Hello, this part of the course on measuring distances is dedicated to trigonometric leveling. Trigonometric leveling allows, from measuring distances and angles, to determine height differences. The question that must be asked in topographic measurements is: how to define a distance unambiguously? I take here a small example with a station P an aiming point Q, a slant distance S' measured here as well as an angle of height beta (β). We can reduce this distance to the horizontal here S horizontal (HZ) and calculate a height difference. The question that we must ask: at what point should we consider the distance like a spherical distance? And also the question is knowing that the measurement is done at a certain altitude in this case here Hp = 1000 meters and that a distance measured at a certain altitude is not necessarily the same as a measured distance at a lower or higher altitude. So we must proceed with what we call reductions and these are the elements that we will see in this part of the lesson with the reduction due to altitude as well as the reduction in the projection plane in order to have distances that are compatible with the national coordinate system. Flat Earth, round Earth at what point do we take into account the roundness of Earth in the measurements of distances? We take here an example between point P and point Q with the chord which is the straight line between these two points and the arc that follows Earth's terrestrial reference surface. If I have a chord of 10 000 meters my arc will be 10 000,001 meters a difference of about 1 millimeter. So, in this case, for our topographic measurements the arc and chord will be undistinguishable. On the contrary the arrow, which is the distance here in the center, between the arc and the chord - I have here my arc and my chord - in this case, the arrow is equivalent to about 2 meters. To illustrate this principle we must stop for a moment at this figure, fundamental to understand the problems with reduction of distance. Firstly, we have here our station P and the point Q which has been measured. At station P I have a vertical here with an altitude HP. At station Q I have another vertical. These two verticals are not necessarily parallel. From P I can consider the distance perpendicular to this vertical namely the horizontal distance S as well as the arc Sp here, which will connect P to Q'. I can also consider Cp as the chord that connects P to Q'. I have thus three types of distances to be considered in this figure. How about for our topographic work? We recall here our three concepts of distance: namely the arc, chord and the horizontal here at P. So we have Sp, Cp and S. Cp - Sp = -1mm for Sp = 10 km and then S - Sp = 1mm for Sp = 5 km. So in topographic work where we have some kilometers of surface we can neglect this effect. Here we come back to the effect of the roundness of Earth. In this figure, we have first of all the angle at the center of Earth which is equal here to 2 ɣ (Gamma). We can consider the triangle OPQ' and its relation to the radius of Earth R x 2 ɣ = distance S between P and Q'. Gamma ɣ is expressed in radians. Then we will consider the triangle PQ'Q'' in this case we can write that E = ɣ x S. Finally, our value E, by combining these two equations, E = S²/2R (radius of Earth). For a distance of 10 km, if you do the calculations, you will get a value of E = 8 meters. After having considered these spherical models we now get closer to a model called flat Earth that we will use for most topographic works and construction sites. We have in this figure a station P at a certain altitude Hp. We measured here an oblique distance S' with a altitude angle Bêta (β) we can reduce at the horizontal the distance S. We also have to consider the height here of the aiming point, namely Z as well as the height of the instrument namely i. We have thus all geometric informations to calculate the height difference. In the case of flat Earth the trigonometric formulas are relatively simple. We have here the Delta (Δ ) Δ h which is equal to S' times sine of the altitude angle β. If we look at the figure we can calculate the altitude HQ = Hq + Δ h which I already calculated + I (height of instrument) - Z (height of shown signal) I start from P, I have the height of the instrument + Δh - Z which gives us Hq. We will now consider the different steps leading to the reduction of distances to finally have an unambiguous definition. The first step is the reduction due to altitude. We see in this figure that the distance considered at altitude Q or at altitude P is not the same. We must therefore bring it back a distance called S zero to the sea level. To calculate the reduction factor we will simply take the proportional ratio between the spherical distances and the altitudes. I consider here Sp/S0 and I take the relation of the distance to the centre of the Earth so r + altitude Hp and then r (radius of Earth) at altitude H=0. I can write that Sp, so at altitude Hp = r + Hp/r x S0. This is significant I will here take an example with Hp = 500 meters and a distance S equal to 1000 meters Sp - S0 = 8 centimeters, in this case. For trigonometric leveling the following problem involves the question of measuring the altitude angle. We have actually in this figure at station P, measured an angle beta with respect to a horizontal Or, we are interested here in the reference surface, namely the sphere at altitude P, here the reference P Q'. In addition, we have a refraction effect which will impact the measured angle. In topography we must, in fact, consider the phenomena of atmospheric refraction when measuring altitude angles. We have here our angle Tau (τ) which is the angle of refraction which has the effect to curve the aimed beam It is accepted in practice that τ = a factor K that multiplies the angle at the centre of the Earth so the farther away the aiming point is the more important is the effect of refraction. Gamma (ɣ) is obviously expressed in radians and with experience, K is a value that we fix to 0.13. If I bring back my figure measuring the altitude angle I can consider the following different parameters: Firstly I have my measured angle beta (β) β: measurement of altitude angle. I have Gamma (ɣ) here, which is half of the angle at the centre of the Earth. I also have here Tau (τ) which is the angle of refraction and finally, what interests me is the reduced angle beta, namely β bar = = β + ɣ - τ (beta bar = beta + gamma - tau). Beware of the units, as we have measures in general gon and then the values of Gamma and Tau which are expressed in radians. We must therefore convert to the right units to do this addition. To calculate the reduction we now consider this geometric figure. We recall here that chord and arc are coincident which allows us to consider here a triangle with points P Q' and Q as well as the reduced angle beta bar and the slant distance measured here S'. We recall here that Gamma [expressed in radians] so the effect of the Earth's roundness is equal to S/2 times the radius of refraction Tau [in radians] = K times Gamma = 0.13 x Gamma. Considering our triangle we can apply the sine theorem and find the value of Sp which is equal to S' x cosine of (β bar + ɣ) / cos ɣ ΔH = Sp x sin β bar / cos β bar + ɣ We thus have the formulas for calculating the reduction of distances on the horizontal and the height difference. You find here a summary of the formulas for reduction of distances and I encourage you to look at the handout for more details. The last phase of reduction consist of applying a scale factor of the projection of our reduced distance at sea level. For this I refer to the chapter dealing with the Swiss projection and give you here the formula for linear alteration which is a function of the distance to the neutral axis. The more distant I am, the more important is the scale factor. The value here, reduced S bar = S0 (my reduced distance at sea level) x 1 + the distance squared with respect to the neutral axis / two times the radius of Earth squared. I take here an example for l'EPFL: I am located -47 km from the neutral axis the value is equal to 28 ppm ie 28 mm per 1 km. To summarize, this problem with reduction of distances contains a lot of elements. I recommend you to do the exercise that allow you to step by step understand all the stages in order to get a reduced distance in the projection system.