The spherical radius Вс of a small circle with с¹с²с³ is called the theoretical range of the visible horizon.
The value of the spherical radius depends on the height of the observer's eye above sea level.
So, if the observer's eye is at point A1 at a height BA¹ = e¹ above sea level, then the spherical radius Bc" will be greater than the spherical radius Bc.
To determine the relationship between the height of an observer's eye and the theoretical range of its visible horizon, consider right triangle AOc:
Ac² \u003d AO² - Os²; AO = OB + e; OB=R,
Then AO = R + e; Os = R.
Due to the insignificance of the height of the observer's eye above sea level in comparison with the dimensions of the Earth's radius, the length of the tangent Ac can be taken equal to the value of the spherical radius Bc and, denoting the theoretical range of the visible horizon through D T, we obtain
D 2T = (R + e)² - R² = R² + 2Re + e² - R² = 2Re + e²,
Rice. eight
Given that the height of the observer's eye e on ships does not exceed 25 m, a 2R = 12 742 220 m, the ratio e/2R is so small that it can be neglected without compromising accuracy. Hence,
since e and R are expressed in meters, then Dt will also be in meters. However, the actual range of the visible horizon is always greater than the theoretical one, since the beam coming from the observer's eye to a point located on the earth's surface is refracted due to the uneven density of the layers of the atmosphere along the height.
In this case, the beam from point A to c does not go along the straight line Ac, but along the curve ASm "(see Fig. 8). Therefore, to the observer, point c appears to be visible in the direction of the tangent AT, i.e., raised by an angle r \u003d L TAc , called the angle of the earth's refraction. The angle d \u003d L HAT is called the inclination of the visible horizon. And in fact, the visible horizon will be a small circle m", m "2, mz", with a slightly larger spherical radius (Bm "\u003e Sun).
The value of the earth's refraction angle is not constant and depends on the refractive properties of the atmosphere, which vary with temperature and humidity, the amount of suspended particles in the air. Depending on the time of year and the date of the day, it also changes, so the actual range of the visible horizon compared to the theoretical one can increase up to 15%.
In navigation, an increase in the actual range of the visible horizon compared to the theoretical one is taken by 8%.
Therefore, denoting the actual, or, as it is also called, geographical, range of the visible horizon through D e , we get:
To get De in nautical miles (assuming R and e in meters), the earth's radius R, as well as the height of the eye, e, is divided by 1852 (1 nautical mile equals 1852 m). Then
To get the result in kilometers, enter a multiplier of 1.852. Then
to facilitate calculations to determine the range of the visible horizon in Table. 22-a (MT-63) shows the range of the visible horizon depending on e, in the range from 0.25 to 5100 m, calculated by formula (4a).
If the actual height of the eye does not match the numerical values indicated in the table, then the range of the visible horizon can be determined by linear interpolation between two values close to the actual height of the eye.
Range of visibility of objects and lights
The visibility range of the object Dn (Fig. 9) will be the sum of two ranges of the visible horizon, depending on the height of the observer's eye (D e) and the height of the object (D h), i.e.It can be determined by the formula
where h is the height of the landmark above the water level, m.
To facilitate the determination of the visibility range of objects, use the table. 22-c (MT-63), calculated by formula (5a): To determine from this table from what distance an object will open, it is necessary to know the height of the observer's eye above the water level and the height of the object in meters.
The visibility range of an object can also be determined by a special nomogram (Fig. 10). For example, the height of the eye above the water level is 5.5 m, and the height h of the sign is 6.5 m, in order to determine D n, a ruler is applied to the nomogram so that it connects the points corresponding to h and e on the extreme scales. The point of intersection of the ruler with the middle scale of the nomogram will show the desired visibility range of the object D n (in Fig. 10 D n = 10.2 miles).
In navigation manuals - on maps, in sailing directions, in descriptions of lights and signs - the visibility range of objects DK is indicated at an observer's eye height of 5 m (on English maps - 15 feet).
In the case when the actual height of the observer's eye is different, it is necessary to introduce the correction AD (see Fig. 9).
Rice. nine
Example. The visibility range of the object indicated on the map is DK = 20 miles, and the height of the observer's eye is e = 9 m. Determine the actual visibility range of the object D n using Table. 22-a (MT -63). Decision.
At night, the visibility range of a fire depends not only on its height above the water level, but also on the strength of the light source and on the discharge of the lighting apparatus. Typically, the lighting apparatus and the strength of the light source are calculated in such a way that the visibility range of the fire at night corresponds to the actual range of visibility of the horizon from the height of the fire above sea level, but there are exceptions.
Therefore, the lights have their own "optical" range of visibility, which may be greater or less than the range of visibility of the horizon from the height of the fire.
Navigation manuals indicate the actual (mathematical) range of visibility of lights, but if it is greater than the optical one, then the latter is indicated.
The visibility range of coastal signs of the navigation situation depends not only on the state of the atmosphere, but also on many other factors, which include:
A) topographic (determined by the nature of the surrounding area, in particular, the predominance of a particular color in the surrounding landscape);
B) photometric (brightness and color of the observed sign and the background on which it is projected);
C) geometric (distance to the sign, its size and shape).
An observer, being at sea, can see one or another landmark only if his eye is above the trajectory or, in the limiting case, on the very trajectory of the beam coming from the top of the landmark tangent to the surface of the Earth (see figure). Obviously, the mentioned limiting case will correspond to the moment when the landmark is revealed to the observer approaching it or hidden when the observer moves away from the landmark. The distance on the surface of the Earth between the observer (point C), whose eye is at point C1 and the object of observation B with the apex at point B1 corresponding to the moment of opening or hiding this object, is called the visibility range of the landmark.
The figure shows that the range of visibility of the landmark B is the sum of the range of the visible horizon BA from the height of the landmark h and the range of the visible horizon AC from the height of the observer's eye e, i.e.
Dp \u003d arc BC \u003d arc BA + arc AC
Dp \u003d 2.08v h + 2.08v e \u003d 2.08 (v h + v e) (18)
The visibility range calculated by formula (18) is called the geographical visibility range of an object. It can be calculated by adding selected from the above table. 22-a MT separately for the range of the visible horizon for each of the given heights h u e
According to the table 22nd we find Dh=25 miles, De=8.3 miles.
Hence,
Dp \u003d 25.0 +8.3 \u003d 33.3 miles.
Tab. 22-c, placed in the MT, makes it possible to directly obtain the full range of visibility of the landmark from its height and the height of the observer's eye. Tab. 22-c is calculated by formula (18).
You can see this table here.
On sea charts and in navigation aids, the visibility range Dn of landmarks for a constant height of the observer's eye equal to 5 m is shown. The range of opening and hiding objects in the sea for an observer whose eye height is not equal to 5 m will not correspond to the visibility range Dk shown on the map. In such cases, the visual range of landmarks shown on the map or in manuals must be corrected by a correction for the difference between the height of the observer's eye and the height, equal to 5 m. This correction can be calculated based on the following considerations:
Dp \u003d Dh + De,
Dk \u003d Dh + D5,
Dh \u003d Dk - D5,
where D5 is the range of the visible horizon for the observer's eye height of 5 m.
Substitute the value of Dh from the last equality into the first:
Dp \u003d Dk - D5 + De
Dp = Dk + (De - D5) = Dk + ^ Dk (19)
The difference (De - D5) \u003d ^ Dk and is the desired correction to the visibility range of the landmark (fire) indicated on the map, for the difference between the height of the observer's eye and the height equal to 5 m.
For convenience on a cruise, it is possible to recommend to the navigator to have on the bridge corrections calculated in advance for different levels of the eye of an observer located on various superstructures of the ship (deck, navigation bridge, signal bridge, gyrocompass pelorus installation sites, etc.).
Example 2. The map near the lighthouse shows the visibility range Dk = 18 miles. Calculate the visibility range Dp of this lighthouse from an eye height of 12 m and the height of the lighthouse h.
According to the table 22nd MT we find D5 = 4.7 miles, De = 7.2 miles.
We calculate ^ Dk \u003d 7.2 - 4.7 \u003d + 2.5 miles. Consequently, the visibility range of a beacon with e = 12 m will be equal to Dp = 18 + 2.5 = = 20.5 miles.
By the formula Dk = Dh + D5 we define
Dh \u003d 18 - 4.7 \u003d 13.3 miles.
According to the table 22nd MT by the return entrance we find h = 41 m.
Everything stated about the range of visibility of objects in the sea refers to the daytime, when the transparency of the atmosphere corresponds to its average state. During transitions, the navigator must take into account possible deviations of the state of the atmosphere from the average conditions, accumulate experience in assessing visibility conditions in order to learn to anticipate possible changes in the visibility range of objects at sea.
At night, the visibility range of beacon lights is determined by the optical visibility range. The optical range of visibility of the fire depends on the strength of the light source, on the properties of the optical system of the beacon, the transparency of the atmosphere and on the height of the installation of the fire. The optical range of visibility may be greater or less than the daytime visibility of the same beacon or light; this range is determined experimentally from multiple observations. The optical range of visibility of beacons and lights is selected for clear weather. Typically, light-optical systems are selected so that the optical and daytime geographical visibility ranges are the same. If these ranges differ from one another, then the smaller of them is indicated on the map.
The range of visibility of the horizon and the range of visibility of objects for the real atmosphere can be determined empirically using a radar station or by observations.
Chapter VII. Navigation.
Navigation is the basis of the science of navigation. The navigational method of navigation is to navigate the ship from one place to another in the most advantageous, shortest and safest way. This method solves two problems: how to direct the ship along the chosen path and how to determine its place in the sea based on the elements of the ship's movement and observations of coastal objects, taking into account the impact on the ship of external forces - wind and current.
To be sure of the safety of the movement of your vessel, you need to know the position of the vessel on the map, which determines its position in relation to the dangers in a given navigation area.
Navigation develops the basics of navigation, it studies:
Dimensions and surface of the earth, methods of depicting the earth's surface on maps;
Ways of calculating and laying the path of the vessel on sea charts;
Methods for determining the position of a vessel at sea by coastal objects.
§ 19. Basic information about navigation.
1. Basic points, circles, lines and planes
Our earth is shaped like a spheroid with a major semi-axis OE equal to 6378 km, and the minor semiaxis OR 6356 km(Fig. 37).
Rice. 37. Determining the coordinates of a point on the earth's surface
In practice, with some assumption, the earth can be considered a ball rotating around an axis that occupies a certain position in space.
To determine points on the earth's surface, it is customary to mentally divide it into vertical and horizontal planes that form lines with the earth's surface - meridians and parallels. The ends of the imaginary axis of rotation of the earth are called the poles - north, or nordic, and south, or south.
Meridians are great circles passing through both poles. Parallels are small circles on the earth's surface parallel to the equator.
The equator is a large circle whose plane passes through the center of the earth perpendicular to its axis of rotation.
Both meridians and parallels on the earth's surface can be imagined innumerable. The equator, meridians and parallels form a grid of geographic coordinates of the earth.
Location of any point BUT on the earth's surface can be determined by its latitude (f) and longitude (l) .
The latitude of a place is the arc of the meridian from the equator to the parallel of the given place. Otherwise: the latitude of a place is measured by the central angle enclosed between the plane of the equator and the direction from the center of the earth to the given place. Latitude is measured in degrees from 0 to 90° from the equator to the poles. When calculating, it is considered that the northern latitude f N has a plus sign, the southern latitude - f S minus sign.
The difference in latitude (f 1 - f 2) is the meridian arc enclosed between the parallels of these points (1 and 2).
The longitude of a place is the arc of the equator from the zero meridian to the meridian of the given place. Otherwise: the longitude of a place is measured by the arc of the equator enclosed between the zero meridian plane and the meridian plane of the given place.
The difference in longitudes (l 1 -l 2) is the arc of the equator enclosed between the meridians of the given points (1 and 2).
Prime meridian - Greenwich meridian. From it, longitude is measured in both directions (east and west) from 0 to 180 °. Western longitude is measured on the map to the left of the Greenwich meridian and is taken with a minus sign in calculations; east - to the right and has a plus sign.
The latitude and longitude of any point on earth are called the geographic coordinates of that point.
2. Division of the true horizon
The mentally imaginary horizontal plane passing through the eye of the observer is called the plane of the true horizon of the observer, or the true horizon (Fig. 38).
Let's assume that at the point BUT is the eye of the observer, the line ZABC- vertical, HH 1 - the plane of the true horizon, and the line P NP S - the axis of rotation of the earth.
Of the many vertical planes, only one plane in the drawing will coincide with the axis of rotation of the earth and the point BUT. The intersection of this vertical plane with the earth's surface gives a large circle P N BEP SQ on it, called the true meridian of the place, or the meridian of the observer. The plane of the true meridian intersects with the plane of the true horizon and gives the last line of the north-south NS. Line ow, perpendicular to the line of true north-south is called the line of true east and west (east and west).
Thus, the four main points of the true horizon - north, south, east and west - occupy a quite definite position anywhere on earth, except for the poles, due to which, with respect to these points, various directions along the horizon can be determined.
Directions N(north), S (south), O(East), W(west) are called the main points. The entire circumference of the horizon is divided into 360°. The division is made from the point N in a clockwise direction.
Intermediate directions between the main points are called quarter points and are called NO, SO, SW, NW. Major and quarter rhumbs have the following values in degrees:
Rice. 38. True horizon of the observer
3. Visible horizon, range of the visible horizon
The body of water visible from the vessel is limited by a circle formed by the apparent intersection of the firmament with the surface of the water. This circle is called the visible horizon of the observer. The range of the visible horizon depends not only on the height of the observer's eyes above the water surface, but also on the state of the atmosphere.
Figure 39. Object visibility range
The boatmaster must always know how far he sees the horizon in different positions, for example, standing at the helm, on deck, sitting, etc.
The range of the visible horizon is determined by the formula:
d=2.08
or, approximately, for an observer's eye height of less than 20 m by formula:
d=2,
where d is the range of the visible horizon in miles;
h is the height of the observer's eye, m.
Example. If the observer's eye height h = 4 m, then the range of the visible horizon is 4 miles.
The visibility range of the observed object (Fig. 39), or, as it is called, the geographical range D n , is the sum of the ranges of the visible horizon with the height of this object H and the height of the observer's eye A.
Observer A (Fig. 39), located at a height h, from his ship can see the horizon only at a distance d 1, i.e. to point B on the water surface. If, however, an observer is placed at point B on the water surface, then he could see lighthouse C , located at a distance d 2 from it ; therefore, the observer located at the point BUT, will see the beacon from a distance equal to D n :
Dn=d1+d2.
The visibility range of objects located above the water level can be determined by the formula:
Dn = 2.08( + ).
Example. Beacon height H = 1b.8 m, height of the observer's eye h = 4 m.
Decision. D n \u003d l 2.6 miles, or 23.3 km.
The visibility range of an object is also determined approximately according to the Struisky nomogram (Fig. 40). By applying a ruler so that the heights corresponding to the observer's eye and the observed object are connected by one straight line, the visibility range is obtained on the middle scale.
Example. Find the visibility range of an object with a height above sea level in 26.2 m at an observer's eye height above sea level of 4.5 m.
Decision. D n= 15.1 miles (dashed line in Fig. 40).
On maps, sailing directions, in navigation aids, in the description of signs and lights, the visibility range is given for the observer's eye height of 5 m from the water level. Since on a small boat the observer's eye is located below 5 m, for him, the visibility range will be less than indicated in the manuals or on the map (see Table 1).
Example. The map indicates the visibility range of the lighthouse at 16 miles. This means that the observer will see this beacon from a distance of 16 miles if his eye is at a height of 5 m above sea level. If the observer's eye is at a height of 3 m, then the visibility will decrease accordingly by the difference in the visibility range of the horizon for heights 5 and 3 m. Horizon visibility range for height 5 m equals 4.7 miles; for height 3 m- 3.6 miles, difference 4.7 - 3.6=1.1 miles.
Consequently, the visibility range of the beacon will not be equal to 16 miles, but only 16 - 1.1 = 14.9 miles.
Rice. 40. Struisky's nomogram
Visible horizon. Given that earth's surface close to a circle, the observer sees this circle bounded by the horizon. This circle is called the visible horizon. The distance from the location of the observer to the visible horizon is called the range of the visible horizon.
It is extremely clear that the higher above the ground (surface of water) the observer's eye is located, the greater will be the range of the visible horizon. The range of the visible horizon at sea is measured in miles and is determined by the formula:
where: De - range of the visible horizon, m;
e is the height of the observer's eye, m (meter).
To get the result in kilometers:
Range of visibility of objects and lights. Visibility range object (a lighthouse, another ship, a structure, a rock, etc.) at sea depends not only on the height of the observer's eye, but also on the height of the observed object ( rice. 163).
Rice. 163. Beacon visibility range.
Therefore, the visibility range of the object (Dn) will be the sum of De and Dh.
where: Dn - visibility range of the object, m;
De - range of the visible horizon by the observer;
Dh - range of the visible horizon from the height of the object.
The visibility range of an object above the water level is determined by the formulas:
Dp = 2.08 (√е + √h), miles;
Dp = 3.85 (√е + √h), km.
Example.
Given: the height of the navigator's eye e = 4 m, the height of the lighthouse h = 25 m. Determine at what distance the navigator should see the lighthouse in clear weather. Dp = ?
Decision: Dp = 2.08 (√e + √h)
Dp = 2.08 (√4 + √25) = 2.08 (2 + 5) = 14.56 m = 14.6 m.
Answer: The lighthouse will open to the observer at a distance of about 14.6 miles.
On practice skippers the visibility range of objects is determined either by a nomogram ( rice. 164), or according to nautical tables, using maps, sailing directions, descriptions of lights and signs. You should be aware that in the mentioned manuals, the visibility range of objects Dk (card visibility range) is indicated at the height of the observer's eye e = 5 m and, in order to obtain the true range of a particular object, it is necessary to take into account the correction DD for the difference in visibility between the actual height of the observer's eye and the card height e = 5 m. This problem is solved with the help of nautical tables (MT). The determination of the visibility range of an object according to the nomogram is carried out as follows: the ruler is applied to the known values of the height of the observer's eye e and the height of the object h; the intersection of the ruler with the average scale of the nomogram gives the value of the desired value Dn. On fig. 164 Dp = 15 m with e = 4.5 m and h = 25.5 m.
Rice. 164. Nomogram for determining the visibility of an object.
When studying the issue of visibility range of lights at night it should be remembered that the range will depend not only on the height of the fire above the sea surface, but also on the strength of the light source and on the type of lighting apparatus. As a rule, the lighting apparatus and lighting strength are calculated for lighthouses and other navigational signs in such a way that the visibility range of their lights corresponds to the visibility range of the horizon from the height of the light above sea level. The navigator must remember that the visibility range of an object depends on the state of the atmosphere, as well as topographic (color of the surrounding landscape), photometric (color and brightness of the object against the background of the terrain) and geometric (size and shape of the object) factors.
Question number 10.
Visible horizon distance. Object visibility...
Horizon geographic range
Let the height of the eye of the observer located at the point BUT" above sea level, equal to e(Fig. 1.15). surface of the Earth in the form of a sphere with radius R
The rays of sight going to A" and tangent to the surface of the water in all directions form a small circle KK", which is called theoretically visible horizon line.
Due to the different density of the atmosphere along the height, the beam of light does not propagate in a straight line, but along a certain curve A "B, which can be approximated by a circle with radius ρ .
The phenomenon of the curvature of the visual beam in the Earth's atmosphere is called terrestrial refraction and usually increases the range of the theoretically visible horizon. the observer sees not KK", but the line BB", which is a small circle along which the surface of the water touches the sky. This observer's apparent horizon.
The earth's refraction coefficient is calculated by the formula. Its average value:
Refractive angler is defined, as shown in the figure, by the angle between the chord and the tangent to the circle of radiusρ .
The spherical radius A"B is called geographical or geometric range of the visible horizon De. This visibility range does not take into account the transparency of the atmosphere, i.e., it is assumed that the atmosphere is ideal with a transparency coefficient m = 1.
Let us draw through the point A "the plane of the true horizon H, then the vertical angle d between H and the tangent to the visual beam A" B will be called horizon inclination
In the Nautical Tables MT-75 there is a table. 22 “Visible horizon range”, calculated by formula (1.19).
Geographic range of visibility of objects
Geographic range of visibility of objects at sea Dp, as follows from the previous paragraph, will depend on the value e- the height of the observer's eye, magnitude h- the height of the object and the refractive index X.
The value of Dp is determined by the greatest distance at which the observer will see its top above the horizon. In professional terminology, there is the concept of range, as well as moments"open" and"closures" a navigational landmark, such as a lighthouse or a ship. The calculation of such a range allows the navigator to have additional information about the approximate position of the vessel relative to the landmark.
where Dh is the visibility range of the horizon from the height of the object
On marine navigation charts, the geographical range of visibility of navigational landmarks is given for the height of the observer's eye e = 5 m and is denoted as Dk - the visibility range indicated on the map. In accordance with (1.22), it is calculated as follows:
Accordingly, if e differs from 5 m, then in order to calculate Dp to the visibility range on the map, an amendment is needed, which can be calculated as follows:
Undoubtedly, Dp depends on the physiological characteristics of the observer's eye, on visual acuity, expressed in resolution at.
Angle resolution- this is the smallest angle at which two objects are distinguished by the eye as separate, that is, in our task - this is the ability to distinguish between an object and a horizon line.
Consider Fig. 1.18. We write the formal equality
By virtue of the resolution power of y, an object will be visible only if its angular dimensions are not less than at, i.e., it will have a height above the horizon line of at least SS". It is obvious that y should reduce the range calculated by formulas (1.22). Then
Segment CC" actually reduces the height of object A.
Assuming that in ∆A"CC" the angles C and C" are close to 90°, we find
If we want to get Dp y in miles, and SS "in meters, then the formula for calculating the visibility range of an object, taking into account the resolution of the human eye, must be brought to the form
Influence of hydrometeorological factors on the visibility range of the horizon, objects and lights
The visibility range can be interpreted as an a priori range without taking into account the current transparency of the atmosphere, as well as the contrast of the object and background.
optical range- this is the visibility range, depending on the ability of the human eye to distinguish an object by brightness against a certain background, or, as they say, to distinguish a certain contrast.
The daytime optical range of visibility depends on the contrast between the observed object and the terrain background. Daytime optical range represents the greatest distance at which the apparent contrast between the object and the background becomes equal to the contrast threshold.
Night optical range is the maximum visual range of the fire at a given time, determined by the intensity of the light and the current meteorological visibility.
The contrast K can be defined as follows:
Where Vf - background brightness; Bp is the brightness of the object.
The minimum value of K is called contrast sensitivity threshold of the eye and is equal to an average of 0.02 for daytime conditions and objects with angular dimensions of about 0.5°.
Part of the luminous flux of lighthouse lights is absorbed by particles contained in the air, so the light intensity is weakened. This is characterized by the transparency coefficient of the atmosphere
where I0 - light intensity of the source; /1 - light intensity at a certain distance from the source, taken as a unit.
To 
The transparency coefficient of the atmosphere is always less than unity, which means that geographical range- this is the theoretical maximum, which in real conditions the visibility range does not reach, with the exception of anomalous cases.
Evaluation of atmospheric transparency in points can be made on a visibility scale from tab. 51 MT-75 depending on the state of the atmosphere: rain, fog, snow, haze, etc.
Thus, the concept arises meteorological visibility range, which depends on the transparency of the atmosphere.
Rated visual range fire is called the optical range of visibility at a meteorological visibility range of 10 miles (ד = 0.74).
The term is recommended by the International Association of Lighthouse Authorities (IALA) and is used abroad. On domestic maps and in navigation manuals, the standard visibility range is indicated (if it is less than the geographical one).
Standard line of sight is the optical range at a meteorological visibility of 13.5 miles (ד= 0.80).
The navigation aids "Lights", "Fire and Signs" contain a table of the horizon visibility range, a nomogram of the visibility of objects and a nomogram of the optical visibility range. You can enter the nomogram by the intensity of light in candela, by the nominal (standard) range and by meteorological visibility, as a result of which you can get the optical visibility range of the fire (Fig. 1.19).
The navigator must experimentally accumulate information about the opening ranges of specific lights and signs in the navigation area in various weather conditions.