WO2001059410A2 - Bladed water meter with 50 millimetre bore and 54 mm external diameter for domestic and commercial use - Google Patents

Abstract

The invention concerns a bladed water meter with a 50 millimetre bore and a 54 millimetre external diameter for domestic and commercial use whereof the first inventive characteristic consists in a function of three variables z=hx + sy3/4 + gu + i with h, s, g, coefficients x, y, g of the variables, i a constant, the variables represent respectively the weight, the revolutions, pressure level, 3/4 (three quarters an exponent), the function Z its determination and its resolution is the object of the main disclosure, said function determines the weight x for which the blade is constant in a single iterative procedure, that is z=1, the blade is pressed by minimum water as concerns the part of the spot system by maximum water as concerns the maximum system; a second inventive characteristic represented by the spots: 4,5,6,7,8,9,10 pertaining to the applications A, B, C, each spot representing a blade; a third inventive characteristic two cubic spots: cube 1 and cube 2, their length should be equal to the perimeter (alpha revolutions) when z=1, the sum of their length should be equal to the length of minimum pressure, the two cubes are in the relationship not distorting the count of the meter; and a fourth inventive characteristic a maximum system starts when spot 5 is reached which is triggered by an amount K1, the system is maintained open by an amount K1=K2, the system is stopped by the drop in the capacity level of the pipe having a 3.04 cm diameter; x the weight is determined by first selecting a weight x, a height (u), revolutions generated by said height (y), and then by keeping the variables, u, y, with z=1 and the unknown, the weight x=z-sy3/4-gu-i/h. The embodiment and the explanation for the operating conditions of the meter are provided by the drawings 1/6 to 6/6 referred to the main disclosure.

Description

 WATER METER WITH 50 MILLIMETER PROPELLER AND 54 MM OUTER DIAMETER FOR DOMESTIC AND COMMERCIAL USE

This is a template counter 50 (fifty millimeters) propeller commercial industrial use, domestic whose base a 1 st invention a function of three variables ¹⁵ z = hx + sy _^3/4 + gu + I whose determination and resolution below:

-The initial idea comes from a slope materialized in the form of a triangle, the two ends of which are the top ul and the bottom one, this slope whose measurement - algebraically a certain number of segment, each segment expresses the rate of equal descent r, the last term is obtained by the following formula:

Un = ul. (nl) r The contractor to establish the slope of the soil towards the salt water pipe he uses an instrument a toyau whose two ends are two bottles sealed with a toyau filled with water, his work consists in laying the concavity of the toyau on the different parts of the distance which separate the two ends, when there is existence of the slope one of the two ends causes water to flow a (quantity) _/ ^> this quantity has a weight, we suppose that this quantity it has a weight 0 ⁺ (zero plus), we suppose that it is in the form of a sphere of perimeter zero plus (0 ⁺ ), in water very terms a ball, if we put this ball on the part of the slope starting point or the vertex and the ending point an algebraic measurement segment T (tau), when the ball completes this Segment the ball makes a course expressed in a

"^ ° certain number of perimeters called α (Alpha), after completion of this

1st segments its weight is varied, when it completes a second segment successively its weight still varies, a conclusion is obvious:

Let = P1 cover the 1st segment

Let P2 be the journey of the second segment 5 Let Pn be the journey of the last segment which gives.

P4> P3> P2> P1

P2 = PI x r1 geometric reason

And rm noted average reason P3 = p2 x π * Which generates a basic exponential function rm, which expresses the sum of the paths: α = e " ^gr _ ^m _e " Lo rm _ j

The exponential function is of resolution by two functions of three variables: n≈z≈hx '+ sy ^ gu ^π, + I ^l * ° un = t = Ex ^t + Fy ^d + or ^m + L Function z = n expresses the number ^of iteration in that ^a ball is in a constant speed ^that is a function of vaπables composed of 1 degree and less than 1 degree.

The idea of determination comes from ^a mathematical mind that is the departure to its location and its resolution in ^other words a set of mathematical ideas that are:

* The displacement of the center O ^of a coordinate system (0, I, J) to a point MB M the coordinates (xi, yi) in ° coordinates (xo. Yo) and (xι, yι) ed by (Mo, Mi):

By ef n t we are co-ordinated e o and i, in the base (ij) are those of Mi in the coordinate system (o, ι j) therefore v = (xl 1 + ylj) - (\ o i + yoj)

= (xi - xo) I + (yι-y) j

The point Mo and M i are parallel

* The displacement ^of a right have yi xi + bi to a straight O y2 = a2 + b 2 which generates y2 = a ^ H b "resolution is done by

II ^the linear adjustment method of least-carrrés ≥L [y2-a3Xi-b3] ² that

1 = 1 this difference is minimum and deriving according to the variable a3 or b3.

In both ideas mathematics in the 1 st the two points are parallel and in the second the two lines are parallel (yl is parallel to y 2) which ^ leads us to think in our function based ^invention n = z = hx + sy ⁴ + gu

+ 1, ^yet whose variables are all of degree greater than 1, the functions ^of a degree greater than one variable are represented by a curve, a question is posed is that the curves of degree greater than one functions are parallel to each other or ^that there is division of functions whose ^υ parallelism is different in response to these two questions have been studied and shown the functions of degree greater than 1 included in the intérvaval [1 -5] see table page table values and functions drawing 6/6 an observation is made on this chart, a separation of functions those which converge towards plus ^infinity y-axis and which converge towards more Linfini ^^ x axis. we can easily characterize them, we take two following functions and we analyze their curve x -x ^" the curve x ^" is more distant from the y axis than the curve x \ let us analyze a function of degree less than 1 for example x this function whose curve converges towards plus ^infinity x-axis, a conclusion is a clear separation of functions those of higher degree 1 converge more ^infinity y-axis and those of less than 1 degree converge more ^infinity axis x 6/6 see right drawing ^of equation y = x illustrates blue although this separation, the drawing also shows the parallel function of displacement, the functions that are below the line ^of x = équationy are parallel one to 0 above ^the other in ^{chronological} order and the other for which or are above the straight line ^of equation y = x.

This parallelism generates moving the curve \ 'to the curve x ³⁷⁴ with contributions of values between [0 l] and a zero, the result see page 10 ^that there is parallelism by these values of ^the _interval> ^> [ng lower values better fit the displacement called more zero (o) zero and one more, the contribution values and these values (o ⁺⁾ causes the displacement ^of a curve to another is true, the displacement is true also for non contributed values, which will lead us to the next simulated value paraghraphe or dispersed or projected by a 0 plus zero dispersion before ^the address basic feature of our invention z = - hx + ^{sy, / 4} + gu + i = n = hx ¹ + sy + gu '"+ i whose exhibitors vaπables t = 1, d = _^3/4, m = s choice comes uU phenomenon and ^the idea of starting recall the one respectively a ball and a slope materialized in the form of a rod, a ball which ach Evé segment a part of the slope is a parcour: i) The variable x represents the weight ^that is an arithmetic phenomenon that can be represented in the form of line y = x see drawing 6/6 dioite blue. 2) y X variable represents the number ^of interation or parcour generated by the weight x and a part of the slope, this phenomenon can be repésenter by a curve which embodies L weight xd ^a ball which completes different slopes of o ⁺ ( zero more), there is a com be whose shape Simulator to those below the right side ^of equation v - ^~ x. we chose y ³ / in ^that it is close to the right side ^of equation y = x 6/6 see drawing it obvious and clear that it takes the value of the towers, we ask, ".

4- 1 ⁴ which generates that () ¹ ^ L = t ⁴ f 'that the values of y ie the turns are reduced by f'. reduced to their initial value by the two coefficients s and g, by the function sy ¹ + g and by their identification determined by values simulated or projected by the geometric reason o (zero plus) and by their contribution in the calculation of the coefficients hs, g, i, sy + g is a function which exists in the right ^of equation y = x. then regulated by the relationship of h and i i0 to express LL ^that is a function which exists in the (o, i. J. K. L).

* 3) The choice of u representing a third variable heights its goal to show déscrmation ^that is to say the only differentiation by the Applications Eres J FK xi. i.ui) and 12 (2, 2, ιπ) with xi = x? dC i = v: different from 112 A with condition m and 112 results in v 1 = y The second applications

Fi ( ^χ ι, y π) and 4 (\ 4, y4. 114) With x di ièrent of U "differ from 114 1-ty, - v 4 different from 114 entrains y" = v4

What explains in common language in the applications l are identical, the iterations are identical there is differentiation in the heights, in other words the height u involves a single and single relation between x and y, its second objective to calculate pυinds the shaft of ^the propeller gear semicircle without interpolation or rule of three two-step see below. Its resolution by simulated or projected values by a dispersion equal reason, in our function there are variables of degree 1 which correspond to an ^• arithmetic phenomenon for which is treated simulated or projected values by o ⁺ (zero plus) equal arithmetic result and there are two variables a is of degree less than 1 between 0 and 1 ^that is there ^ι and a second n = z which corresponds to a geometric phenomenon which is treated simulated values or projected by a dispersion o ⁺ (zero plus) = geometric reason, this paragraph ^which aims to contribute values to calculate HSGI coéficients

* 1) Values siililécs or projected by an aritmetic dispersion reason zero plus equal standard deviation:

Either a sequence # 2 Schemat ized below on ^the axis below:

1 3 5 7, u l u2 u3 ιι4

With i ^'= 2 with 2 = 114 - a = 1 + (4- 1) 2 = 1 be a sequence aritmétique ^N' I schematically on ^the shaft below

0 1 '4 1 2 3/4 1 u l u2 u3 u4 u5

W ith ^the I = 'in - ^~ a - 0 - (5- 1)' iun = 1

An arithmetic N 3 schematically on ^the shaft below with π

- 4 __.

1 0 1 4 26 θ With u ^ = one with u- one. 11- 1) 4 26 Or an arithmetic sequence N4 schematized below which groups together the sequence N ^C I, the sequence N ° 2. the continuation N ^

t 1? I 1

4 '2 4 5 7 10 14 18 22 26 ^' with r =! . one = 26

2 (> = o + (iι-l) 1/4

26 = o + n / 4 26+ 1.4 l / 4 = n / 4 n = 105 \ = 4 n = 105 4 ^that is a sequence number 3 the number of terms n = 105 ^{l The} analysis shows if we divide the reason for the sequence N ° 1 by the number of this last sequence (ie n = 105), we obtain r is less than ri and we constitute another sequence of 1st term r and of reason r. n ^'th term of the series # 3 becomes: ^n' th tarnished = r + (n-1) r> Whatever the arithmetic result equal reason gap tyme =

Note that ^the gap. ,

Type =, so a = ui + (n-ι) and that the simulated values or projected as they appear on ^the axis above are 3-5-7-10-14-22-26 a dispersion equals 1 / 4, if we divide r / n the values will be simulated, so tends that r tends towards o ⁺ (zero plus) the same values will be simulated, a conclusion is obvious the values of the dispersion are the values o (zero plus), reason o (zero plus), this analysis is valid for the values of x the weight.

* 2) Values simulated or projected by a geometric dispersion reason equal zero plus.

Let be a geometric sequence N 2 schematized below:

•>

3 5 7 il with r'- 11/3 rm = 11/3 ^m = 1.34

Let a second geometric sequence N "1 of reason r m = 1st term, un = nth term

1/32 1/16 ^~ 1/8 1/4 1/2 5/4

5/4 = u 1.ri 54 = 132 ii m)

40- il m ilinn = 2.09

Either the sequence N ⁿ 2 increases ^to an end with u5 = a = 13 r3m is decreased with respect to πn ** l ** 1, 14 is less than 2.09

More N 'value of 1 o ^"(zero plus) due rm and number of terms n, with n plus ^the inlïnym tends towards zero more (3, was added two values in this sequence:

A - 4000 n = 80 001

4000 = 1 rm ^{<S ,, (,,,,,} . Rm = 1 28000 '^' O "

32 rnι = 1 .000147 This value of 1 000 147 ^that is the desired dispersion contribute to the values determined using the two values called path or marker to the location of the geometric ratio 1 +0.000147 or due o ⁺ (plus zero) since 1 is ^the neutral element, this marker is used every time when the phenomenon is geometric ^that is:

Un = 4000 N = 80 001 following the phenomenon

Values simulated or projected by a geometric reason equals zero plus by a quick formula:

Yim a grateful sign

- Plus v if mil is greater than rm with nu l. πn2. RM3 ... different reasons and kl The .k2.k3% of ^the increase in mark No 1%

2% .... 3% .... etc.

With K 1 = m -rm I. k2 rm - rm2 rm rm. kv = rm - rmv ⁴ rm

v = v -1 + 1 = n mark + v

- Less v if nu l is less rm

rm

K2 = = rm - rm2 r

K3 = = rm - ιτn2 0 rm K3 = kl xi ^: avcci = k3 / kl ^{| /:}

With kv = rm-rmv Rm kv = kl .i ^{1 "1} ^ kv = i '"'

^• , sare solution by 2 zh.x -sv ⁴ ~ uu -i)

h = -Σ (-x -kv ^' ' -Lu-k ut -1k -pl-kp8) (z ^' -λy 4 -oi λBu -z "-f -po-p8λ) _ (-.x-ky - lu-kβu-tf -pl-kp6) ²

C-Λ

i = l /> ι∑ [z ^' -hx-sy ¹ -gu]

After calculating the coefficients, its resolution is done in two stages: * 1) first take a weight x included for example in Finlérval 1/50, and 0.55, u height included in Finterval 1mm, 11mm, the weight x is constant in Zéde iteration which entails in 2nd period to keep the same values and to determine x the weight Z (zéde prime) is constant in a "single iteration what causes that Z ^" = 1

h -The task system qui'.n-; a second and a third invention, the second invention the cross ^flow containing the following tasks see drawing S l / 6c which are 4-5-6-7-8-9-10 applications belonging to ^the application A contains spots 4-5 and ^application B 6-7-8 spots, the application contains the C 9-10 spots. s ^adds thereto two n spots belonging ^to any of the applications that are cited précedament Cubel and CuBe2 and all these tasks are organized into scheduling on the graph drawing l / 6g, and a third 0 based stains invention both Cubel CuBe2 and their length must be equal to the length of the minimum pressure, the stains are selected 4-5-6-7-8-9-10 priorities spots ^and prior art, priorities spots are separated spots ^of prior art by Cubel and CuBe2. separation and lajéléction their goal to solve the problem between application in ^the application of counting ^the flow of a chronologically and ^independence and not simultaneously:

Alpha rounds Cube width =

1 xh The spots system injects a quantity of 283.34 cubic centimeters in 8 thirds, the total quantity injected by the counter is 452.91 cubic centimeters in 1 1.134 thirds, the difference of 452.91 - 283.34 or 168.57 cubic centimeters, the latter quantity its injection is the maximum system utility. -The maximum system resides in the counter in a 5th division called Application D, operation begins when the spot 5 is achieved by a triggering of the system caused by kl amount which rotates ^the propeller see drawing 3/6 (No. 4009 ) of the trigger helix that moves from left to right making the back pad 1 and buffer 2 see dessin3 / 6 (N4020, 4008) in the spring clamp mêmetemps see drawing 3/6 (N4003) ^at a distance equal to the displacement ^propeller trigger, the maximum system is maintained open by an amount k2 = kl which rotates ^the propeller see drawing 3/6 (# 406 l) maintains, when the buffer 1 and buffer 2 are injected recede the entire capacity of the 3.04 cm diameter pipe (spot> 1 1) see drawing 3/6 (N4001) either a capacity of 168.59 cm ³ or five times the maximum quantity of 33.7 18 cm ^' (cubic centimeter ), from the latter is taken the quanity k2 added second quan maximum tity adds maximum amount 3rd ^Until the n ^'th maximum amount, a remark is obvious the one maximum amount era is not minus and the plus th is a regularization ° is automatic injection quantity by the judgment the system generated by ^the judgment of Ihélice maintenance due to the decreased level of the capacity of the capacity of the diameter 3,04cm see drawing 3/6 (No. 4001), in other words it will neutrality increase and decrease generated by the system shutdown. î> ._ system The trigger injects the 1st amount of 33.718 cm which presses ^the propeller motor see dessin4 / 6 v elocity constant in a single iteration Called Z = 1. -The 50millimétre class counter is ^a pipe diameter capacity

3.04 cm, and ^a flow time of 8.78 cm. with a capacity of 452.91 cm in ^ ^û 1 1, 134 thirds.

Ana is mathemati.iue of the neutral elements o and 1 and interpretation of a negative number:

This analysis will help to understand the logic of Z our function ^when it equals a negative number simply a variable learned ^the identity element 0 and 1 or a negative number, because Z c ^'is a function of values: ^ - L equality of degree functions [1. + 2] is true for neutral elements:

0 0 1 - 1 fx = xx. x = x = I

ι ^{l / l0} = MT = ι

0 = 0 zero power zero equals zero

IF X yl = y2 = y3 = 0

1.837117 + 1.099807 -0.75896 +0.8347095 0.5760202

With b = 2.4354397 = 1, 3342322

1.8253573 a = 0.25140 (x ^V4 } = 1.3342322 x ^{l / 2} - 0.25140 = 0.27 / 20 x 1/2 5/20

Claims

1 ^Cie Claim

The present invention is a water meter with a propeller with a bore gauge of 50 mm and an outside diameter of 54 mm, the basis of the invention having a function of three variables z = hx + sy ^{3 4} + gu + i, the determination of which is the subject of the main thesis, this function determines the weight x for which the propeller is constant in a single iteration (a single turn) squeezed by the minimum amount of water, with regard to the system part of the spots, squeezed by the maximum water with regard to the maximum system part, this iteration lets the propeller rotate in a number of turns called (alpha), a second base of invention and a third base of invention which are the object of task system and a fourth invention which is the object of the maximum system, the second invention, the third invention and the fourth invention are the object of the second part of the thesis, the second basis of invention a flow center of containing the following stains are d e numbers 4,5,6,8,9, 10,1- belonging to applications. Application A contains spots 4 and 5, application B contains spots 6,7,8, application C contains spots 9 and 10, in addition to these two spots not belonging to any of the applications mentioned previously which are cube 1 and cube 2, all these spots are organized in order, the third basis of invention the two cubes (spots) 1 and 2 their length must be equal to the perimeter (alpha) turns, the sum of their length must be equal to the length of the minimum pressure, this condition explains that the two cubes are in the relationship not to distort the count of the counter, ie to count in a chronological way, which generates the determination of the cubic widths 1 and cube 2, the volume of a cube is equal 1 x L xh is equal (alpha turns), so L = turns / 1 xh with 1 = length, L = width, h = height.

And a fourth invention, a maximum system begins when the spot 5 is reached causes its triggering by a quantity Kl which rotates the propeller of the trigger which moves from left to right by making back the buffer 1 and the buffer 2 at the same time tightens the spring by a distance equal to the displacement of the tripping propeller, the system is kept open by a quantity K2 = Kl which rotates the retaining propeller when buffer 1 and buffer 2 move back cause the entire capacity to be injected the pipe diameter of 3.04 cm, a capacity of 168.59 cm * (cubic centimeter) is 5 times the maximum amount, the latter is taken the amount K2 added to the second maximum amount, added to 3 ^rd maximum amount etc ... up to n-1 maximum amount a remark is obvious to the ^cre amount is undervalued and the nth amount is increased, adjustment is automatic by shutdown begotten é by the stop of the propeller of maintenance because of the drop in the level of the capacity of the pipe of the diameter 3,04 in other tarnishes there is neutrality of the increase and the reduction generated by the stop of the system.

The characteristic dimensions and elements of the meter are the portrait of the meter product, caliber 50mm, overall dimensions 30 cm, height 14.07 cm. box width B 9.50 cm, box diameter A 8.30 cm, box length B 16 cm without the connection pipes and the applications mentioned above in the task system, each of which contains a motor which records the units of the cubic meter and an application D mentioned previously in the maximum system which contains a motor of the units of cubic meter which functions using the rod, the propeller, each of these motors manipulates the gear of the applications which registers the cubic meters with for number 8 in the drawing boards. At the end of this third part, other elements of yields or flow and of the maximum system's driving capacity characterizes the product a necessary quantity of 452.91 cubic centimeters in a time of 1,134 thirds, a flow of water in 1 hour of 8.78 cubic meter, and a pipe capacity placed in the system maximum diameter 3.04 cm. 1st .... VENDICATION

A popularization and an illustration of the functioning of the counter are made by a flowchart flowchart drawing board 2/6 A and a flowchart explaining the reasoning of the turns of the propeller drawing board 2/6 B

1) z = hx + sy V * + gu + 1 its resolution is done in two stages

The coefficients of the function are: h, s, g

Variables: x weight y turns u pressure height i constant

¹ in place selecting a weight x, u height, towers generated by this height z translates a value for which the weight is constant in x z towers is the l ^st step.

The second step, just keep the variables u, y with z = 1 and the unknown the weight:

- {[(z -z) + | A (-x- "-i ')]} ₊ ËS ^, -F) I (-"-' ^, ) -lg = -

∑ (- u - P) '

2) Calculation of the widths: by making the equality between either V the volume of the cube or z = 1 carry out a certain number of turns called (alpha) volume of the cube = 1 x L x h = α turns with 1 = length

L = width = α ALL

\ Xh

H = height

, EMH CLAIM

This is an invention of a water meter with a caliber of 50 millimeters of bore and an outer diameter of 54 millimeters with a propeller, for which I have been obliged to identify the new technique or its basis for invention and its operation. which are developed in a plan, then popularize the - understanding of its operation by an example called selected opening illustrated on a drawing, other drawings from 1/6 to 6/6 are attached which complete the portrait image of the counter, the development plan is as follows:

I) Its basis of invention a function of three variables.

II) Its basis for invention a system of spots:

.-λO a) Spots belonging to applications b) Cubes are spots not belonging to any of the applications.

III) Its basis for invention a maximum system.

IV) Popularization of its operation by an example.

V) Operation of its elements.

Λ "VI) Characteristics of size and shape.

VII) Flow and efficiency characteristics and driving capacity.

It is a counter which functions using the gears which turn using the propellers belonging to applications, this counter comprises four applications application A, application B,

2 υ the application C belonging to the spots system, each application of the spots system comprises a motor of the units of m3 (cubic meter) which operates using three gear cylinders manipulated by a cylindrical gear producing a tight parallel shape, application A and C each containing two propellers and application B containing three propellers and application D belonging to the maximum system which includes a motor of units of m3 (cubic meter) which operates using three

2 J cylinder manipulated by a vertical rod see drawing 4/6 (N ° 4060) and a propeller for counting the maximum quantity see drawing 5/6 (N ° 4022), each propeller of the two systems is enclosed in its support, when the number representing the m3 (cubic meter) is reached by each engine of the applications of the spots system is manipulated by the third cylinder with four teeth of gears there is iteration of the gear of the applications and when the number representing the m3 ( cubic meter) is achieved by "engine D applying the maximum system, the gear applications see drawing 4/6 # 8 is manipulated once the cylinder 3 ^ee see drawing 4/6 (No. 4034) which form with the concentric gear hecoidal gear see drawing 4/6 (N ° 4038), the application gear in turn turns the engine of m3 (cubic meter) containing six cylinders between two cylinders two gears and a third sealed at number 9 and a gear connected to the gear of applications see N ° 650 of

^ 2> ^> drawings, each gear and each propeller either for the maximum system or for the stain system turns clockwise which comes back to him respectively see the appendices, before going to the next paragraph it is important to ask the following question which turns the motors of the systems and the gears of the applications, the answer is the propeller turning by water therefore what will lead us to understand the reasoning of the turns of the propeller either stain system,

H ^û is of the maximum system.

The] ^ae invention of the water meter bore gauge 50 millimeters which is a function of three variables Z = hx + Sy + gu + i whose determination and resolution are the subject of the main thesis, this function consists of variables and coefficients and a constant:

P- - The variables x, y, u

^ - The coefficients h, s, g

A constant (i)

This function is interpreted Z the number of iteration for which x the weight is at constant speed with (y) turns of the l " ⁶ iteration generated, by a height, this I ^th iteration example a ball and a slope, in our case the weight it is the propeller and its components are pressed by water which are expressed / u respectively by a ball and a slope (height), the function its resolution is done in two stages: l ^≈e stage the propeller is pressed by the minimum of water making a certain number of turns it is the value of (y), these turns are expressed by a height of a slope, these turns are equal to the distance traveled by the ball descending from this slope, so the height is (u), in this l " ⁶ step Z is interpreted as the number of iteration is greater 1, (if we use this solution the propeller will never be at speed O • constant whatever the number of iterations, which explains why the ^1st minimum or the maximum iterates the propeller and let the propeller performs a number of tricks that will allow the second iteration starting, the third iteration begins only when the 2 ^ee iteration performs his tricks ... etc which translates the propeller always returns to its original weight , so it is impossible to realize this solution in the gear technique, only one solution is obvious and possible in the technique

9, fl of the gears that the propeller is of the spots system, either of the maximum system or at constant speed in a single iteration, so the function becomes Z = 1, which will lead us to:

2 ^nd step, to solve Z = 1, keeping only the same values of the function and we take as unknown (x) the weight: ³ λ z - Sy - gu - i

% ς x = - h

X the weight of the propeller, of the rod, of the four-tooth gear it is the weight pressed by the minimum of water "stain system" which manipulates the cylindrical gear, and x the weight of the propeller, rod and counting gear it is the weight pressed by the maximum water "maximum system" which manipulates the two parallel gears, the two weights are materialized respectively rod, 1.0 propeller, four-tooth gear see minimum system, on drawings 4/6, 5/6 see annexes 18,19 (N ° 4045, 4022,4035). Before going to the paragraph reserved only for the task system it is important to point out what is said in the 1st paragraph that the counter is divided into a certain number of applications, each of the applications containing tasks, application A contains the spots 4 and 5, application B contains spots 6, 7, 8, application C contains spots 9 and 10, application D ζ contains spot 1 1, each spot represents a propeller and a pipe, all dérigées and gather in a 4 ^th division called water flow center see FIG (1/6 number 25), now ask the question is that the water flow through the various tasks of the system of spots pose a problem that of falsifying the account of the counter for that let us try to verify the problem by two examples which can occur because of the leaks which the

T ° closings and opening of water.

¹ example is the closing of the square head valve 0 (gauge) 400 by ONEP (National Office of Drinking Water), when a leak is detected at the place where is connected the size counter 400 which serves as a regularization of account since ONEP is supplier of the Lyonnaise des Eaux, this counter is installed in each sector representing a district which clogs streets, at the end of a

^ - hour the whole sector is in a state of lack of water, taking a building located in this sector of ten floors in each floor three apartments, this building is supplied by a water meter of caliber 50 millimeters, households who are there are in a state of lack of water, while waiting for the opening find the means to warn to leave the taps open, in this case the meter with full opening with pipe connected to the meter is empty, when the

.Λ O leak is repaired and ONEP opens the valve the spots 4,5,6,7,8,9,10 of the counter work at the same time ie there is manipulation at the same time of the gear cylindrical of each application by the stains which come back to it, this simultaneity does not let it turn it is the problem of the application, to solve this problem it is necessary a ^CTe organization of the center of water flow it is the goal of next paragraph.

To illustrate the second example remaining at the complete opening operation of all the spots and • 6 imagine by being located in the places of the propellers in time fraction of a third the organization of the flow center requires an illustration by reasoning which will allow the user to have the idea popularize different stages of the process the apparent solution of the ^ae observation implies that the spots in the center of flow are arranged one above the other forming a single pipe bore of 38 millimeters, in this case the water flow is explained by imagining that ^ .Q it is a glass pipe of length of 30 centimeters filled with water closed at both ends by two hands, opening one end the water is seen escaping realizing a cylindrical form of 1 ^er6 quantity (RXRX3,14), the others flow in the clironological order of a length of 30 centimeters equal to the cylindrical volume, let us pose the question is what the count is good the answer is no because no spot is identified and that in each application two propellers of the same direction 25. clock hands operate by simultaneously counting two units of m3 (cubic meter) in a single unit of m3 (cubic meter) (one unit of cubic meter is not counted), or when two propellers oppose clockwise no units are counted. Now try to keep them away from each other without identifying spots the same problem will now try to identify the spots and select the priorities and stains stains ^• a grandfathering each application provided with priority spots and the prior art remote spots from each other, the selection is as follows 10-7-4 then 8-9-6, then the spot 5, whereas subsequently the 1 ^st spots priorities 10-7-4, check the account it is good the engines function and count in a chronological way, the simultaneity is discarded because of the spots scattered from bottom to top and because of the independence of the applications, it is advisable to?) ->'them. scattered and confused see figure 1/6 (N ° 25) which illustrates that the 1 ^st priority tasks are distant from each other but dispersed and confused in the spirit of the flow of water, the figure indicated last ent illustrates well this dispersion of 3mm (three millimeters) between the spots, the spot 10 of placement N ° 0 is placed from bottom to the top of 6mm (six millimeters), the spot 7 of placement N ° 20 placed from bottom to top of 9mm (nine millimeters), the placement spots No. 21 placed from bottom to

* <12mm (twelve millimeters) high, the problem between applications is resolved.

To illustrate the second example remaining at the complete opening operation of all the spots and let us imagine by being located in the places of the propellers in the fraction of a third time, this time equal to the time of flow of the maximum quantity on the number of times the minimum representing the

-φ maximum see paragraph equal to 8 thirds / the number of times the minimum, this value is zero point of dust, let's reason on the last part of the preceding paragraph and operate

one of the posterior spots without it being identified, let's identify what happens in the three applications because of the relationship that exists between spot 4 and one of the posterior spots 8-9-6 the account is necessary because in a application two propellers turn clockwise counts one unit of m3 (cubic meter) or places of two, two opposite propellers clockwise different two units of m3 (cubic meter) c ' that is to say are not counted, the malfunction comes from the fact that the propeller is iterated in only one time at constant speed is realized a certain number of turns called a (alpha) realized by two propellers in the simultaneity, it is necessary that it There is independence of realization of a (alpha) turns (alpha turns expressed in perimeters), the solution lies in finding a way to shift the beginning of alpha turns from one of tasks 8-9-6, this

- o. offset leaves the spot 4 made its 0. (alpha) turns independently, let's reason by the example indicated above that of the glass pipe filled with water closed at both ends by two hands, when we open one end, l flow is done chronologically with equal quantity XrayonX3 radius, 14X length (30 centimeters) by achieving the interior volume, now let's try to increase the volume by forming a cube in the middle of the pipe below in other words it

A ^→ v separates the two parts by 15 cm, this cube is 1 cm wide, 3 cm long and 1 cm high, let's try to empty the pipe we notice the second part of the pipe is delayed by the length representing the volume of water contained in this cube, this length may be expressed in perimeters, this perimeter may be equal to alpha turns and the second part leads one of the posteriority spots 8-9-6, e referring to drawing 1 / 6 (N ° 25) we notice a cube N ° 16 whose

$, O - dimensions length and height knowno and width undetermined, above the cube appeared the spot 8, the cube is of volume length times heights times width is equal to the perimeter alpha turns which generates the following equality is V the volume of cube and Z = 1 performing a certain number alpha turns = (equal) 1 x L xh = (alpha) turns

(alpha) turns

• • £ - Width = W x H

It is understood that the cube which separates the tasks of priorities and posteriorities these latter identified which are 8-9-6, which aims to eliminate the simultaneity of account, it will solve the problem in the application. Now let's try to see what happens see drawing 1/6 (N ° 25) between spot 6 and 5 which count in independence because of the usefulness of cube 2 but generates another jO. problem with the 1 ^st tasks that of counting in the simultaneity (two units of cubic meters in one unit of cubic meter etc), the illustration of the problem resides we ask you to pay attention by imagining what happens in the fraction of the third of the relation which exists between cube 1 and cube 2, the length of cube 1 delays the operation of spot 8 leaving the helix of spot 9 performs its α (alpha) turns similarly cube 2 delays the operation of spot 5 leaving the propeller of the

? oh. task 6 performs its turns, the anomaly of counting in the error lies in the condition that the sum of the lengths of the two cubes is less than the length of the minimum, which gives rise to the following explanation the priority tasks each comes from an application and the spot 5 belongs to one of the applications mentioned above, they must be delayed in their operation by leaving the spot 4 carried out its C. (alpha) turns a solution is evident in the drawing 1/6 (N ° 25) the length of the minimum 0 which iterates the propeller must be equal to the sum of the length of the two cubes or twice (X (alpha) turns. In summary of this paragraph the second basis of invention and the third base of invention respectively ia selection of spots in priority spots and posteriority spots which are 4-5-6-, cube 1-7-8-9 - cube 2-10 organized in the center of water flow by the scheduling and identified, cubes 1 and cube 2 their width equal to (alpha) turns / 1 xh and the sum of their lengths

-3 of water equal to the length of the minimum allow the engines of the applications to count easily, that is to say in a chronological way not in the simultaneity.

Before going to the second part "maximum system", it is important to specify its usefulness and its objective for this one is obliged to indicate only and not the way of determining the useful quantity of this counter, the useful quantity sa determination will be specified in a paragraph below _* Λ ° below, a useful quantity of 452.91 cubic centimeters in 11, 134 thirds with quantity which comes back to the spots system of 283.34 cubic centimeters in 8 thirds, the difference being 168 , 57 cubic centimeters which will be injected by the maximum system operation.

Its operation begins when the spot 5 is reached by a triggering of the system caused by a quantity Kl which turns the propeller see drawing 5/6 (N ° 4009) propeller of the trigger which

- λ D. move from left to right by moving buffer 1 and buffer 2 see drawing 5/6 (N ° 4020, 4008) at the same time tighten the spring see drawing 5/6 (N ° 4003) by a distance equal to the displacement of the triggering propeller, the system is kept open by a quantity K2 = Kl which rotates the propeller see drawing 5/6 (No. 4061) holding propeller, when the buffer 1 and the buffer 2 move back cause all the capacity of the pipe with a diameter of 3.04 cm see drawing 5/6 (No. 4001) or a capacity 9 of 168.59 cm3 (cubic centimeter) or 5 times the maximum quantity of 33.718 cm3 (cubic centimeter), of this last is taken the amount K 2 added to the second maximum amount, added to 3 ^rd maximum amount. .. Until the nth maximum quantity, a remark is obvious the ^CT maximum quantity is reduced and the nth is increased an automatic adjustment of the quantity injected by the shutdown of the system generated by the shutdown of the sustain propeller due to the decrease in

2 level of the capacity of the pipe diameter 3.04 cm see drawing 5/6 (No. 4001), in other words there is neutrality of the increase and decrease generated by the system shutdown. The triggering of the system injects the quantity \ ^ae of 33,718 cubic centimeters which presses the propeller of the engine see drawing 5/6 at constant speed in a single iteration called Z = 1, Z = 1 is determined by a function of three variable variables determined and resolved in the main brief -

% t? Z = h x + Sy + gu + i

With x the weight the unknown chosen arbitrarily u height Pressure y the turns expressed in perimeters i a constant 35 • With h, s, y, i coefficients determined in l main memory, with Z is the unknown:

The calculation consists of ¹ to determine location Z which translates a value for which the weight is constant in x Z towers with Z is greater than 1.

In 2 ^nd place the unknown is the weight: it suffices to keep the variables u, y with Z = 1, the weight found is the weight of the propeller of the rod, of the 4-tooth gear respectively in the drawing 5/6 (N ° 4022), 4/6

Lf o (N ° 4039, N ° 4045).

Z - Sy - gu - i X =

Before going to the next paragraph it is wise to materialize the difference between the propeller pressure by the minimum and by the maximum quantity, this difference lies is that the quantity> minimum and the maximum quantity would be during operation of the counter in the lower direction, the minimum can be defined as the lowest quantity close to the threshold Opening that person psychologically agrees to open but equal to the sum of the lengths of the cubes may be lower but never higher because of the constant speed in a single iteration while the maximum quantity can be neither lower nor higher because it is a constituted quantity which

J_ 0 presses the propeller when the maximum system is triggered, this quantity is part of the other quantity when they have become available, i.e. formed by the stain system and injected by the capacity of the pipe with a diameter of 3.04 cm (centimeters) and available when the time for the flow of one cubic centimeter of the same maximum quantity injected by the spots equal to the time for the flow of one cubic centimeter of the same quantity injected by the maximum system,

- * S this difference is popularized in the following paragraph

Choice of the maximum quantity for which the flow time is minimum: the determination of this quantity lies in practice, that is to say to draw up a test table and to use a conical head pipe of reference r = 0 , 10 arithmetic reason

3.0 n = 7 number of terms ul = 1.20 ^1st term un = 0.70 last term f (n) = R + (nl) r ul = radius = R

__ see table and drawing of the conic 1/6 B, with tl in t3 X Qcm / 3394.92 X60 tl and t2 in thirds fraction of - * seconds, the analysis of this table reveals that there is a choice to enter between line 3 a quantity in cm3 (cubic centimeters) of 43.96 cm3 and a time of 10.92 thirds and line 6 brings out a quantity in cubic centimeters of 41.54 in a time of 8.09 thirds, the choice lies in calculating the minimum time to inject a cubic centimeter, for

10.12

- line 3 the time is = 0.23 thirds V 43.96

8.09

- line 6 the time is = 0.19 thirds

41.54 it is obvious that the interior time is 0.19 thirds which corresponds to the quantity in cubic centimeter of 41.51 cm3

3 'Calculation of the maximum quantity representing N times the quantity of 168.59 cubic centimeters, its objective is explained if we divide the quantity of 166.59 cm3 (cubic centimeters) by 41.59 cm3, we find 4.053 times, the quantity of 41.59 cm3 which presses the propeller whose rod is vertical of the maximum system has four times an amount equal to 166.36 cubic centimeters, the differences being 168.59 cm3 minus 166.36 equal to 2.23 centimeters which represents 0.053 times, this difference is injected without being tf counted imagine the total error by counting one m3 (cubic meter) or 1000 000 cm3 (cubic centimeter) multiplied by 2.23 cm3 (cubic centimeter) we obtain 13404.665 cm3 ( cubic centimeter) per cubic meter, this is the reason why I have to reduce the quantity by 41.59 cm3 and let’s

the question that the whole number greater than 4.053 times multiplied by the reduced quantity gives the quantity of 168.59 cm3 (cubic centimeter), the whole number closest to 4.053 is the number 5, we find the reduced quantity of 33.718 cm3 (cubic centimeter) whose length is equal to the radius of the conical pipe squared X3.14X length equal to 33.718 cm3 (cubic centimeter):

0.70 x 0.70 x 3.14 x 1 = 33.718 cm3 (cubic centimeter) 33.718

1 = = 21.91 cm3 (cubic centimeter)

1.5386

-Λ O the time for the flow of the maximum quantity is equal: 33.718 x 0.19 thirds = 6.40 thirds

Comparison of the flow time of the maximum quantity through the conical section of the reference pipe entered see drawing 5/6 (N ° 4001) bore 3.04 cm and the outlet see drawing 5/6 (N ° 4024) of conical reference reason (-0.10) (n = 7), (u 1 = 1, 20) and (un = 0.70 and the time of the same quantity

Λi> - constituted by the minimum injected by the spots: This comparison makes it possible to better appear the functioning of the caliber 50 counter, that is to say to count without error, is interpreted as a major element of its invention which gives priority when the condition is carried out with both systems, task system and maximum system of counting either the maximum quantity when it is constituted the priority is with the maximum system, when the quantity is lower than the maximum quantity f0 the priority is with the task systems, this comparison is seen when the quantity that will be injected by the maximum system is greater than 7 x 40.62 cm3 equal to 284.34 cm3 (cubic centimeter), the comparison will be made by calculating the time of flow of a cm3 (centimeter cube) included in the maximum quantity, i.e. 33,718 cm3 (cubic centimeter) at its full injection by the maximum system and the time of flow of one cm3 (cubic centimeter) e) of the same amount injected by

25 spots:

6.40

- By the maximum system = 0.195 third parties

33,718 8

By the stain system: = 0.196 thirds

30 40.62

A remark is obvious the two are identical the priority is automatic belongs to the maximum system because of its pipe diameter 3.04 cm placed at the bottom of the meter, this priority is unique and exclusive when the maximum quantity has elapsed at full it to say at 33,718 cm3 (cubic centimeter), an obvious conclusion that can be drawn when the quantity becomes less than 3 ⁾ 33,718 cm3 the time is $ “Aft.rieur à 6,40 / 33,718 and becomes equal only when the maximum quantity is constituted, an observation in the test table see as long as the value of the diameter of the conical (output see drawing 5/6 N ° 4024) tends towards 1.60 cm, 1.80 cm, 2 cm, the quantity of water decreases and the time increases, it goes without saying when the maximum quantity is not made up the time of a cm3 becomes greater than 0, 195 thirds the priority is automatic belongs to the system of spots. ko Usefulness of the different parts that make up the maximum system: 4/6, 3/6, 5/6 401 1 a horizontal rod that rotates clockwise when the maximum system is

triggered and anti-clockwise when the system stops, the movements are made by the tripping propeller and the holding propeller.

4009 the tripping propeller this starts to rotate clockwise V when spot 5 is reached, the propeller moves from left to right.

4061 holding propellers this works clockwise and moves from left to right using a quantity Kl = K2 taken from the maximum quantity when buffer 1 and buffer 2 move back inject the maximum quantity, the propeller of maintenance also controls the stop of the system when the level of the pipe of the diameter 3,04 cm3 see drawing 19/22 (N ° 4001) drops that is yλ o ie of the spot 1 1.

4019 propeller support: it contains the propeller

4007 it seals the maximum system is called the fastener.

4003 the spring is tightened when the triggering propeller starts to rotate and kept tightened by the holding propeller, its usefulness is to stop the maximum system.

At S 4001 the pipe with a diameter of 3.04 cm3, the capacity of which to inject five times the maximum quantity of 33.718 cm3 (cubic centimeter), ie an amount of 168.59 cm3 (cubic centimeter). 4022 the propeller of the engine: it turns clockwise, manipulated by the triggering of the maximum system, pressed by the maximum quantity of 33,718 cm3 which makes it at constant speed.

% JD 4050 the engine has three cylinders: between two cylinders three gears, the utility of which is to handle the cylinder which is placed on the second, one is sealed in the center of the cylinder handling the second is sealed to the cylinder plate, the third is sealed to the place of iteration of its perimeter, the cylinders whose perimeters are equal determined on the basis of how many times the maximum quantity of 33,718 cm3 (cubic centimeter) exists in a cubic meter sit 1,000,000 / 33,718

2 5 _so it 2 657.75 times by making the equality which expresses how many times the cylinder must be manipulated to handle the second cylinder either:

29 xt = 29657.75 with 3 which expresses three cylinders t = 10.075 times This value of 10.075 and the value of 4mm representing the length of the rectangle which reveals the 3 ^ figure imposes the dimensions of the cylinder, in particular its perimeter, ie the following equality :

4mm x 10.075 = (radius + radius) x 3.14 40.30

2 radii = = 12.834 mm representing the diameter of the cylinder:

3.14 the diameter of the cylinder: 1.28 cm 6 To illustrate the operation of its installation utility is applicable to a building whose number of apartments is 30 at most, so it has several operations:

1) For the task system operation of one task in one application operation of two tasks in two different applications, three etc. tf ζ). - operation of all stains.

2) For the maximum system.

Task 5 is reached, maximum system triggering and:

maximum system operation and two task operation in _ two different applications, three etc

- * - operation of all tasks

We chose to popularize the illustration of its operation the complete opening those of all the spots which are respectively for the maximum system and for the system of spots is spots 4, 5, 6, cube 1, 7, 8, 9 , cube 2, 10, 1 1, and adds to them an unidentified spot "the threshold" it is a quantity less than or equal to the minimum all organized and schematized in the graph of A o order it see figure 1 / 6 g including a second objective to calculate the time later and the earlier time of the water flow, time is the fraction of a third, 1 hour etc, the opening is materialized by two flowcharts in loops

The ^1st illustrates the reasoning of the propeller turns see drawing 2/6 B and

AS annex 10 A

The second the different opening possibilities see drawing 2/6 A appendices 10 B, 1 1, 12

The flow and performance characteristics, driving capacity with the determination of its diameter of 3.04 cm are calculated

1) The debits are materialized in Z * ® - time at the latest in time at the earliest time in time

A necessary amount of usefulness of the counter

The l ^CTe solution is an algebraic solution materialized and calculated a second time 2.S »graphically a) Calculation of the quantity required Before approaching its calculation it is wise and important to specify a logic which will be the basis of its calculation, a logic that is in the building or is installed the meter caliber 50 mm (fifty millimeters), this meter supplies 30 apartments each time needs water it opens a tap of bore pipe 1.5 3 ^ cm (diameter interior) which flows a quantity of 40.62 cm (cubic centimeter) when it is fully opened, these two elements will be the basis for calculating the necessary quantity, let's reason in calculating

* The area of the 1.5 cm bore circle

1.5 (0.75 x 0.75) x 3.14 = radius x radius x 3.14 with radius = = 0.75 cm

3> 5

Area = 1.76 cm3 (cubic centimeter)

* The area of the bore circle 50 mm = 5 cm

5 radius x radius x 3, 14, 2.5 x 2.5 x 3.14 with radius = = 2.5 cm

2

We ask the following question in 19.625 cm3 (cubic centimeter) how many beans l, 76 cm3, answer just divide 19.625 cm3 over 1.76 cm3 or 1.15 times is the number of pipes, the quantity required is equal 1 1, 15 x 40.621 cm3 (cubic centimeter) or an amount of 452.91 cm3, this amount comes back to the stain system and to the maximum system. - Quantity of the stain system.

7 spots x 40.62 cm3 or 284.34 cm3

- ^ "- The difference is injected by -

The maximum system is 452.91 - 284.34 = 168.57 cm3 b) Calculation of the different flow times

Their calculation is done using two solutions the] ^ae is an algebraic solution and the second is a graphical solution its result is taken from an observation which is a graph path - 5 and an iteration carried out in the fraction of a third 1 °) algebraic solution a) The time of the flow of the quantity of 40.62 cm3 by a spot belonging to the applications A, B, C (is also called maximum quantity) £ The bases of calculation will be

The maximum quantity of the 3/2 cm bore pipe is 40.62

A cylindrical shape with the characteristics Radius 12 cm 3.14 = 3.14 / length 10.17 cm

The cylindrical shape is filled with water by the maximum quantity in a period of 15 seconds, the cylindrical shape contains 12 x! 2 x 3, 14 x 10.17 or 4,598.46 cm3 filled in 15 seconds, in one second the cube is filled with 4598.96 cm3 divided in 15 seconds:

4598.46 cm3 (cubic centimeters)

15 seconds = 305.56 cm3

3 ^ Then the quantity of 306.56 cm3 represents how many times the quantity of 40.62 cm3 or 306.56 / 40.62 equal to 7.54 times, we know that in one second there are 60 thirds, so the quantity of 40.62 cm3 has elapsed for 60 / 7.54 or 8 thirds b) The time for the flow of the reduced maximum quantity is: from 33.718 cm3 x 0.195 thirds or 6.57 thirds see table and paragraphs calculation of the maximum quantity

35 representing N times the quantity of 168.59 cm3 and choice of the maximum quantity whose flow time is minimum

One remark is obvious that the two results are the maximum amount of 33.718 cm3 6.57 thirds and the overall amount of 284.34 cm3 8 Third, the time of ¹ result "differs from the time the two ^em lésultat, logical calculation lies in increasing the time by 6.57 5 thirds up to 8 thirds and the proportional quantity which corresponds to the increased time which generates a rule of three For 6.57 thirds a quantity of 33.718 cm

For 8 thirds a quantity of Q?

The product of means is equal to the product of extremes

- *> O 8 x 33.718 cm

'Q = = 41,056 cm (cubic centimeters)

6.57 third parties

A global quantity of the two systems iterated in 8 thirds is 284 cm3 + 41.05 cm3 or 325.40 cm3

452.91 x 8

J Ç c) The flow time of 452.91 cm3 is = 11.134 t thirds

325.4 d) Calculation of the time of a flow of one cm3 in a cubic meter il 1000,000 cm3 (cubic centimeter), it suffices to ask the following question how many times the quantity of 452.91 cm3 in a cubic meter is

1000000 = 2207.94 times, just multiply this result by 1 1.134 thirds or 2207.94 x

452.91 1 1,134 or 24,584,204 third parties

24584.844

Conversions in minutes = 6.829 minutes

60 x 60 e) Calculation of flow in hours (60 minutes) this result is obtained by 5 converting 60 minutes to thirds

60 x 60 x 60 = 216,000 thirds, just ask the following question in 216,000 how many times 1 1,134 thirds is

216,000

= 19400.036 times and multiply this result by 452.91 cm3 (cubic centimeter)

1 1.134 or 8 786 470.30 cm3 (cubic centimeter), this result is to be converted into m3 ^ (cubic meter) or

8,785,470.30

= 8.78 m3 in 1 hour, this result is obtained from the time of a flow rate of cubic meter

1,000,000

60 i.e. = 8.78 m3 (cubic meter).

6.829

^ 2 °) Graphic solution: this solution will calculate the time later and the earlier time of the flow of the required quantity, ie 452.91 cm3 (cubic centimeter). a) Calculation of time later: its calculation lies in the observation of the general graph! for the scheduling see drawing 1/6 g which reveals the path of the spots I, 7, cube 1, 9, cube 2, 5, 1 1, 7, cube 1, 9, cube 2, 5. i Q _ Time: 8 thirds / DX minimum + 8 thirds / x times cube 1 + 8 thirds +

8 / x times cube 2 + 8 thirds + 6.57 thirds + 8 thirds + 8 / x times cube 1 + 8 thirds + 8 / x times cube 1 + 8 thirds + 8 thirds / x times cube 2 + 8 thirds. b) The earlier time: this is the time calculated after completion of all the tasks becomes repetitive, repetitive time and quantity of 452.91 cm3, to calculate it you have to materialize it

4 * 5 by two graphs which illustrates the iteration of the maximum quantity (maximum system) and the iteration of the minimum quantity (stain system) see drawing 1/6 g the stains 10,7,4 to cube 1 and the spots 8,9,6 towards cube 2 towards spot 5 (spots system) injects a quantity of 284.34 cm3 in the simultaneous loop after 8 thirds and see drawing 1/6 g maximum system, spot 1 1 towards spots 10.7.4 an iteration of the maximum quantity of 33.718 cm3 every 6.57 thirds, the analysis of

I∑O two results reveal that the iterated quantity of the spots system is 284.34 in 8 thirds differed from that of the iterated quantity of the maximum system in 6.57 thirds is 33.718 cm3, the calculation of the earliest date is observed in the second iteration of the two quantities which are respectively 284.34 cm3 and 33.718 cm3, which generates the increase up to 8 thirds and the proportional quantity which corresponds to the increased time which generates a rule of three see paragraph algebraic solution (b ).

, λ 3 °) Calculation of the diameter 3.04 cm of the pipe placed in the maximum system see drawing 5/6 (No. 4001): This pipe will allow, using the reference conic, reason - 0.10, n = 7, ul = 1.20, un = 0.70 connected to it to inject the quantity of 168.57 cm3 in good time the fraction of seconds in other words to make it appear by forming a length of 5 times the maximum quantity for this it suffices to establish the following equality:

& 0 Department x Department x 3.14 x 23.09 cm = 168.57 cm3

R = 1.52 cm

Diameter = R x 2 = 3.04 cm

In conclusion of this claim is indicated in the appendices a lexicon of the different drawings, flowcharts, graphs

1000 * Opening

* Threshold = task I Input handling

* Minimum

1010 - Has the minimum been reached (test) 1010 No

1020 Task I Handling

Task input 10 1040 Threshold element processing 1070 Proportionality of revolutions processing (recording by motor)

* Operation of the counting gear (task 10)

* Functioning of the counting gear (spot 10) 1050 Test is what spot 10-7-4 are reached

1080 Yes task 10.7.4 Input handling

1110 Treatment mq Ql = Q2 = minimum

1140 l) m x towers for application A

* Operation of the application engine A units of cubic meter

2) m x towers for application B.

* Operation of the engine application B units of cubic meter.

3) m x towers for application C.

* Operation of the engine application C units of cubic meter. 1060 No vγΛection Q 1 = minimum

^ Task 10 is reached

1090 processing of the minimum 1120 Registration by the application engine C:

* Operation of the spot counting gear 10

* Lap recording 1100 Nonsense around 1130

1130 is it closing

1130 Test not around 1000

1130 Yes forward 4 towards end

1100 Yes reference 1 to 1320 _ff Yl- _'

1150 Test is the number 50,000 reached

1150 Ouivérs 1320

1160 Test yes return 5

1,160 No around 1060

1150 Yes around 1320 forward 2

1320 Handling the application gear

1330 Registration of cubic meter

1340 Handling of the cubic meter unit

1170 Is cube 1 reached

1170 No reference 20 to 1190

1190 Has the number 50,000 been reached

1,190 Yes, return to 1320

1190 No around 1220

1220 is this test close

1220 No reference 3 to 1080

1220 Yes to forward 30 to end

1170 Test yes return 21 to 1200

1200 Are spots 8-9-6 reached

1200 No around 1190 Annexa 17 - Complement of the system there side B

- 4025 Additional system maximum side B

- 4030 Stains 4-5-6-7-8-9-10-11

- 2 Line 2 or spot 5

- 3 Line 3 or spot 6

- 5 Line 5 or spot 8

- 1 Line 1 or spot 4

- 6 Line 6 or spot 6

- 7 Line 7 or spot 10

- 4 Line 4 or spot 7

- 4024 Outflow pipe 1

- 8 Line 8 or spot 11 see drawing 4001

- 4022 Maximum system counting propeller - - 4031 Co'rtp height 4.07 cm

- 4026 width plate

- 4033 height pipe plate

- 4001 Inlet flow pipe

'- 24'.7 cm in length of a pipe either the inlet or the outlet, the two ends of the meter

Annex 18 Complement of the maximum system side B

- 611 Maximum side B system supplement

- 605 Bearing

- 8 Application gears diameter 4.30 cm

- 4039 Rod of the helical gear

- 4050 3rd counting engine cylinder (above) from the gear

- 4046 Helical, clockwise. "- 4036 Tm /> gear one above the other

- 4037 Above clockwise

- 4042 Clockwise in the middle

- 4040 Below clockwise

- 4047 2nd cylinder of the clockwise counting engine

- 4035 Two parallel gears, one connected to the other, for handling the cylinder clockwise

- Rod management

- 4060 Rod

- 4045 Counting gear in the shape of a clockwise cylinder

- 4034 3rd cylinder of the engine clockwise

- 4043 rod bar (4060) and helical gear rod (4039)

- 4044 Bar fixing plate (4043)

- 4041 account engine.

Annex 4 - Flow center -25 flow center

- 14 spot 6 to place 25mm

12 cube 2 to be placed at 25 mm width to be determined 11 thickness 2mm

- 21 spot 4 to place at 12 mm 15 water outlet bore 40mm

- 13 applications

16Cube 2 width to be determined ⁷⁰ applications C spots 9 and 10

17 Cube 2 to be placed at 12 mm pipe No. l outside diameter 6 mm

18 spot 8 to place at 21 mm ² pipe n ° 2 outside diameter 6 mm 0 spot 10 to place at 6 mm ³ pipe n ° 3 outside diameter 6 mm

20 spot 7 to be placed at 9 mm 4 pipe n ° 4 outside diameter 6 mm

21 spot 5 to place 12 mm 5 pipe n ° 5 outside diameter 6 mm

68 application of stains 4 and 5 ⁶ pipe n ° 6 outside diameter 6 mm

69 applications B spots 6,7,8. 7 pipe n ° 7 outside diameter 6 mm

. inexe 10 A Flowchart explaining the Reasoning of the propeller turns

1) Quantities worked Ql = minimum = Q2 = Q3

2) Test is it a quantity above the threshold

- No around 6 or - Yes around 3

3) Test is what equal to the minimum

- No around 5 or - Yes around 4

6) An element of the threshold towards 8

8) recording proportionality of laps.

4) Quantity 1 = minimum 10) recording (Alpha tours)

5) Is x times the minimum

- Yes around 9 or - No around 7

9) recording x X Q 1 l) Display x X laps.

7) Processing x times the minimum + one element of the threshold.

12) Display x X (alpha laps) + proportionality of laps.

13) Test is closing the quantities worked

No around 6, around 7

No towards 3 or Yes towards end

Annex 19 Maximum system seen from front A

4032 maximum system seen from front A

- 4011 horizontal rod 9.20 cm in length 4010 Connection ring

4061 clockwise holding propeller 4009 clockwise tripping propeller

- 4008 Buffer 1

- 4020 Buffer 2

4007 maximum system tie 4027 Leak ring, leak seal 4006 maximum system length

- 4028 maximum system maintenance operation 4005 veltage of the 1mm trailing ring

2; line 2 or spot 5

4004 conical flow line

4003 spring

- 4002 veltages connection ring

- 4001 inlet flow pipe.

- 4024 outlet flow pipe

- 4023 outlet pipe ring outlet 1

4022 maximum system count clockwise

- 4029 Stash of pads diameter 2.50 cm

- 4021 Veltages conical pipe

- 4018 Buffer ring 2

4017 This is the distance between the maximum system and the gear of the applied actions.

- 4016 Veltage buffer ring 2

- 4015 buffer hiding cover

- 4014 buffer hiding thickness 0.20 cm

- 4013 buffer hiding diameter 2.10 + 0.40 thick

- 607 Outlet 1 pipe ring and ring seal

- 4019 Propeller support

- 610 Opening Veltage. 1210 Injection - minimum stains 8-9-6

1230 Operation * of the application A engine

* from the application B engine

* from the X Processing application engine

M X laps for application A F X laps for application A M X laps for application B F X laps for application B M X laps for application C F X laps for application D

1240 Is cube 2 reached (test)

1240 Test not around 1250

1240 Test yes around 1270

1250 Test is the number 50,000 reached (Provisional number since Alpha is not

1250 Test not around 1260 determined ⁾

1260 Test yes to forward 9

12 ^* 60 Test is it near end

1260 No around 1210

1270 Test is task 4 reached

1270 Test yes around 1280

1270 Test not around 1250

1280 Injection of minimum stains:

-10-7-4

-8-9-6

-5.

1290 M X A

F X B

M X B

F X C

M X C

M X C

1300 is it closing

1300 Test yes return 50 towards end

1300 Test not around 1310

1310 Test is the number 50,000 reached (Provisional number because alpha no

1310 Nonsense test around 1280 determined)

1350 Is the number 999 999 reached

1350 Yes Test around 1360

1360 Handling by the application gear the number 999 999,

1370 Meter processing

1380 Counter stop

1320 Handling the application gear

1330 Processing of the cylinder recording of the cubic meter

1500 Does maximum system trigger

1500 Around 1300 test not towards return 11 around 1080

1500 Test yes

1350 Test no

1510 Injection of the maximum quantity of 33,718 cm (cubic centimeter)

1520 Maximum system operation: propeller and engine rod - handling gears of the ^1st cylinder unit of the cubic meter.

1530 Registration of the unit of cubic meter

1540 Is the number 29,697.75 reached

1540 no around 1510

1540 Yes reference 10 to 1320

1 22 cm

Radius 7 cm

Volume 3384.92 cm

The last two lines express the same values rounded to one digit after the decimal point to the upper digit