agencies, and from Caltrans as well. We wish you well in your study toward becoming a member of California's professional land surveying community. Hi, welcome to the world of astronomy. Well, maybe not astronomy. Maybe better said, welcome to the world of astronomical asthma determination. Astronomy is a subject like that big, and we're only going to talk about maybe that much of it. Only that part that applies to the land surveyors examination and what you need to know to answer a problem in astronomy. The purpose of this video is to help you answer an astronomy problem in the land surveyors examination. And all that we're going to talk about is going to be structured toward that end. One thing you want to remember that in the examination, it is not the examination that's your enemy, it's the clock. The examination is an open book examination. And if you had all the time in the world and you got an open book, let's, hey, you would pass the test, right? And the test, once you pass it, you're going to have your license to survey. So the test is your friend. It's that clock that's your enemy. But you are limited. So again, we're going to structure how we deal with the land surveyors examination and how we work on astronomy problem doing that. Now, my name's Jack Sands. By the way, Jack's the nickname for John. And I'm going to be your tour guide for this video. Now, I work in the San Diego County Surveyor's Office in the Department of Public Works. Now, in this video, we're going to cover the PZS Triangle. We're going to talk about how the sun, the earth, and the stars interact. We're going to talk about latitude, longitude, Greenwich hour angle, declination, local hour angle. These are all the things that you need to work the PZS Triangle. We're going to be doing this using graphics and also a three-dimensional model. Now, the three-dimensional model is going to help you a little better understand this interaction of the earth and the sun, where sometimes graphics just doesn't quite get it done. And then we're going to take a recent land surveyors examination and we're going to go through it step by step. So you'll see just how an astronomy problem can be worked. As we move through the video, we'll be naming and describing various terms that are associated with astronomy. You're going to find these terms in your workbook in Unit 4, and that unit is titled Asmuth Determination by Celestial Observation. So let's take a look at a PZS Triangle. The PZS Triangle is a spherical triangle, and in a spherical triangle the parts, the three sides, the three angles are all measured in degrees, minutes, and seconds. Now, the PZS takes its name from the letters assigned to the three corners. That's P for pole, Z stands for zenith, and the S stands for sun or star. The PZS Triangle can have many shapes. It's dependent on where Z is and where S is. Now, P is the north pole. It stays where it is. If it doesn't stay where it is, then we've got a problem a whole lot bigger than this Land Surveyor Examination, and that'd be a whole other ballgame. But it stays where it is. It's the pole and Z and S move, and that's what changes this shape. The PZS Triangle that you see is a morning observation with the Z, the observer, somewhere in California. You know, it's going to be a California Land Surveyor Examination, so that's why California. And the S, the sun, is somewhere in the morning because it's coming up. It hasn't got to our meridian yet. The PZS Triangle shown here is on the Earth's surface. The PZS Triangle used to compute the azimuth is on the celestial sphere. The celestial sphere is a sphere of infinite radius with its center at the Earth's center. It's north, it's south pole at the extension of the Earth's north-south pole, and it's equator, an extension of the Earth's equator. The sun and the stars are all presumed to be on the surface of this globe regardless of their distance. We will show the PZS Triangle on the Earth for the sake of understanding a little better its relationship to the Earth. Let's take the display, showing the PZS Triangle and build it one step at a time. All the lines of latitude and longitude need a starting place and origin. For latitudes we use the equator, and for longitudes we use the line of longitude that runs through Greenwich, England. Latitudes are measured north or south from the equator, and the longitudes are measured either east or west from Greenwich. The reason we use Greenwich, the Greenwich meridian as the zero meridian, is because, hey, the English were here first. They wrote the book. You think they're going to put an Ireland? No, sir. Now let's add a line of latitude and a line of longitude at z. Remember that the location of the observer is at z. And next we're going to add the line of latitude and longitude at s. That's where the line from the sun to the center of the Earth touches the Earth's surface. Now, when we're talking latitude and longitude for the sun, we don't call it latitude and longitude. We call it declination, and we call it Greenwich hour angle. But it's still a measure, declination of latitude and longitude. Now we can put in the legs of our PZS triangle. P down to z, P down to s, and z over to s. And now with a little magic, we can make our triangle shine. Ain't TV wonderful? Now let's add the symbol for west longitude and the symbol for latitude. The angle of longitude measured west from the Greenwich meridian to the longitude of the observer is called west longitude. The symbol for longitude is the Greek letter lambda. Just think of a giraffe going west. The angle measured north or south from the equator to the latitude of the observer is called phi spelled P-H-I. That's a little circle with a line through it. Next, we're going to put in G-H-A, declination and L-H-A. The angle of longitude measured west from the Greenwich meridian to the longitude of the sun or the star is called Greenwich hour angle. The angle measured north or south from the equator is called declination. The angle of longitude measured west from the longitude of the observer, that's all the way around west on our picture, to the sun is called local hour angle. And the key here is the first word when we're talking from Greenwich around to the longitude of the sun in this case, that's Greenwich hour angle. When we're going from the zenith or the local, the place of the observation we go around west, local's the key, that's local hour angle. Now all that's left to do is identify the parts of the triangle. Let's start with the angles. The angle P, I should say, the angle at P is called little t, that small letter t. And it's related to LHA. Whenever LHA is more or less than 180 degrees is how t relates. You can see that in our picture LHA is very large because it goes all the way around and what's left is t. So you can see the two together make 360 degrees. Now this is a morning observation. If this were an afternoon observation then you could see that the local hour angle would be rather small and that t and it would be the same thing. So now you can see how the key to t is LHA. When LHA is 180 degrees or less t is the same thing. When LHA is greater than 180 degrees then to get t we subtract LHA from 360. But the nice thing about it is you don't have to worry about that at all because we just don't use t anymore. Just know that it's there and that's why we identify it at that part of the triangle. All of our equations now just use LHA. The azimuth angle, that's the azimuth, that's the angle we're after. So it can be observed but in this situation because we're talking about the Land's Surveyors examination allow that that's the angle that we're after computing. And then the other one, the one at where the sun is that's called the parallactic angle. Why the parallactic angle? Why do we call it that? Hey, you got to call it something. As stated earlier the sides of the triangle are measured in degrees. The side from P to Z is 90 minus the latitude co-latitude co-stands for complimentary angle. The side from P to S is 90 minus the declamation, co-declination again complimentary angle. The side from Z to S is related to the vertical angle. It can be determined by measuring the vertical angle to the sun's star or it can be computed from the time. The letter H is used to identify this side, sometimes referred to as the altitude. The side itself is 90 minus H, co-H, again co, or excuse me, complimentary angle. And there you have it. The PZS triangle with all of its supporting information. Now that we have developed the PZS triangle, let's find out how it gets started. Let's take a three-dimensional model that we have which is going to help us a little better understand this and just show how we can get that PZS triangle. Here is our three-dimensional model. This is the earth, this is the sun. Now contrary to what you've been told, the sun is not this big fiery ball in the sky that we go around. We do go around it, but it's just what you see. I mean when you look up there and you see the sun, it's a little round ball. That's what you've got. A little round ball sun. I lie. But for the sake of the three-dimensional model, this is what we have to have. Now another thing that we're going to do here is we're going to I would like to say pretend we're God, but the Caltrans people say I can't do that on the video. So we're going to be astronauts. And an astronaut, we're going to be up in the above our solar system as an astronaut looking down with the idea of what is our perspective. Looking down as our astronauts, as we as astronauts are doing everything is counterclockwise. The sun may even turn counterclockwise. We turn counterclockwise. We revolve around the sun counterclockwise. This is one year that you're looking at. One complete revolution of the earth going around the sun. So think of it. Everything clockwise. We turn clockwise. We go around the sun clockwise. The sun may even turn clockwise. Now what you're looking at right now is the sun shining on the top part of the earth, the northern hemisphere. That would be our summer here in California. Right now because an engine also see that the pitch of the earth has maintained itself. That's a pitch of about 23 and a half degrees and that stays that way. Even though we're turning, you can see that that still stays at that pitch. As we move around and get to this position, you can see that the sun will be shining directly at the equator. And because we're going from summer toward winter, that would be what we call the octominal equinox or autumn equinox. As we continue on for another three months and we get to where the sun is shining on the lower hemisphere, that would be our winter. That's what we'd call our winter soltice. And then again on around to where as the sun shines to the equator coming from winter to summer it would be the vernal equinox and then again one year later the sun shining on the northern hemisphere back to the summer soltice. Now, we need to build a PZS triangle and where does it come from and how? Well first of all we're going to have a P. P stands for North Pole. Z stands for us. Where are we? Zenith. Where is the observer? Because again we're talking about the California land surveyors and it's going to be in California so let's just say that right here is California and that's our Z. Now an S in the graphics display PZS triangle we saw, how did that get started? It started because we were doing our thing like this in the summertime and let's first of all put in Greenwich. Remember we had to have a zero longitude? Well here's G for Greenwich. Let's say that Greenwich is right here. Now when the sun's shining on Greenwich what is it in Greenwich? Right, noontime. And as we continue on eventually the sun gets to our Zenith it would be what? Right, it would be our noontime. Let's go back. Let's just say that it's noontime at Greenwich and we're moving along and all of a sudden up we stop the earth. Now astronauts cannot stop the earth. Only God can stop the earth. So now regardless of what Caltran says we must pretend we're God. As God and stopping the earth when the sun was shining directly on the earth toward the center boom there was S. And because we stopped it we created the three points of our PZS triangle P, North Pole, Z where we made our observation and S where we stopped the earth. Now naturally this kept going and we stopped it. That's where they were and now you can see the PZS triangle. Let's just take a little piece of thread here and you can see the triangle very similar to the one in the graphics. PZS triangle. And that's how we got it. Now this can have many shapes. The only reason that the shape is this way is because of where Z and where S are. Now because it's California land surveyors examination and that's what we're working on we're going to keep Z where it is and we don't want to change P. When that changes again we got a problem that's a whole lot bigger than this examination. So P stays the same Z stays the same. But let's say that this was an afternoon observation. Let's say that here we were say at Greenwich, moving along the sun eventually passed our zenith and continued on until we stopped the earth again the line from the sun right to the earth boom. Again we have a very similar triangle except that now P and Z have reversed themselves. Here's our triangle again very similar looking again except Z and P are on the, I should say S and Z are on different sides. Let's say that it's a winner observation. Let's come around over here to where we would be in the winner. And again remember looking down from the top everything is counterclockwise. Let's say that the sun starts at Greenwich and it might be a morning observation again. Well we stopped the earth and aligned directly from the sun to the earth would be where S is. But now you can see we're going to have a very long elongated PZS triangle. Because now the sun is below the equator. That would be a minus declination. Remember when we talk about the latitude or the star it's declination, it's not latitude. So this would be a minus declination. And one thing to be thinking about in that examination is when we're talking about declinations they can be plus or minus. In the Little Ephemeral book that you'll be probably having a copy of in the Land Surveyor's Examination the minus will be shown the absence of minus is a plus. Don't do that in the examination. In the examination when it's plus you write down a plus. When it's minus write down a minus. And that way if it was minus and you forgot to write it in and you just figure it was a plus because of the absence of it you're going to get the wrong answer and there goes the question. Anyway you can see how the PZS triangle can have different shapes. Let's go back to the summer and so that we work with a triangle similar to the one in the graphics where we had Greenwich and we finally had a morning observation that the sun was directly shining to the center of the earth and our PZS triangle looked like this. Sort of. One other thing I want you to remember or know about is there's a thing called sidereal hour angle or sidereal time and what is sidereal time versus solar time or civil time, sun time. And that is this. Right here we are at our zenith looking directly at the sun. The stars work a little different in this regard in that the stars are all out here on the celestial sphere and if you'll see we have a star. Our star and our sun are on direct line with each other and remember everything's counter clockwise. Now let's take one 24 hour civil solar day and we'll just go around I'm going to exaggerate the movement here so that you can see what we're talking about. Here we are back to I'm going to stop and the zenith is looking right back at the star. We've gone 360 degrees through the star but you can see we have yet to get back to the sun. A little bit more actually about 4 minutes and we're back to the sun and there's 24 hours so you can see that measured in solar time the time on your clock, civil time measured in that time a sidereal day is less. It's 23 hours and approximately 6 minutes the solar day 24 hours so you can see sidereal time, sidereal hour is a little bit less. One other thing to be aware of is we can do and create a PZS triangle without ever making an observation with an instrument because all you have to have is a clock as soon as you say stop the world there's your PZS triangle, it's done but we don't, we have an observation we have an observer, we have a zenith and at that exact time an instrument sighting the sun stopped its motion. Now things went on but we had stopped it right here and at that time when we were observing the sun let's allow that we took our instrument and looked to an observation like that. Now we have the PZS triangle from which we will later see how we can compute this azimuth angle and having measured an angle to some fixed point on the ground we can now add the azimuth angle to that angle and we can have an azimuth to a fixed point that means we can pick our instrument up and go away and come back at some other time and use the azimuth established on that line. So there you have it the PZS triangle where it comes from and how we do it. Now let's take what we've learned here and go back and apply it to a land surveyors examination problem. We'll use the problem given in the 1989 California land surveyors examination. It was a 20 point problem. Now 10 points of this were for determining the latitude and longitude at station Ruck and 5 points were for computing the azimuth from a line from a solar observation and then 5 points were for computing the angle of the closure. We're going to deal with only that part that had to do with determining the azimuth from observing the sun. The problem asked that we use either the hour angle method or the altitude method and we're going to use the hour angle method. The equation for the hour angle method and the equation for the altitude method are displayed on the screen. They are also given in your workbook. An important consideration at this point is which will take the longer the hour angle method or the altitude method. Remember, time is the enemy. In today's world of solar observation the hour angle method is preferred. It is more accurate. It is easier to observe because only the vertical M is observed. But this ain't the real world. This is the test. And in the test its pass or fail and again times your enemy. The altitude method in this problem will take less time because the hour angle method requires we compute LHA and declination. The altitude method needs only the declination and that's one less thing to do. The other component H and latitude, the other components H and latitude are given. If we had to correct H for refraction and parallax then it might be a trade off. You're going to have to decide which is going to take the least amount of time. In this problem the altitude method would be the one to use because H is given corrected for refraction and parallax. Now after all of that why are we using the hour angle method so that you'll get an instruction in how to compute LHA. The equation for the hour angle method requires that we know LHA of the sun the declination of the sun and the latitude of the observer. So let's develop these items one at a time and then put them together into the equation. To compute the LHA we need to know the time of the observation. Not just any time we need coordinated universal time UTC sometimes called Greenwich time. UTC is then corrected to get the base time on the actual rotation of the earth. This correction is called DUT and the DUT correction is given in the problem. This corrected time we then call UT1. The reason we need UT1 is because the ephemeris data is for UTC. So here we go. To get LHA we first must convert the pacific daylight standard time excuse me, saving time pacific daylight saving time to UT1. The display shows a way to deal with the information given to arrive at UT1. The display shows a way to deal with the information most LS problems of this type can be answered in this style of format. We list all the necessary information and then deal with it algebraically as a group. It can be done each one by itself but it's probably a little better that we deal with it as a group. First the time 523.35 Now the reason we know this is PM because we're going to add a PM correction is because of the picture that you're going to have to look at in your workbook of the problem. They didn't tell us this was PM now whether that's deliberate or not we don't know. We do know that it's PM by just observing that the sun was observed in the afternoon. The PM correction 12 hours now the correction for time zone 7 hours. Why 7 hours? Trust me when it's Pacific Daylight saving time you add 7 hours to get to UTC when it's Pacific Standard time you add 8 hours. Fortunes have been won and lost betting on whether you add or subtract spring back fall ahead I don't even remember the little jargons just allow that with this problem it's Pacific Daylight saving time and we're going to add 7 hours. Next we put down the watch correction now when you've got a fast watch you subtract the correction this we subtract because the watch is fast why do we subtract the fast watch for the same reason that we add a slow watch because sand says so. You can figure it out for yourself later. Right now let's just trust Jack. Our last correction is the DUT correction this is given in the problem it's a minus 5 seconds. This is the correction needed to change that UTC to UT1. Now we will take the time of the observation 523 and add all these corrections algebraically the result is UT1. As you can see we have more than 24 hours at UT1 24 23 34 2 and you can almost bet that any astronomy problem that you get of this type in the examination they're going to do this it's going to go over 24 hours because they want to know that you know what to do with it that you have to go into the next day when you're looking into the ephemeris information. Now we can compute GHA the ephemeris will give us the GHA for zero hours on the day of observation we just we have just previously determined that the UT1 of the observation we have just previously determined that the UT1 of observation the GHA of the observation is the GHA at zero hours plus the amount of GHA for the elapsed time from zero hours to UT1. To compute GHA for the elapsed time we need to first compute the total amount of GHA for the day of observation we do this by taking the difference between the GHA for zero hours today that's the 180 50 for and the GHA for zero hours tomorrow the 180 52033 and then we add 360 and the reason we have to add 360 is because there was a change in the elapsed time but it took a whole 24 hour period for it to happen and that's a 360 degree addition the result can be more or less depending on whether the GHA at the beginning of the day and the end of the day if you will the next day was more or less we then multiply this total GHA for the day by the UT1 divided by 24 to get the GHA for the elapsed time if this all sounds very complicated well welcome to the club as you do it it will start to make sense each day must be considered separately because the total amount of GHA changes every day here we might want to talk a little bit about equation of time fortunately you don't have to worry about it anymore because all the ephemeris data is in GHA but the equation of time is that because the sun does not travel excuse me the earth as it travels around the sun does not travel at a constant speed it has a speeding up and a slowing down and this causes the real sun what they call the apparent sun let's just talk about noon time at noon on our clock it may say noon but when you look up the sun may not be right on the meridian it may be before it may be after because of the speeding up or slowing down and that's what the equation of time is all about it can be very confusing very complicated and again fortunately you don't have to worry about it because your ephemeris tables will take care of this for you the GHA corrects for that and with our problem all together the GHA for the elapsed time is 523 341 and then we add this to the zero hours and this gives us the GHA at the time of observation 186 44 32 5 now we can compute the LHA the formula LHA equals GHA minus west longitude you can find in your little ephemeris leaps tables which is a real good little study text that also has the ephemeris in it 119 46 54 5 and now we've got the first part of our problem subtracting that we get LHA 66 57 38 0 now let's do the declination computing the declination you're going to find is a lot similar to what we did when we computed GHA but generally the same procedure we take the difference between the declination for zero hours today 16 32 17 5 and the declination for zero hours tomorrow 16 48 59 7 multiply this by ut 1 divided by 24 add the answer to the declination for zero hours today sound familiar? the declination for the time of observation 16 32 39 as you can see the procedure is very similar to that for getting GHA one thing I might bring to your attention here is that there is a correction to the declination that this problem didn't address and it can be as much as 3 or 3.4 seconds the problem didn't ask for it so you're not going to have to worry about it if it did give you a declination correction they'd want you to attend to it you'll find there's an equation for this in your little again the little leach's ephemeris tables because our ut 1 was so close to zero hours the correction would be zero this problem did not supply this correction and did not require it but it's a good thing to know about it the formula for computing this again is found in your little leach's ephemeris handbook now the last of our three items that we need for this equation is the latitude and that's a given that's 36, 48, 57 if you'd worked the first part of this problem that's where that would have come from but we're not dealing with that now let's take these three factors that we've been able to put together and let's put them into the equation and solve the equation shown on the screen is also given in your workbook in unit 4 this is the hour angle equation so go ahead work the equation you can stop the video while you work the equation well how did you do? if you had to work the equation about 6 times to get the answer then you're par for the course the good thing is that it isn't for the gold you're just practicing for the real thing you probably made dumb mistakes like you know 2 plus 2 is 5 you didn't change the say the degrees minutes and seconds to decimals of a degree most mistakes are because you're in a hurry and you will be in the exam remember time is your enemy there but the more you practice and the more in control you will be you will be I know you're always hearing this you know practice practice practice but it's true be prepared and you're going to pass that test I can just about guarantee it the answer to our equation is minus 89 47 48 2 when you have the answer to the equation you need to do a normalization so that the result will reflect the azimuth reckon from north goes clockwise I guess for you that would be this way and is positive and less than 360 use the box in the display to normalize the answer this box is also given in your workbook the LHA is between 0 and 180 in our problem and the answer to the equation was minus that means that we have to add 360 and then that makes our answer the azimuth to the center of the sun 270 12 11.8 however we did not cite the center of the sun remember we cited the left edge of the sun to compute the azimuth to the edge of the sun we need to compute what we call the dh correction sometimes dz correction we're going to say dh the formula for dh is the dh equals the semi-diameter divided by the cosine of h if the left edge is cited the correction is subtracted and if the right edge is cited then we would add it why? you guessed it because jack says so if you draw a picture you'll see why it is so why can't we just use the semi-diameter without dividing by the cosine of h good question it's because as the sun moves from the location to the local meridian the spherical triangle changes its shape this changes the size of the dh correction angle it's not important that you understand this it is important that you know about it in our problem the semi-diameter which is 1552.7 and the h which is 2805.49 are given the semi-diameter changes very little to only two-tenths of a second and because our observation was so close to zero hours we will use 1552.7 h is the vertical angle corrected for parallel action refraction it can also be computed if lha and latitude and declination are known but that takes time and time again we don't have so we'll just use h given the computed dh is 18 minutes and because the left edge was cited we're going to subtract the azimuth to the left edge is 269.5411.8 now let's take a look at the display and start at the top and compute each factor first we take the computed azimuth the 270.1111.8 then subtract the dh correction the 18 minutes this gives us the azimuth to the left edge of the sun 269.5411.8 from this we subtract the angle measured at station ruck from the left edge of the sun two station rots 187.3056 this angle is given in the problem all this results in the astronomical azimuth from ruck two rots 82.2316 and that's our answer we're done and you thought this was going to be hard the rest of this particular problem is that we compute the angle of closure at ruck but this is not an astronomy question that's getting into the conversion of astronomical azimuth to geodetic azimuth and then converting to geodetic azimuth excuse me, to grid azimuth this will be covered in the video dealing with the california coordinate system in what it's worth department that answer is 15 seconds we all know that a picture is worth a thousand words so let's take a picture and see, just go through this again now north in the picture is true north or astronomical north and the reason it's astronomical north is because we just solved the pzs triangle which is an astronomical triangle the p in the triangle is the north pole you can see where we had the azimuth to the center of the sun that was computed from the equation to 7211.8 our little correction of 18 gave us that azimuth to the edge the leading edge or the left edge of the sun 269, 54, 11.8 we then subtract the angle measured at station ruck from the left edge of the sun to station rots again, this is the angle given in the problem 187, 30, 56 and that gives us the azimuth from ruck to rots 82, 23, 16 now if this were a Polaris problem we'd deal with it just the same way you could use the same equation everything would be the same except you would not have to compute the dh correction for obvious reasons you wouldn't have to be citing the left edge of the star the equation shown on the screen can be used for computing the azimuth when you cite Polaris this equation is a little simpler this equation is also given in your workbook this equation should not be used for the sun or other stars it works for Polaris because Polaris is so close to the north pole less than a degree in the equation h is needed if h is not given you would be better off using the hour angle equation because the time it would take to compute h would be more time than just working the hour angle equation the computed azimuth using the Polaris equation is small either east or west of north it is east of north when the lha is between 180 and 360 it is west of north when lha is between 0 and 180 when it is west of north and the azimuth is reckoned from north in a clockwise direction well that would be like this for you it must be subtracted from 360 what we've covered so far is the kind of problem more than not in the Lancer Veiers examination on astronomy there are times when you're going to get what I'll call a word problem you should burn a candle that you get one of these offer your first born to the goddess of examination questions I call these gimmies these kinds of problems do not take time they can be answered without time consuming computations remember that time is your enemy let's take a look at a gimme problem go to unit 4 in your workbook and find problem B3 from the 1988 California Lancer Veiers examination this problem is a 10 point problem that means in this examination you were given 48 minutes to work this problem measure that against the time it would take to work the solar problem we just did that was only 5 points this problem has 17 questions 15 true or false to multiple choice if you did nothing more than guess the probabilities are that you could get half of them right and that would be 5 points and maybe 5 minutes leaving you as much as 43 minutes to work on another problem now you can see why I call these friendly problems you're in an open book examination if you just search for the answers you should be able to get maybe 7 or 8 points in 20 minutes maybe you should offer not only your first born but your spouse also that you get one of these let's take the questions one at a time the problem states that if the question is false you should give a brief explanation and the key word there is brief don't spend a lot of time so let's start right out number one astronomic azimuth is based on true north true nothing else to do number two grid azimuth is based on true north well we just said no to that so it's false and we can say c number one for a statement geodetic azimuth is astronomic azimuth minus mapping angle plus second term I don't know and I'm the expert you go to a book and you find a formula and you'll find this formula in a publication 253 remember you've got an open book exam and you've brought every book you own to this thing so just look it up and you're going to find that the correct is grid azimuth equals geodetic azimuth minus mapping angle plus second term number four to determine true north from observations on Polaris the latitude of the observer must be known no false look at the Polaris equation with no latitude in there and again because you're going to be making just a brief statement just write it down see Polaris equation and then maybe show the Polaris equation with no latitude in it number five gha equals lha minus west longitude well we've just gone through this but again you don't try to remember equations go to your little ephemera's book and there it is lha equals gha minus west longitude you could say false and nothing more than just write the equation down on to number six the best time to observe the sun to determine azimuth using the altitude method is one half hour after sunrise or one half hour before sunset no false because when you're looking at the altitude method and you're measuring vertical angle you're going to find refraction is just too much to deal with but again in the test you're just going to make a brief statement you're going to say false too much refraction period number seven exact time is more important when using the hour angle method this is true period nothing else to say number eight the hour angle method requires both horizontal and vertical observations to be determined excuse me to determine the azimuth no we only have to look at just the vertical observation so you're going to say false horizontal only number nine it is not necessary to know the latitude of the observer when using the hour angle method false see the hour angle equation just write it down see we're flowing right through we're halfway through this and I'll bet we haven't spent even talking about it more than three or four minutes number ten I think we just did number ten no let's do it anyway even if we didn't the best time to observe the sun for determining azimuth using the hour angle method is just after sunrise or just before sunset that's true because you're going to have the sun lower in the sky and the lower the better with the hour angle method again I'm just kind of telling you why but all you got to do is write true and get on to number eleven standard time must be converted to local time to determine Greenwich time false again don't try to remember equations just go into the book and find it and we're going to say that the Greenwich time equals standard time plus in this case for California be eight hours depending on what time zone you're in and here we're in the Pacific standard time zone number twelve local time is increased in California by eight hours to determine Greenwich time no it's the standard time not the local time and again you could just write out the little equation for that thirteen true solar time is local civil time minus the equation of time now again there's that pesty old equation of time which again we don't have to worry about because our ephemeris tables take care of it but here they're asking us about it so what do we do again you don't remember formulas go right to that little Leitz ephemeris handbook and you'll find about maybe the third column over it'll talk right at the top it'll tell you the little equation for the equation of time and just copy it down fourteen a level line at sea level is parallel with a level line at eight thousand feet now what the heck that got to do with astronomy I don't know but it's here you know you gotta answer it and I'm just gonna say false and I'll just tell you that because you're out there further and centrifugal force is working harder on it you're gonna have more of an elliptical path than you would at the sea level so it's not gonna be parallel and that's about as much as you'd want to say about that in fact it doesn't even belong in there but again what are you gonna do about it fifteen one sidereal day is longer than one solar day measured in civil time remember our three dimensional model when we were looking at the star and we could come all the way around and we would face the star a little bit before the end of the solar day so we know that the solar day is shorter excuse me that the sidereal day is shorter the solar day longer so number fifteen would be false sidereal day is twenty-three hours fifty-six minutes and then we got two more to go and I'll bet we still haven't used seven or eight minutes and again right here let's say you take a hat pin and you just pick A, B, C or D maybe your chances aren't as great with true or false but you could do it but let's go ahead and go through them anyway at what time will the effect of a small air in the determination of the observer's latitude be minimized when making azimuth observations on Polaris? I will peek at the answer and it says C C the Polaris is at culmination okay just allow that at culmination you're right online with the star so it wouldn't matter if you're way down here or way up here it ain't gonna change the azimuth any so naturally at culmination would be the answer moving right on to the last one number seventeen at what time will the effect of a small air in the determination of the time of the observer's excuse me of the observation be minimized when making azimuth observations on Polaris? well again this would be because when the star which is making its circle and I may be showing it wrong it's you know I'm looking at you you're looking at me just allow it's going around in a circle when it's at the outer edges it appears to be only moving vertically and time is not very critical so elongation would be the answer let's see if I got that right number seventeen says B haha look at that Polaris at elongation and we're done we've used up I doubt it we've used up ten minutes and again you know there's 43 minutes to work another problem well that covers the kind of problems that you're going to get in this LS examination there are going to be either be a problem or you're going to have to work out some numbers that you hope you don't get or a word problem like you say you're going to be praying to get and let me add just a little bit more on how to maybe approach the examination whether it be astronomy or other questions again remembering the time is your enemy take the examination when it's given to you open it up look at the first question and say hey do I know this is the time fair let's say you look at the first one you know what they're talking about go right on to the second one it doesn't matter about time the second problem you look at and this is hey I know how to answer this but it probably take me a day and a half to do it go right on to the next one put little check marks by these problems that you're not going to work number three it's a problem you know and you say hey the time is about fair I think I can do this work problem number three problem number four same thing I don't know what it is problem number five hey again time's fair I know how to work it I work problem number five you can see what I'm getting at here I'm working the problems that I'm going to get the best points from I can go through the whole examination like this and I may end up only working let's say it's a ten point excuse me a ten problem exam and I've only worked six of them I go right back to the beginning again and number one don't know how to work it don't bother with it on to number two okay again I know how to work it but it takes too much time I don't work this one because I remember one in the back was the same thing except I could work it the time wasn't quite as bad I go back and I work that problem let's say I go through the whole examination like this and I end up only maybe answering eight of the ten problems but I picked the best eight the two that I didn't work are two I shouldn't have been working with anyway spinning my wheels where I could have been getting good points so you see what I'm talking about you want to enhance your chances of getting the best chance of working the problems that are going to give you the best points now there's no magic but the better prepared you are and again I know that's been beat to death but it's a fact the better prepared you are the more selective you are about the problems you work first the more you enhance your chances of getting a passing score so guys, gals, good luck and break a leg