BE473463A - - Google Patents

Description

       

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 EMI1.1
 



  "REVERSIBLE AND IRREVERSIBLE, MONO- OR BIHILICOIDAL GEAR WHEELS, IN BOTH CASES ENGAGING BETWEEN PARALLEL AXES ..



   The reversible wheels covered by this patent have the following characteristics:
1 The diameters of the controlled wheels are appreciably reduced compared to what they should be in relation to the gear ratios, so that., Between these gear ratios and the corresponding diameters, there is not, in a fairly large range of limiting links.



   The main object of this is to achieve high speed ratios with only two wheels, and consequently to reduce the size, weight and cost of the whole assembly, compared with the gears known hitherto.

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   2 The contact between the dantmms is extended over almost the entire surface between the arcs abe and cda in fig. 1,
It is obvious that the realization of a large contact surface as described in paragraph 2 aims to reduce the unit pressure to very low values, and consequently to obtain a high efficiency, despite the sliding existing between the teeth. as a result of the reduction in the diameters of the controlled wheels, as described in paragraph 1.



     It should be noted in the meantime that the reduction in unit pressure is very remarkable, or in relation to the gears already. known, or compared to the patented gears in.

   Italy by the plaintiff on June 6, 1929, July 7, 1930 and December 18, 1930, and in England, Germany and the United States of America - and in 14 other countries during the years 1929 to 1932, In erfet, it is said in these patents that the contact between the teeth is limited only to the leading wires of the teeth of the wheel and to the leading wires of the pinion teeth and other lines, but always to lines, never to surfaces, and consequently the practical applications made with the ideas of the said patent have never had good success thus completely frustrating all the content of paragraph 1 *, which is also included in the claims of the said patents.



   It is true that in claim 7 of the patent obtained on December 18, 1930, the case is envisaged of rounding, faceting or dulling only the edges of the heads of the teeth, in order to try to transform said lines and wires into surfaces, but it is clear that such a claim is insufficient and lacks definite indication. It is for these reasons that we have never had good practical results and that we have never been able to carry out practical applications.



   We will immediately review the process to be used to obtain the meshing between the dantmm of a pair of reversible gears according to the paragraphs oitée.



     If we arbitrarily choose the pitch diameter of a pi-

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 gnon in 32 millimeters and that we want to achieve a speed ratio of 1 to 8, according to the known theory the driven wheel should have the pitch diameter of 32 x 8 = 256. On the other hand, we reduce this diameter as follows and we establish * the following data Piston: internal diameter 136, pitch diameter 32, diameter
 EMI3.1
 be external 38, number of teeth bibélîooldaleo 3, axial pitch of the gear 19.05, axial pitch of the propellers 19.05 x 3 - 57, lys.



  Impeller: internal diameter 136, pitch diameter 144, external diameter 150, number of bi-heliooidal teeth 84th axial pitch of the gear 19.05, axial pitch of the propellers 19.05 x 84 - 45'1.3cl (see fig. 1) .



   However the inclination of the helicon of the pinion with respect to the outer circumference and to the normal plane 4 the axis is given by:
 EMI3.2
 57.15 dng. tg ..: ###### - 0.47873 258 34 521 38 x 3.1416
 EMI3.3
 and the izalination of the helios of the wheel with respect also to the outer circumference and to a plane normal to the axis, is t
 EMI3.4
 45720 Ang. t. 457 # 20 - o, 9asa 44. z.



  150 x 3.1416 The axial pitch of the gears being 19.05, the normal pitch of the
 EMI3.5
 wheel is 19.05. opes. 44 88 - 13.67.



  Having determined the elements below, we can draw fig. 8, which shows the partial plan view of two toothed teeth of the wheel, where the lines yy and xx are respectively the axes of the propellers of the pinion and the wheel, while the distance between the lines of the middle of the two teeth corresponds to the normal pitch previously established, namely 13.67.
 EMI3.6
 



  L ", 6paienou, v of the heads of the teeth will be fixed below.



  In addition, the projection of point a fig. 1 on construction line NN, zig. 8, determine points B, n, 8, R, H.



     Obviously, if the axes of the propellers had had the same tilt, points D and H would have coincided and the profiles of the teeth could have been determined according to known rules; But,

 <Desc / Clms Page number 4>

 
 EMI4.1
 given the reparquable difference in the angle of the said axes, and further expected that we want the contact between the teeth also at
 EMI4.2
 point! fig. the in the space between points.! 1: and I3 (fig. 2) will be contained half the thickness of the heads of the teeth of the two elements.



   After which, we find the value of 1 $ space between the points. !! - and D as follows
The axis of a propeller of the pinion which has its origin at d on the
 EMI4.3
 fig. 1 and in Ru in FIG. 2, by traversing the arc .aTa-nce axi aie- ment of the line BD; however (see fig. 1)
 EMI4.4
 882 + 19 - 1S2 Ang. oos 3t4? ' vs ---------------- m t'al6û 4a- 7 '49 "2 x 88 x 19 If a propeller of the pinion moves axially, during a whole
 EMI4.5
 full, 57.15, in 4Z "71 490 it advances by:
 EMI4.6
 5 ?, 15 x 42. 'l' 490 ## --.-..- .. # <. = z66. BD trait.



  360 * In a similar way the axis of a propeller of the wheel which has its
 EMI4.7
 origin in (fig. 1) or in (fig. 2), by traversing taro ba advance axially of the line BH.
 EMI4.8
 



  So: 753 + e8 "p -19 Ang. Cos & 0'0 = -" - ## - #### 0 $ 98546 - 9 "4? ' ad a x 76 x 88 If a propeller of the wheel for a full revolution advances axially
 EMI4.9
 lamant of 457e2Dp in 90 47e 8 "it advances by:
 EMI4.10
 457.20 x 90 47 '8 "------- - --------- - = 12.42 - HF3 trait.



  36 # -
 EMI4.11
 But (see fi. 3) 19.05 naked = x 0 "- ##. # - f! 953 it results from it:
 EMI4.12
 Bn = BH - nE - 120 42 - 9.5S <2.90 and nest - BD - Bn == $ 6 68 - 2.90 ** Z, 78. So the thickness of the tooth heads of the pinion should be
 EMI4.13
 larger than the thickness of the heads of the teeth of the wheel, since the first ones also do a greater job, so 3.78 + 0.23 4 - thickness of the heads of the sprocket teeth

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 normally seen at a generator.



   3.78 - 0.22 = 3.56 = thickness of the heads of the teeth of the wheel normally seen on a generator.



   And the play of the hollows of the wheel at its outer circumference, also seen normally at a generator, or according to the arrow F in fig. 2, results from 19.05 - 3.56 = 15.49, hence;
 EMI5.1
 15.49 bone L = ###, -. #. a '' "/ 4r 2
We now divide the height of the teeth of the pinion into 4 parts spaced 1.50 millimeters apart (see fig. 1 and 3) and then we lead, from the points of intersection of the arc 1, 2 and 3, as many parallels to the line co'c 'until interoepter, in frame 2, the same lines which were intercepted by line a HDS n B.



  (To avoid complicating the figures by too many lines, we have lowered only line 2, between said parallels, for the necessary demonstration of the problem for which we are looking for the solution, while, for the others, we limit ourselves only to decrying them. ).



   By examining Figures 2 and 3, we notice that, if the thickness of the heads of the teeth of the pinion is found to be 4 millimeters, at height 2 the thickness of the teeth is found to be twice the distance between the points 6 and D, fig. 2, distance which is given by the projection of point 2, fig. 1.



   Of course, the same reasoning applies to the projections of points 1, 3, b.



     We must therefore obtain the lengths of the lines BD which each belong to said projections.



   We will start with the projection of point 1
The axis of a propeller of the pinion which has its origin at d in fig. 1 and at o "in fig. 3, during the course of the arc subtended by the angle 100 ', axially advancing the line BD (see fig. 1)
 EMI5.2
 So: 888 - <- 1?, 3a-- 7S p Ang. ooa. 100 '=.-¯ -.-. ¯¯¯¯.¯¯¯¯¯¯¯¯¯¯ De78741 - 389 3rd Z10 3x88x17.50
If a propeller of the pinion, during a complete revolution, advances axially by 57.15, during 38 3 '21 it advances by:
 EMI5.3
 57.15 x 380 3 '810 # -.-.-- .. # 6 * 04 = BD tra-it.



  360. 4

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 In a similar way, the axis of a propeller of the wheel which has its
 EMI6.1
 origin in b fig. 1 and also in bone fig. 2. during the course of the arc subtended by the angle 10'0, advance axially of the line BH.
 EMI6.2
 



  : 752 + 882 - 17,502 Years. these lo 'o ###. ##### - ##' = 0, 98960 + S. 16 '11 "a x 75 x 88
If a wheel propeller, during a copplet turn. axia advance. lament of 457.20, in 8 '16' 11 "it advances by:
 EMI6.3
 457.20 x 80 16 '11 "#.-.-. ### -. # .-. # - # 10.50 m line BH.



  3600 But 15.49 SH = o "L - - ## - # -. - 7.74 a the result
 EMI6.4
 BS - BH - aH 10.50 - 7.74 276 and SD = BD - BS = 6.04 - 2.76 = 3.28
 EMI6.5
 3.28 x 2 - 6.56 * = thickness of the teeth of the P1gno the height 1 seen normally L a generator.
 EMI6.6
 Now to determine the thickness of the same teeth related 4 to the height 2, faithfully following the process applied for
 EMI6.7
 the. height 1: hatu 88 + lS - 75 Ang. c0 s. 200 'm ##. #. # .- # - # - Oj84339 - 33 * 30 "ax 88 x 16 5?, 15 x 32 * 30' - ##, -.--. # -. ### 5 ..15 - BD line.



  160.



  After:? 5 g + 882 - 162 Ang. aos 20'0 - - #######. # -. 0.99340 - S. 35 '9k 2 I 75 x 88 46'1.20 x 6 * 35' 9 "## -. ### .-. ## 8 36 1111 trait BH 360 * Corn
SH = 7.74 this results in:
 EMI6.8
 BD - BH - 8R m 8.36 - i, 74 Oe 62 et, 8D BD - BS - 5.15 - à, 62 = 4.53 4.53 x 2 = 9.06 m thickness of the teeth of the pinion at the height 2
 EMI6.9
 normally seen at a generator.

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 EMI7.1
 



  To the. dozer 3, we have: 88 + 14.50 - 758 Ang. cob 3oo '== ### -. ##.-.-. #. # oeglll4 = 849 20' 9 "2 JI; 88 14.50 51.15 x 24. 20 9.



  --------------------- 386 = BD line 5h + 88 2 - 14.502: ng. Coa 30 o "--- - - ax 7th x 88 99687 a '8 3a 4" 2 x 7 I 88 45', 8ti x 49 381 z ¯. "... ¯¯¯¯¯¯ ... .¯, ........ ¯ ... ¯., ...,. A 5.75 relates to BH 3608 but
SH = 7.74 this results in:
BS = SH - BH = 7.44 - 5.75 = 1.99 and SD = BD + BS = 3.86 + 1.99 = 5.85 5.85 x 2 = 11.70 = thickness of the teeth of the pinion at height 3 seen normally at a generator.



   And finally to 'la. height h the thickness of the teeth of the pinion, always seen normally with a generator, is o "L x 2 = 7.44 x 3 15.48
According to the dimensions marked in fig. 3, it will be noted that this figure represents the drawing, in natural size, of two consecutive denta of the pinion seen normally at one generatriae; and, from the above mathematical calculations, it follows that with the profiles of said teeth, the contact between the aro ab of the wheel and the teeth of the pinion is obtained.



   Of course it is the same for the aro bc.



   Let us now determine the figures of the spaces and the teeth of the wheel, remembering that the thickness of the head of the teeth of the pinion seen normally with a generator, is 4 millimeters.



     FIG. 4 is drawn, which is a partial reproduction of FIG. 1; we divide the height of the teeth of the wheel into 4 parts separated from each other by 1.50 mm (see fig. 4 and 6) and we determine on the arc ad the points a, 1, 2, 3 , d.



   From each of said points we lead a parallel to the line

 <Desc / Clms Page number 8>

 
 EMI8.1
 oo'R "until intercepting, in fig. 5e the line which indicates the unknown inclination of the fictitious axis of an orals of the wheel at the point of division that is envisaged; the line yy, which is l 'axis of a pinion tooth with respect to the outer circumference, the-
 EMI8.2
 which, as we have seen, is inclined by 250 341 m 5 the line AA, which represents the leading edge of a tooth of the same pinion; and
 EMI8.3
 finally the construction line NN, also determining the points e, 11, i, n. The distance between parallels AA and A'AS represents the thickness of the head of the pinion teeth.
 EMI8.4
 



  To avoid overloading the figures with too many lines, only the. first and second do. say paral.lèlea.



   In order to facilitate the understanding of the process which will be followed to establish the figure of the hollows of the wheel to obtain the
 EMI8.5
 contact all along the aro adc, we start by assigning to the line xx the inclination of the fictitious axis with respect to the outer circumference of the wheel which is, as we lt, 440 8 'In this way, we observe immediately, in fig. 5, thanks to the projection of point a fig. 4 that the distance of points 1 and e between them along said protection, is exactly equal to the
 EMI8.6
 distance 8H = o "L, corresponding L to the projection of point a of fig. 1 in fig.

   2, and which, being doubled @, corresponds to d j., As can be seen from the amplitude of the orals of the wheel, that is to say to 15.4q, with respect to the outer circumference, and always seen normal-
 EMI8.7
 lying to, a generator. The above is confirmed, as it is intuitive, by the mathematical calculations which have been omitted for abbreviation.



   In point 1, fig. 4, the amplitude of the troughs of the wheel will therefore be double ie, which is deduced from. projection of point 1. The same reasoning can obviously apply to points 2, 3, d.



   We will therefore start by deducing the length of the line ie which
 EMI8.8
 belongs b. the projection of point 1.
 EMI8.9
 



  Ang. cos. 100'e - 882 + 19 - 73a 5fi8 z80823 366 4- * 3511 Ang. these. loo * "####.-.-.-. # -.-.- .. # = 0.80833 - '36' 35" 2 x 88 x 19 57.15 x 36 4 '35 "#. ## - - .. - 5th? 2 = line 7ah 360 *

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 EMI9.1
 Ang. cos joeo 73J502 + B88 w 19 0 * 98834 451 Ang. oos lo'o = ¯ ---- ¯¯¯¯¯¯¯¯¯¯¯¯¯¯,., 0.98834 m 88 45 '37 "8 x 7350 x 88 457J20 x 8 45' 97M ¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯.¯.¯¯¯¯ 11.13 "line ne 36Q But
 EMI9.2
 ha - nd - nh - 11, la -. $ 5 7 2 - 5.40 4 hi m On R = ----- * 2
2 it follows:
 EMI9.3
 1 $ =: hi + he - 2 + 5440 - '?, 4ti' 7, 9.fi1 x 8 14.80 - amplitude of the Creux de la rousàà, the depth corresponding to point 1, seen normally at, a generator .
 EMI9.4
 



  In point 2, we have: 88 2 19 2 ... 72 2 Ang. cos. 200 '= ### -. ### -.- .. 0 * 87 350 a 290 74 573 3x88x19 57.15 x 894 71 57 "## -. ##. - .- # - #. M 4th 68 = line nh 360102 a 72 + 88 - 19 Ang. cos. 2O'O "#. #. ####### 0399171 - 70 22t 65" ax 72 x 88 457e8O x 70 22 '53u #. # -. - ####.-.-. #.-.- <{;), 37 - trait ne 3600 maie
 EMI9.5
 he = ne - nh = 9.37 - 4.63 == 4.75 hi = o "R = 2 it follows:
 EMI9.6
 ie = hi + he = 8 +4.75 - 6, '5 6 75 x 3' '13.50 - amplitude of the troughs of the wheel at the depth of point 2, seen normally at a generator.
 EMI9.7
 



  At point 3 we have 888 + 19 7f, 5f7a Ang. go $. 3oo> m #. #. ####### a. Q, 93? 43 * 20. 23 '38 "2 x 88 x 19 5'7, J.5 x 80 33' 38" -------------------- - 3sa3, = line nh J60 * Ang. cos. z 70.502 + 88a - 19a Ang. oos. 30 o - <###. # .. ### 0> 99558 - 5. 33 '19 "2 x?' C, b0 x 88 457,20 x se 23 '19" -. # -. # .- .-.-....- M se 6.84 "line no 3609

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 But he = ne - nh = 6.84 - 3.23 = 3.61 hi =. o "R = 2 it results from it: ie = hi + he = 2 + 3.61 = 5.61 5.61 x 2 = 10.22 = amplitude of the troughs of the wheel at the depth corresponding to point 3, especially seen at a generator.



   Finally at the point d the amplitude of the troughs of the wheel, always seen normally with a generator, obviously results from o "R x 2 = 2 x 2 = 4.



     Now, it should be noted that the teeth of the rouafig. 6, are drawn taking into account the dimensions in width of the corresponding recesses oi established above, and by means of these teeth it is possible to obtain contact between said detms and the arc ad freezes 1 of the pinion. It is the same, symmetrically., For the arc dc.



   It should now be noted that the profiles of the teeth of the pinion, fig. 3, also although the profiles of the teeth of the wheel, fig. 6, happen to be conoanes. It is precisely this concavity which limits the contact only between the arcs abc and oda. And it is precisely, among others, the aim of the invention, for this contact to also be extended to a very high percentage of the area between said arcs, by considerably reducing the unit pressure.



   To arrive at. this, we will follow this process we find the radius of the arc which connects the 3 points of intersection of the profile of the hollow of the wheel, fig. 6, in lines 1, 2 and 3. Using this radius, the profile of the pinion teeth is modified. fig. 3, from thickness about 9.06 up to the head ,, but so that the thickness of the latter is as great as possible, without increasing the thicknesses which are above the head up to odds 9.06. In this way, as can easily be seen, the teeth of the pinion will be in constant contact with almost the entire surface included in the rtie of the crown circilaure defined by the spokes oa and 02 fig.l, in addition to the contact

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 which remains on the arc line 3, b, c.

   Once the modification has been made, the profile will be found, as can be seen in fig. 7.



   Then the radius of the arc passing through the 3 intersection points of the profile of the pinion hollow is determined, fig. 3, with lines 1, 2 and 3. With the same radius the profile of the teeth of the wheel is modified, fig. 6, following exactly what was done previously for the teeth of the pinion, starting at dimension 5.55 approximately; and the result is that a profile is obtained as shown in FIG. 8. It is evident that, same law, the contact is extended to almost all the part of the circular crown defined by the radii o'a, 0'2 fig. 4, in addition to the contact that remains on the arc line 2, d, c. Finally, we will complete Figures 7 and 8 by thinning the edges to a suitable height in order to increase the contact surface.



   Of course, it is intuitive that, following the modifications made according to figures 7 and 8, the centers of the two gears must be brought together by a few tenths of a millimeter, and this is why the hollow has been brought to the bottom to 1 millimeter, while half would have been sufficient.



   It can be said that, thanks to said amplification of the contact surface, carried out after years of practical experience, the applicant has succeeded in constructing reducers whose very great utility for industries has been demonstrated, while being satisfied. fully respecting what has been said in paragraphs 1 and 20, at the beginning of the present description.



   Consequently, since, as we know, the teeth are seen normally at a generator, while for work it is necessary to have normal views to the respective propellers, we can add, but only as a complement (being given that this is already theoretically and practically acquired) that. in order to obtain said views normal to the helices, it suffices to multiply the ootes of figures 7 and 8 by the aosinue of the angles formed by the corresponding helioes, and to replace the values thus obtained.

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   Figure 2 shows a numerical example of how to proceed, on which .. starting from. dimension of 19.05, normal for a generator, we obtained dimension 13.67, normal for the propellers.



   DESCRIPTION 8'A COUPLE OF NON-REVERSIBLE GEARS ENGAGING BETWEEN TWO PARALLEL AXES. - The reversible wheels specified above have the following characteristics:
1 - They are generally bi-helical, but can also be simply helical.



   2 - The diameters of the driven wheels are remarkably lower than they should be for the corresponding speed ratios; but the axial pitches of the soft gears (pitch of the propellers) are equal, so that the unevenness of the angles at the top, more or less, sum the reverable wheels that we have just discussed.



   3 - The profiles of the teeth must have a distinctly trapezoidal shape, inclined as much as possible towards 45.



   4 - Contact between the teeth must be made simultaneously and continuously over the entire line.! LE of fig. 9.



   5 - The same line nr must be inclined by about 18 in relation to the plane of the axes.



   A pair of non-reversible wheels, already built in practice and satisfying all the requirements, has the following characteristics Pinion: internal diameter 32.60, pitch diameter 38, external diameter 41.60, a single bi-helical tooth and axial pitch 12 , 70.



     Impeller: internal diameter 218.60, pitch diameter 224, external diameter 227.60, number of bi-helical teeth 20, axial pitch of the propellers 12.70 x 20 + 254 = Gear ratio 1:20, precisely in relation to the number of teeth. Distance between centers: 131.



   It emerges from the above data (see fig. 9) that the height of the teeth which take part in the transmission of the work is 3.60, and that there is a clearance of 0.90 at the bottom of the oral .

 <Desc / Clms Page number 13>

 



   It follows that the inclination of the propellers of the pinion, compared to the internal diameter, increased by said clearance and with respect to a noimal plane at the axis, is given by:
 EMI13.1
 12870, Anise. t 8- - ---------------- - - 0.11751 - 60 42- '7Q 32.60 + 2 x 0.90
And the inclination of the helios $ of the wheel .. reported, the external oirconference and also with respect to. a normal plan to
 EMI13.2
 the axis, is: 354 Ang. tg. ### <. ## -.- <. ## .-. - 0 * 35583 - 190 33, ZZO as, 60 x 3.1416 -
Since the axial pitch of the gears is 12.70, the normal pitch of the wheel will be 18.70. cos. 19 33 '22 "= 11.96.



   By means of the preaitéa elements, it is possible to draw the pin 10 which represents the partial view, in plan, of two consecutive teeth of the wheel, and in which the lines yy and xx are respectively the axes of the helioes of the pinion and of the wheel, while the distance between the centers of the two teeth corresponds to the normal pitch which has been obtained before, i.e. 11.96.



   8 is now deduced the thickness of the heads of the denta; we proceed as follows:
We trace two consecutive teeth of the wheel, fig. 11, seen in elevation and normally with the propellers, taking into account what we know that the normal pitch is 11.96; since we must satisfy condition 3, the angle abc = 45.

   This will then result in the thickness of the heads of the teeth of the wheel and that of the heads of the teeth of the pinion which, as we know, penetrate into the hollows up to a depth of 3.60:
 EMI13.3
 IIP (5.60 tg. 458 X a) 11.6 - 7aO ze 36 11.96 rw. "Ww.r., .. n..m.rr.lsr.ryn, rrrr 3C m4rwrr.r-lrrlrsnwuirYi 7iT 2j3B
But, since it is clear that the thickness of the heads of the teeth of the pinion must be greater than that of the teeth of the. wheel, this thickness is reduced to 3.16 millimeters.

   Consequently, the anplitude of the troughs of the wheel at the outer circumference, and seen normally at the propellers, is:
11.96 - 2.16 = 9.80

 <Desc / Clms Page number 14>

 And the amplitude of the same hollows at the outer circumference,
 EMI14.1
 but seen normally at a generator, will be:
 EMI14.2
 9.80 # .- .. ## .-. ###. <- M 10.39 Oos. 19- 33 22 "Hence 10..39 0" L = ¯¯ - ¯¯¯a = 5j19 (see fig. 10) 3 [however, it follows from the graph that the contact line nr fig.



  1, is about 3 millimeters long, so that at point r the thickness of the teeth of the pinion is given by (see fig. 11)
 EMI14.3
 3 x tg. 450 x to 6. To this oh1tffe must be added the thickness of the heads of the teeth of the pinion, corresponding to 2.38 + (2.38 - 2.16) = 2.60.



   So, at point 1: the thickness of the teeth of the pinion is 6 + 2.60 + 8.60.
 EMI14.4
 



  By leading, in fig. 10, a parallel la, the line 1z which 8.60 is k a distance of where it-c1 equal #.-.-.- this parallel @ 2 crossed the leading edge of the teeth of the. wheel at point a which, graphically, happens to be slightly less than 4 millimeters from the plane of the axis. From said point a we take a parallel to la.
 EMI14.5
 line o'.o '0 until the points a, h, 1, .1 are determined in fig.



  10, and the point. ± on fig. su All the data included in the square parenthesis are a.pproX1ma.ti ves .. The same data was used only to establish that the point r, fi. 1a is about 4 millimeters away from the plane of the axes, as has been said above *. but this distance must be determined with mathematical accuracy, with all the other data. 0-ballast for this reason that we deduce beforehand the heights and the corresponding thicknesses that the teeth of the pinion should have to obtain the contact is to the right-
 EMI14.6
 you either at. left from point r to the under-indicated diatanoa r'itrairaa aprbamuo: L, by suitably thinning the profile, the contact is reduced to a single point.



   How this result is achieved will be explained below:

 <Desc / Clms Page number 15>

 
We must first of all specify that we are seeking to deduce the length of the line in fig. 10, which, once doubled, will give, as is clear, the thickness that the teeth of the pinion must have in order to obtain the contact at point r which results from the projection of (point a fig. 9.



   Consequently, we lead from point r, which we consider on the line co0, a perpendicular, and we have the half-chord rz. Furthermore, if we imagine that we are still lowering, still on the outer circumference of the wheel, an arbitrary number of other half-strings having their origin in ± and spaced from the plane of the axes of the following lengths: 4.60 - 4.30 - 4 - 3.70 - 3.40 - 3.10 - 2.80,
At point r, 4.60 from the axis plane, we have:

   o'z = # 113,802 - 4,602 = 113,706 oz - 131 - 113,706 - 1è, 294 o r = # 17,2942 + 4,602 17,895
By subtracting from the outer radius of the pinion the value of gold determined above, we obtain the height of the tooth of the same pinion at point r, namely: 30.80 - 17.895 = 3.905
It is obvious that the half-thickness which must exist at said point r for there to be contact corresponds to the length of the line ae FIG. 10, which we deduce as follows:
The helix of a pinion tooth which has its origin in the plane of axes 0 with radius or fig. 9, axially advancing the line he fig.



  10 during the course of the arc subtended by the angle roz.



   In a similar way, the fictitious helioe of a hollow of the wheel which also has its origin in the plane of the axes.!. with radius o'r advances awially from the hi line during the course of the ara subtended by the angle ro'z.



   However, the lengths of the lines ho and hi are determined from the
 EMI15.1
 as follows: Ang * son roz - 4.60 Ou, 25705 - 140 53 "Ang. san roz m --------- * 0.25? 05 * 14 * 56 '43' 17.895 Si, during a full turn, the propeller with a pinion length advises axially by 12.70, in 14 53 '42 ", it has:

 <Desc / Clms Page number 16>

 
 EMI16.1
 12 70 x 14 53 '42 "¯ line he 360.



  And: / 4.60 Ang. sen zo'r = ---------- -, p4a 2 * 19 '113,80 If the fictitious helioe of a hollow in the wheel advances axially
 EMI16.2
 of 254 during a turn plotted in 2. 19 'she advances by:
 EMI16.3
 254 x 20 19 '.n, "... r.¯..¯.¯..ā¯¯¯ 10634 - hi line 36Ct'" But (see mig. 10)
 EMI16.4
 ai = 0 L =, 19 (il where ah = a.1 - hi * 5.19 - 1.634 = 3.556 and a ah + h 3.556 + 4, G85 4.081 4.081 x 8 8.163 - thickness, normally seen L a genyatrioej, that the teeth of the pinion must be at a height of 2.905 to obtain the oontaot at the point ± 4.60 distant from the plane of the axes.



   In the meantime, we faithfully follow the procedure applied above., To determine the corresponding heights and thicknesses that the teeth of the pinion must have in order to obtain contact, at the other distant points, as said above, of 4.30 - 4 - 3.70 - 3.40 - 3.10 and 2.80 from the axis plane.
 EMI16.5
 



  However, in the description, the account of these determinations is omitted, since it was only a question of dreaming up the ideas already developed; we won't. that calculated them mathematics.



   Consequently: at point r, distant by 4.30 from the plane of the axes, we have:
 EMI16.6
 o'z z 13.80a - 4.302 - 113.718 oz = 131 - l13,? 18 "17.% 282 or 1l.?, 8888 + 4, 60 * 17,808 20.80 - bzz 8, 9. -¯- height of teeth.
 EMI16.7
 



  In addition: 4.30 Ang. a8ïc1 king = --- 4.30 --- = 0.241'46 - 134 58 '210,, 17.808 hence l? 0 x 58' 31 "--------------- ----- - = Op 492 trait ho 360.

 <Desc / Clms Page number 17>

 
 EMI17.1
 



  Then: don ro $ z --- 4 # 30 0.03778 ng. sen ro'z = ---------- 0.03778 = 2- 53 ", 113 * 80. This results in: 254 x 3 9 '53." ----------------- m 1,527 = hi line 3600 May
 EMI17.2
 ai - 0 "L 5.19 this results in ah - ai - hi = 5.19 - 1, 5â? 3, 663 and ae = ah t ho = 3, 663 + 0, 492 - 4.155 4.155 x; 3 == a, 310 \ III thickness, seen normally with a generator, that the teeth of the pinion must have at the height of 2,992, for
 EMI17.3
 obtain the contact at point g 4.3f from the axis plane.



  At point r, dictating 4 millimeters from the plane of the axes, we have: o'z = 1133 802 - .4a 11,? A9 oz - 131 - 113.729 = 1?, A71. or = 'V17.2712 + 42,. 17.728 20.80 17.738 "3073" height of the teeth. In addition :
 EMI17.4
 Ang. son roz - w-¯ -., .... ¯ pa563 130 3 '240 3', â8 hence 13.70 x 13. 2 '24 "-..-.-. ###. #. -.-.- 0, M0 # trait he 3609 Xnauit a Ang. Sen ro 'z ---------. 0 03514 = 80 49 "115.80 resulting in
 EMI17.5
 254 x 20 490 .... #. #., - ####. 1,420 # trait hi 3600 Maia
 EMI17.6
 aï .. 0 "t 'L - 5, 19 this results in ah = ai - hi - 5.19 - 1.420 = 3.770 and
 EMI17.7
 ae = aü + he = 3,770 = 0,460 = '4th 830 4,34 x 2 8,460 ..

   thickness, seen normally at a g6néxatricelo

 <Desc / Clms Page number 18>

 that the teeth of the pinion must have at the top = 3.072 in order to obtain contact at the point ± 4 millimeters from the plane of the axes.



   At the point r at a distance of 3.70 from the plane of the axes, we have:
 EMI18.1
 o * z = l13.80 - 1.70? m 113.73g oz - 131 - 113.739 = - 17.261 or = 17.2612 + 3070? - 17.653 20.80 - 17.653 * Z, 147 = height of the teeth.
 EMI18.2
 



  In addition: Zig. sen roz 3,70 Ouzo959 120 5311 À; g, son roz m -------- * 0.20959 12 5 '53 "17.663 d' or 70 zig * sufi 12.70 x 12.5 '53" . # -.-.-.- ## -.-.-, # .- .. n: 0.426- trait he 3600 Then: 3.70 Ang. sen ro 'z = ##. ## - =. 0,, 03251 - 11 Sl '46 "the result is llZ, 254 x 10 51' z6" - - --------------- - l 314 = line hi 360. but ai = o "L = 5.19 it results
 EMI18.3
 ah a ai - hi - 5le - 1, Z14 = 3876 and ae = ah + he = 3,876 + 0,426 - 4,302 4,302 x 2 = 8,604 = thickness, seen normally at a generator, that the teeth of the pinion must have at the height of 3,147 to obtain the contact at point r at a distance of 3.70 from the axis plane.



   At the point r at a distance of 3.40 from the plane of the axes, we have:
 EMI18.4
 o'-z :: = t l13.80 - 3.40 m llô, 749 oz 131 - 113.749 = 17.251 or V 17.7492 + 3e402 3- 17.583 20.80 - 17.582 - 3.218 = height of the teeth.
 EMI18.5
 



  In addition, Ang. sen roz = 3.40 - 0.19337 # Il 8 '5811 17.582 or 12.70 x 11. 8th 58 "------------------ - - = , 0.393 = 'line he 3600

 <Desc / Clms Page number 19>

 
 EMI19.1
 Then: S 40 Ang. sen ro 'z ---------- # 0102987 - 10 4al 42N 113.80 i it results in 254 x 19 488 480 - # .- .. ###### 1.807 - line M 360 but
 EMI19.2
 ai - 0 "L m 5.19 this results in ah ai - hi - 5.19 - 1> 807 3, 983 and ae * ah + he = 3, 983 + 0, 393 - 4.376 4.376 x 2 * 8,? 52 shoulder, seen normally with a generator, that the teeth of the pinion must be at the height 3.218 to obtain the contacts at the point ±. 3.40 distant from the plane of the centered.
 EMI19.3
 At the po int r di tart of 3.10 from the axis plane, we have:

   o'z V 113.80 -3.10 - <113.757 oz - 131 - 113.757 - 17.243 or 1:11 Ô7 B4J '+ Z, 102 m 17.519 30.80 - 17.519 "3.381 height of the teeth.



  In addition :
3.10
 EMI19.4
 Ang. sen roz - #### .-. <. 0.17695 10. '11 * 31 ", 17.519 from or
 EMI19.5
 13.70 x 10. 11 '31 "# #### -.-.-. <-. #. 0.559 line he 360.



  Then: 3.10 Ang. sen ro'z m #####. M 0.0373 = * 1 '33'39u 113.80 resulting in 254 x 1.33' 39 "#, -. # -. #. --- ..- #. = 1.101 = hi line 3608 May a
 EMI19.6
 have . 0 "L 5.19 it results
 EMI19.7
 ah = ais hi - 5, 19 - le 101 - 4.089 and ae == ah + he = 4.089 + 0.359 = 4.448

 <Desc / Clms Page number 20>

 
 EMI20.1
 4,448 x 8 = 8,896 - thickness, vbe normally at a generator, that the teeth must have at the: height 3,281 to obtain the
 EMI20.2
 contact at point r 3.10 from the axis plane.



  At the point ± 2) 80 from the plane of the axas, we have 4 'za J;. 13.802 - 380 "113.765 os 131 - 113.766 =' 17.335 00 17.838 + 2.88 17.460 30.80 - 17.460: ci: 5.340 = c height of teeth.
 EMI20.3
 



  Again: 3.80 Ang. sound xoz m ---------- * 0.16036 m 90 13th 39 "17460 from on 18>? 0 x 9 '13' 3911 ------------ ¯¯ ¯¯ - ¯M - 0..325 = line he 360.



  Then 380 Ang. sen roz -------- "0.03460 = * 10 24 '340 resulting in 115.80 254 x 10 241340 -------------- - OeOD4 - hi line 360 * But
 EMI20.4
 z1 - 0 "h 5, 3.L it results ah = ai - hi - 5.19 - 0.994 = 4,, 196 and ae = ah + he = 4.196 + 0.325 - 4.521
 EMI20.5
 4531 x 2 9.042 x thickness, seen normally with a generator, that the teeth must have at the height 3.340 to obtain the contact at the point r distant 2.80 from the plane of the axes.



   Now, we are going to draw the sketch of a tooth of the pinion, fig.
 EMI20.6
 



  12, always seen normally at a generator, taking in consideration the height and the thickness of the same tooth, with respect to the point r distant of 3.70 from the plane of the axes, so that there is contact in this point. By the way, on said figure we find the dimensions
 EMI20.7
 3,147 and 8,604 which were precluded, deduced previously at the point r distant from 3.70 of the plane of the axes, In addition, by realizing the thickness of the head of the tooth 2,32, one gives to the profile a

 <Desc / Clms Page number 21>

 inclination as such excludes contact 4 to the right and to the left of said point ±. We will explain the reason
It should be noted above all that (see fig. 12):

   
 EMI21.1
 1/2. 8604 1/2. z # zz Ang. tg. apb '<## -. # - # .-. # -.-. ### 0.99841' "440 57 '150 3.147
After which, by establishing the following table the statement results analytically, given that at distances 4.60 - 4.30 - 4 - 3.70 - 3.10 - 2.80 from the plane of the axes, we have:
A 4.60) 2.32 + (2.905 x 0.99841 x 2) = 8.120, <8.162 dei 0.042
A 4.30) "+ (2.992 x" x 2) - 8.394 <8.310 of 0.016
A 4 "" ++ (3.072 x "x 2) = 8.454 <8.460 of 0.006
A 3.70) "+ (3.147 x" x 2) = as fig.

   12
A 3.40) "" + (3.218 x "X 2) = 8.745 <8.745 <8.752 of 0.007
A 3.10) "+ (3.281 x" x 2) = 8.870 <8.896 of 0.026
A 2.80) "" + (3.340 x "x 2) = 8.988 9.042 of 0.054" + (4.50 x "x 2) = 11.30
Reading the table from bottom to top, we find in the first line the known method that we applied to establish the thickness of the tooth at the maximum height.



   On the second line, note that the tooth, at height 3.340, has a thickness of 8.988, while, to obtain the contact at the point ± 2.80 distant from the plane of the axes, this thickness. should have been 9.042, as has been shown precisely before; so there is a difference of 0.054.



   On the third line, we find that at a point ± 3.10 distant from the plane of the axes, there is no contact due to a difference of 0.036.



   On the fourth line, where the point: ± is found to be 3.40 from the plane of the axes, there is no contact either, due to a difference of 0.00 ?.



   In the fifth line, obviously, the thickness of the tooth is equal to that shown in fig. 7, and there is however contact at the point r at a distance of 3.70 from the plane of the axes.

 <Desc / Clms Page number 22>

 



   Finally. at the points indicated in lines sixth, seventh and eighth, there is no contact due to the respective differences of 0.006 - 0.016 and 0.042.



   We have therefore just demonstrated the way of establishing the shape of the teeth of the pinion, and also demonstrate that, with these teeth, contact with the head threads of the teeth of the wheel can be obtained, only at point ±. at a distance of 3.70 from the axis plane.



   Obviously, the outline of the hollows is easily established by following the procedure applied for the reversible torque pinion that has already been treated.



   However, we must first demonstrate how we establish the sketch of the teeth of the wheel e and, then, how we obtain the oontact on a whole line nr, fig. 9.



   We draw fig. 13, which is a reproduction of FIG.



   9, and we join the point n to the centers of the two gears, and from the same point we lower a perpendicular on the line 00 'which determines the half-chord nv.



   Fig. 14 is a schematic and partial plan view of FIG. 13, on which the axis yy of a tooth of the pinion is indicated, inclined according to the inclination of the helix which results in the outer circumference and the thickness of the head of the same tooth, delimited by the lines EE , E'E '(in fig. 14 this head is drawn on a double scale); the fictitious axis xx of a space of the wheel referred to the inclination of the propeller which found at the outer circumference and the amplitude of the same space delimited by the lines AA, A'A '; thus like the projection of the point n, on the axes xx "- yy, on the construction line NN, on the leading wire of the tooth of the pinion indicated by the line AA, by determining the points a i, e, h, a , B.



   Whereas contact between the teeth must be made over the whole line nr, fig. 1 ", it is first necessary to determine the amplitude that the dreux of the wheel must have to obtain the contact only at point n, and then we will see for the rest.



   It should be noted that point G, which is on the wire

 <Desc / Clms Page number 23>

 of the pinion teeth, in the plan view of FIG. 14, falls on the same leading thread at point a. So, in order for there to be contact at this point, the hollows of the wheel must have an amplitude equal to the distance between the points a and i, multiplied by 2.



   Now, we are going to recall condition 5 ,, and, by satisfying it., We will obtain as a result that point n is at. a distance of 5 millimeters from the plane of the axes. But this distance must be established with mathematical accuracy, which must therefore be obtained by first deducing the different amplitudes that the spaces should have to obtain the contact to the right and left 0 of the point n, at the arbitrary distances indicated below, and, later, by suitably tracing the sketch of the same hollows, the contact will be reduced only at point n.
 EMI23.1
 



  Let the above-mentioned distances be 5.7fi - 5.40 bzz 4.80 - 4.60 - 4> 30 - 3, 9ü.



  At point n, 5.70 from the plane of the axes, we have (see f i g. I3): ¯¯¯¯¯¯¯ # fig. 13): ov 0.80 - b, 7C38 ao, 003 o'v = 131 - 20.003 - 110.997 o'2 * llf, 997 + 5.702 = 111.143
If we subtract from rayof. outside the wheel the value of o'n which has been determined above, we obtain the depth of the hollows of the same wheel, at point n at a distance of 5.70 from the axis plane,
 EMI23.2
 c 'e st 4, -di re: 113.80 - 111 143 - 3.667
In the meantime, the helix corresponding to the outer circumference of the pinion which has its origin in the plane of the axes and which is indicated by the point O "in fig. 14, advances axially by the line he during an angular appearance equal to the subtended arc. by the angle nov.
 EMI23.3
 



  8 l'1 :: I.élioe corresponding to the radius oln of the wheel which has ega-. also its origin in the plane of the axes and in.2 !. (fig. 14), axially advancing the line lA !. during a course of the arc subtended by the angle nov.



   The lengths of the above-indicated lines of the

 <Desc / Clms Page number 24>

 
 EMI24.1
 as follows: 5.70 Years. sen nov => 0 É-- "W 27403 .. 150 bzz 15" 20180
If the helix of one tooth of the pinion advances axially by 12.70 for each complete revolution, in 15 '54' 15 "it advances by:
 EMI24.2
 1Z, 70 x 150 54 '15 ", # -. ##.-.-. ##. # -.-. # = 0.661 = line he 3600 And 5.70 Ang. Sen no'v = #. #. ##. # = 0> 05 128 m 20 56 '31 "111.143
If the helix of a tooth of the. wheel advances axially by 254 for each complete revolution, in 2 56 '21 "it advances by:
 EMI24.3
 254 x 2. 5 '210 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 2.073 = hi line 360.



  But
 EMI24.4
 ai - hi - he m 2.073 - 0.561 = 1.512 3.33 1.16 ae o "R = * ------ 1.16
2 it results
 EMI24.5
 al - ae + ei * 1.16 + 1.512 - 20672 2 $ 672 x 2 - 5.344 - amplitude, seen normally at a generator, that the spaces of the wheel must have at the. depth 2.657 p to obtain contact at point n, 5.70 from the axis plane *
At point n at a distance of 5.40 from the plane of the axes, we have:
 EMI24.6
 ov == V 20.802 - 5 $ 4th - 20.086 olv 131 - 20.086 110.914 oln - 110.9m42 + 5> 40? * 111.045 113.80 - 111, G45 - 2e755 - recess depth.
 EMI24.7
 



  In addition, S, 40 Ang. sen nov == ¯ .... ¯ ....... ¯ = '0.25961 "15. 3' 48" 20.80 aU 12,? ax 158 2 '48 "--- f" "0 j 53 Z1I treat he 360- Then;

 <Desc / Clms Page number 25>

 
 EMI25.1
 5.40 Ang. sen no 'v = --.- # ..- 0.04868 - 80 47 * 120 1114045 254 x 2. 47' 126 ##.-.-.--- #. #.! ml 966 8 # trait hi 3600 But
 EMI25.2
 ei * hi - ho m 1.966 - 0.53 w 1.436 ae 1.16 resulting in ai = ae + ei = 1.16 + 1.436 2.596
 EMI25.3
 8.596 x 2 - SJ192 = amplitude, seen normally at a generator, that the hollows of the wheel must have at a depth of 8.755 to obtain contact at point n, at a distance of 5.40 from the plane of the axes.



   At the point n at a distance of 5.10 from the plane of the axes, we have:
 EMI25.4
 ov 30.80 - 5.1t8 20.165 0'v 131 204165 ue llCs, 835 0'n "{ll0.835-t. 5.10 * 110.953 113.80 - llCt, 95a 2.848 - depth of troughs.
 EMI25.5
 



  In addition: 5.10 Ang. sen nov ### .- wt 0.34519 "14- 11 '35" or 30.80 12.70 x 14- 1P, 35 "¯, ¯¯¯¯¯¯ 360 ¯¯¯¯¯ .... ¯.¯¯¯ o 50 M trait he 360 Et: 5 ,, la Ang. Sen no'v = ## .-. ## - t <0.04596. 2- 38 '2 "110.953 it results in 254 x 2- 38 '3' - # -.- # - .., 1.858. hi line 36Ci But
 EMI25.6
 have. hi - he m 1 # 858 - 0.60 = 1.358 ae - 1, 16 resulting in
 EMI25.7
 ai - ae + e1 1.16 + 1.358 "8.518 3.518 x 2 = 5.036 -! amplitude, seen primarily at a generatrix, that the hollows of the wheel must have at the depth 3.848 to obtain the contact at point n, distant from 5.10 of the axis plane.

 <Desc / Clms Page number 26>

 
 EMI26.1
 



  At the y-point being 4.80 from the axis plane, we have 0'11 11: 1 20.238 ov = 30.80 - 4.8Q - = 30.338 o'v - 111 ... 20.238 - 110.762 dn. 1I: t V 110, 76aa- + 4 ,. 802! Id 1104865 113.80 - llÔ, 865 * 2..935: III depth of troughs.
 EMI26.2
 



  In addition: 4..t80 Ang. sound nov * '1 --- "" ----- - Oa3076 - 13 "30 * 39", 20.80 hence la10 x 138 20' 39 "# -.-. # - # .-. # ## at 0 47 = line he 360.



  Then: 4.80 Ang. sen no'v. --------- - 0.04339 2 * 28 '51 11' ?, 888 the result is: 254 x 2 38 '51 "<.-. ## <. ##. #. #. # # 1,, 75. Hi line 360.



  But
 EMI26.3
 ai i he 1.75 0; 47 - 1.88 aa 1.16
 EMI26.4
 it results
 EMI26.5
 ai 1: 1 Be + ai 1,1.6 + 1,28 - 2044 2,44 x 2 - 4,88 "amplitude, seen normally with a generator, that the hollow of the wheel must have h depth 3., 935 to obtain the contact at the point z at a distance of 4.80 from the axis plane.



  At the point n distant from, 6 of the plan of the centered ,, we have: ov *, 80a - 4,, 502 - $ 4 "ie 3 (i '0'" 131 - 20,307 - lioe663 0'ri I16, 893 + 4 $ 502 - 110.784 113.80 - 110.784 - 5.016 "'oral depth.
 EMI26.6
 



  In addition: 4) 50 Ang. sen nov - --------- - e Bi639: la- 29 '38 "\ 1 20.80 d'ou 12,? 0 x 12 29' 38" .-. # .-. # .. # g9,4 trait hie v6Ci Then: z9 5cl Ang. sen no'v = llC, 78 0 04061 = go 19 '39 "110.784

 <Desc / Clms Page number 27>

 it results from it:
 EMI27.1
 254 x 3 1 '39 "3600 1 m 1, 644 - line hi J60 Maize
 EMI27.2
 ai - hi - ho - 1.644 - ou44 1811 1.204 ae = 1.16 it results
 EMI27.3
 ai - ae + ei = 1.16 + 1.304 - 3.364 2.364 x 8 4e7Z8 = amplitude, vuexmo = alement k a generator, that the orals of the wheel must have at depth 3.016 to obtain contact at point n distant from 4, 50 of the axis plane.
 EMI27.4
 At the int n at a distance of 4.20 from the axis plane, we have: ov a0, 8 (i8 -., 3t38 = 3C, 371 o'v 131 - 30.571 "llf, 629 0'tn lla 6a98.;, 28 c 116, 7Q8 115.80 - 110, 7t5s 3.09a- depth of hollows.
 EMI27.5
 



  In addition 420 Ang. sen nov - zoj, 80 030193 - 11 38 '56 "or 80.80 l2.70 x 11 38' 56" '"########.-.- .. 0 4l - trait he 3o0 Then: 4.20 Ang. Sen noev - -.- ### .- .. #. Q, 03? 93 = 2- 10 '34 "the result is 110, 68 F9e x 29 bzz' 'au -. # ..¯ .. l 533 .. trait h,:! 360. 1 * 533 trait Maize
 EMI27.6
 41 - hi - hO -9 le 533 - 0.41 - 1.183 ae - 1.16 resulting in
 EMI27.7
 ai <ae + el - l, 16 l # lz3 - 3.383 3.383 x 2 = 4.666 "amplitude seen normally L a generator that the hollows of the wheel must have, the depth 3.092 to obtain the contast at point n, distant from 4, 20 of the axis plane.

 <Desc / Clms Page number 28>

 At point n at a distance of 3.90 from the axis plane, we have:
 EMI28.1
 olov = * 802 - 3> 902 - 20.431 0 IV on 131 - 20.431. 110.669 o'n- 'V 110.5692 + 3.90 = 110.637 113e8O - llOe637 - 3.163 - depth of troughs.
 EMI28.2
 



  In addition ! 3t90 Axag. sen nov - ##### 0.18750 DI 10. 48 '2511 2> 80 or 12.70 x 10' 48 '25 "-.-. # -.-.-.-.-. #. - # .-. #. # - 0 381 trait ho 360 'Then: 3.90 Ang. Sen no'v \ BI -.- #. ### - 0) 03525 - 2- l * 12'r 110t637 il the result is 854 x 2- 1 '18. " ## .....- .. ## .-. <. ic 1.425% rait hi 360 "But
 EMI28.3
 eî - hi - ho - 1.435 - 0.381 "1044 ae = 1.16 it results
 EMI28.4
 ai * ae + ei lj, 16 + 1044 - '3,304 2,204 x 2 - 4,408 * amplitude, seen normally at a generatrioe .. that the hollows of the wheel must have to the depth 3,163 to obtain the contact at the point n distant from 3 , 90 of the axis plane.



     We must now remember that the amplitude of the oral. wheel at the outer circumference, normally seen in a general, is 10.39 (see figs 10 and 14) and, forming the hollows of this wheel so as to obtain contact at point n at a distance of 4.80 from the plane of the axes, FIG. 15, on the.-. that the angle abc is:
 EMI28.5
 1/2. 1939 - 1/2. 4.88 Ang. tg. -...-.- ¯¯.¯¯, ¯.¯..¯¯¯¯¯., ¯. ", .. 0.93696 = 430 8 '8" 21935
By the way, in said fig. 15, we find the dimensions 2.935 and 4.88 which determine it, which were previously deduced in correspondence with the point, distant 4.80 from the plane of the axes.



   Consequently, the table below is established taking into account

 <Desc / Clms Page number 29>

 account of previously determined data relative to points n at different distances from the plane of the axes (see fig.



  13) nv = 5.70) 10.39 - (2.657 x 0.93696 x 2) = 5.403> 5.344 from 0? 059 nv = 5.40) "- (2.755 x") + 5.219> 5.192 from 0.027 nv + 5.10) "- (2.848 x" ") = 5.045> 5.036 from 0.009 nv = 4.80)" - (2.935 x "" as fig. 15 nv 4.50) "(8.016 x" ") = 4.731> 4.728 from 0.003 nv = 4.20) "- (3.092 x" ") = 4.587> 4.566 from 0.020 nv) 3.90)" - (3.163 x "") = 4.455> 4.408 from 0.047 "= (4.50 x "") = 1.948
Reading the table from bottom to top, we find that the first row was only used to derive the amplitude of the bottom.



   In the second line, we notice that at point. !! , distant from
3.90 of the axis plane, there is no contact, expected that in order to have this contact, the amplitude of the hollows of the wheel, always seen normally at a generator, should have been 4.408, as it resulted precisely from operations which have previously been carried out in relation to said point; but this amplitude is 4.455, so there is a difference of 0.047.



   In the third line, we find that, at the point n distant from
4.20 from the axis plane, there is no contact due to a difference of 0.020.



   In the fourth row, where nv = 4.50, the difference is only 0.003.



   In the fifth line, we see that at point n, distant by 4, sa from the plane of the axes, there is contact since the amplitude of the valleys at this point is la; same as that shown in fig. 15.



   And finally at points n corresponding to the sixth, to the seventh and to. the eighth row, there is no contact due to the differences which are respectively 0.009 - 0.027 and 0.059.



   It has therefore just been demonstrated how the drawing of the hollows of the wheel is established, fig. 15, seen normally at a generator, from

 <Desc / Clms Page number 30>

 one can easily deduce that of the teeth; and it has also been demonstrated analytically that, by means of this purge, one obtains contact only at the point n distant from 4.80 from the plane of the axes.



   But we already know, that there is also a contact only at the point: ± .. So, by joining the points n and r we will obtain the contact line nr which was specified at the beginning. This line of contact can be considered as straight, since, if we determine many other points of contact along said line, according to the theoretical product which has been exposed, then by joining them together, we obtain precisely a straight line as indicated in fig. 13, since the differences that exist are so small that they are invaluable, particularly for the purposes of the irreversibility which must now be demonstrated.



   The presentation will obviously not be extended to the determination of the other contact points, only in order not to overburden a simple patent application with such work.



   Likewise, to make it shorter but also as a complement, we limit ourselves to specifying that, since the teeth of the pinion, fig. 12, as well as the hollows of the. wheel, fig. 15, are seen normally at a generator, while, for manufacture, it is necessary to determine the same figures as seen normally at the respective propellers; for the pirocédé to follow, precisely to obtain these last views, it suffices to use what was said before on this subject when speaking of reversible wheels.



   DEMONSTRATION OF IRREVERSIBILITY, - By consulting the first table, we find that, to the right and to the left of the point r there is no contact due to the differences such that it is correct to think that the mean line of the contact surface is exactly 3.70 from the axis plane. And, by consulting the second table, we find that at point n, 4.50 from the axis plane, there is no contact due to a difference of 0.003 only, while at distance 5 , 10, this difference is 0.009; therefore, we can think that the mean line of the contact surface is probably at point n

 <Desc / Clms Page number 31>

 distant 4.70 from the axis plane.
 EMI31.1
 



  We can obviously obtain a more exact mathematical effect by carrying out operations on values of nv different successively from 0.3, as was done but by a smaller quantity. If the aforementioned differences were limited to 0.1, the dimensions marked in fig. 15 depending on nv, could have been determined above all from the dimensions resulting from the value of nv = 4.70.



   However, to obtain good results in practice, the method adopted is more than sufficient.
 EMI31.2
 



  Entretamps, referring to the point of oontaot r fig. 9, we already know that rz = 3e7o "og - 17.361. And, referring to the point of contact n fig. 13, if nv = 4.70, we have, approximately:
 EMI31.3
 aO307 - 30, 238 ov - 20, 838 + ###. #####, ## '- <20.361. a (the data 20.307 and 20.238 correspond to the values of ov, respectively with respect to nv = 4.50 and 4.80).



   Consequently, in the same figure 13, we see that:
 EMI31.4
 ov = oz = 30.361 - 17.361 3 nv rz = bzz 3à7O - 1.



   By tracing using these sides a right triangle, we find that the line nr is inclined 18 26 '4 "with respect to the plane of the axes.



   Have done (see fig. 16):
 EMI31.5
 Ang. tg. abc = 1: 3 = t ?, 3â333 18 '36' 4 "Of course, the length of line nr is equal to the lon-,
 EMI31.6
 heal of the hypotenuse. !!! , o est-àire: nr = ab = [l2 + 32 - 3.16
Now, since the line nr is not disposed radially, it is, between the teeth, in a plane the inclination of which is different.
 EMI31.7
 different from the inolinaisons of the profiles according to figures 12 and 15.



  Indeed, if we look at a hollow of the wheel normally at a generator, taking into account that the amplitude of this hollow at the point
 EMI31.8
 r is 10.39 and, given that we have considered nv m 4.70, the amplitude

 <Desc / Clms Page number 32>

 at point n is, very approximately, of
 EMI32.1
 4.88 - 4.75 4.88 - ###. ## -.-.-. =. 4.83 3 (Dimensions 4.88 and 4.73 correspond to the amplitudes of the trough to those of the two points respectively 4.80 and 4.50 apart from the plane of the axes. See 2 table).



   Moreover .. since the length of the line nr = 3.16, we can draw figure 17, from which it follows that
 EMI32.2
 1/2. 10.39 - 1/2. 9, $ r Ang. tg. & b4 1 / 2- 10 * 39 - 1/2. 4,83 = Oj? 8'7816 = 41 '17' 17 p. 6
So the contact surface is, between the teeth, in an inclined plane of 41 17 '17 ", and therefore its length is:
 EMI32.3
 3, 6 ### - <# .-- ,. 4.20 cos 419 1? 17 "We now draw Figure 18, which is a sohaematical and partial view, in plan of the wheel and on which the teeth are inclined according to the inclination of the propellers to the original reference, that is to say :

   
 EMI32.4
 224 x 3, 1416 Ang. tg. ### - # .-.,. r? 57. 70 * 9 'Ion 254
As it is intuitive, * fig. 19 is a reproduction of FIGS. 9 and 13.



   If we cut the gear along line nar, fig. 19, and if we look at this section normally with a generator, we obtain fig. 20, which is a partial star view in which the contact surfaces between the danta are found to be inclined, as is known, by 41 17 17 11.



   If we call P the peripheral force transmitted to the wheel (see fig. Le) the contact surfaces are pressed with force:
 EMI32.5
 QI m .w ... ¯ ... ¯¯¯¯¯.¯¯¯¯ ,. (see fig. 18) cos 700 91 low, And replacing P by 100 (kilog2ammes): 100 oj'3g952 '"' * 294.13 kilograms

 <Desc / Clms Page number 33>

 In addition :
 EMI33.1
 F = P. tg. will '9' 10 "= 100 x 2e $ 770 1 = a ''. 05 kilograms.



  That is, the axial thrust, or force normal to the fronts of the wheel, is 277.05 kilograms. In fig. 20, this force is represented by the forces S'Q 'and S'Q ". And naturally, given the inclination of the contact surfaces between the teeth,
 EMI33.2
 we have the two parallelograms: 8'I4 ',' ',' and '.,. 1'.a R ".



  Obviously, while the resultants 8 '' and 8qo are balanced, the components * 8 '' 'and 8' '"produce their effects and by the same we also obtain respectively the forces S'r' and
 EMI33.3
 S'r! '. But the line r''r is paxalléle to the contact line nsc fig. 19. Thus, in the same FIG. 19, the force S'r 'is represented by the force AÏ, and the force 88ru. equal and con-braire L 8 ';' ,, is represented by the force sT ". therefore, the peripheral force, the. measurement of the angle at the top of the wheel, and the inclination of the surfaces contact between the teeth, results in the pair of gears being subjected simultaneously (see fig. 19):
1 - the peripheral force sP which tends to rotate the wheel clockwise;

     2 - to the force sT 'which tends to the same goal; 3 - to. the force sT which tends to. stop the rotation of the gears.



   Meanwhile, in fig. 20, we have:
 EMI33.4
 B'F '= SQ "0013 41' 17 '17" "277.05 x 6,' 5141 = 2t8.17; and 8'r '# S'F' oos (90 '- 41- 17' 1?" ) * 2t78.1 '. 0013 48 # 42 '43 "= 208.17 x 0.65985 137.36 kilograms.



   Obviously, we also have S'r "= 137.36 kilograms.



   Now, in FIG. 19, we obtain the resultant a2 of the forces aP and sT.
 EMI33.5
 



  But sP = 100, eT 137, 36, therefore sR = 1,1002 x 137.362 = 169.90 kilograms. The effect of this resultant, only in relation to the wheel, -It-

 <Desc / Clms Page number 34>

 is sR x o'H.
 EMI34.1
 We can deduce the lever arm ¯o'H in the following way: Point 0 was at. half the distance between the two pointa is r.



   Lowering from the same point s the perpendicular so on the line
 EMI34.2
 ool, and remembering that in fig. 13, nv = * 4.70 and rz = 3.70,
 EMI34.3
 we can write: can er1re 470 -b m? é3 4.20 as = #. # .-. M ..- # ...., 30 2 But we can also recall that, in the same fig. 13,
 EMI34.4
 ov = 20, à6l and oz = 17.261 and therefore (see fig. 19):
 EMI34.5
 2J261 - 17.261 00 = 7.i, 261 + 20,> 261 - 17.261 = 18 .. 761 17.261 2 18.761 hence
 EMI34.6
 0'0 - 131 - 18.761 = 118.23çr o's = olo2 + e02 = V112.2392 + 412e - 112.317
 EMI34.7
 Meanwhile: sa 420 Ang. tg. ao'o = <---- * ### -.--. =. O0374a = 20 8th 35 "0 $ 0 liaez3g
 EMI34.8
 the angle Bso 'is equal to the angle ao'o, hence ang. Esoe = ang. ESB + ang. B803 = 18% 36 '4 "' 20 after 35" = 20 34 '39 "
 EMI34.9
 Then! 137.36 Ang. tg.

   RsP = ------- = 1, 37360 = 530 56 '40 "100 Given that the abgle B! SP == 900, we have.
 EMI34.10
 



  Ang. o'sH = 90 - (ang. Eao * + ang. RsP) = * 908 - (20 34 '39 "+ 53' 56 '4011)"' "1528'41" and the desired lever arm is: o 'H = o's sen 150 38' 41 "- 112.317 x 0.26689 - 29.97 Consequently, the effect of the r6atz.tanta R multiplied by said lever arm, is: sR x o'H = 169, 90 x 29.97 5091.90 kilogram-millimeters
However, the lever arm, on which the peripheral force sP is exerted, is:

   

 <Desc / Clms Page number 35>

 o'A = o's. cos. ang.so'P 112.317 x oos 20 34'39 "
 EMI35.1
 = 113.317 x 0.936afJ 105.15
But sP x o'A = 100 x 105.15 = 10515 kilogram-millimeters it follows that the force sT reduces the effect of the force
 EMI35.2
 peripheral of the product op x o'A = 10515, to the product OR x o'H = 5091,90,080st that is to say that the force OT produces a braking action of:
10515 - 5091.90 = '5423.10 kilogram-millimeters (2)
 EMI35.3
 Entretempa, the resultant aR, if we consider its direction, produces a thrust which breaks through the center of the wheel, also
 EMI35.4
 the to the intensity of the:

  force at. but, by analogy with what happens in all the other wheels meshing between axes pa-
 EMI35.5
 alleles, the sM force opposes the equal op and gong force, which tends to rotate the pinion in the opposite direction to the direction of the peripheral force. On the other hand, the force sT ', equal to the opposite direction to the force sT, tends to rotate the pinion in the direction of the peripheral force.



   Therefore, the resultant of said opposing forces, added to the force specified in formula (2), will provide the magnitude of the final braking action.
 EMI35.6
 



  Now: aF '= SE * SR oob.ang 0 "$ H - 169.90. Oos 150 2S", 41 "- 169.90 x 0.96374 = - 163.73; oD = oo' sen ang so'o 131 sound 2 8 '35 "
 EMI35.7
 v z 131 x 0.03739 = 4089 hence: OP x oD 163.73 x 4.89. 800.63 k11ograms- # 11l1meters CA) Then: As angle 860 = angle zsB * 19. 26 # 4 ", this results in:
 EMI35.8
 00 4j, zo 00 = j .. #. <.-.-. #.-...... ¯ -t Mx ....,. s; la and tg 180 86 "4" 0.33333
 EMI35.9
 00 =. '00 - 00: -' 18.761 - .1 ,, Va 6.161; hence ON. this water 189 26 '4 "= <6.161 x 0.31621 m 1.98.

 <Desc / Clms Page number 36>

 
 EMI36.1
 



  But aT '= 137j36 therefore:
 EMI36.2
 sT 'x oN = 137.36 x z 8? 1. 9i kilograms-mil1Leter (B)
By subtracting the value found in (B) from that found in (A), we have:
 EMI36.3
 800.65 - 271.97 m 528.66 kilogram-millimeters Gold: ¯¯¯¯¯ ¯¯¯ --- # -------- os = r oc 2 + SQ: = 2 18a? 61, ¯ 48 -., 88; further 528.66 19.33 = 2 ?.50 kilograms.



  So the resultant of the forces sF and à results in a force of 2? ,, '<¯5Q: iii.agrmë acting normally on the line nsr in the opposite direction of the direction of the peripheral force sP and, naturally, with o, A as leverage.
 EMI36.4
 Maïa, since o'A, = 105.15, we have - 27.50 x 105.15 - 2891.62 (3) By adding the values found in (2) and in (3), we have:
5423.10 + 2891.62 = 8314.72 (4)
 EMI36.5
 That is to say that it results definitively that the wheel is ensured% - tie simultaneously with the driving action (1) and with braking inaction (4) whose report is:

     
 EMI36.6
 831472. ### -.-.-.- <# -. = p 9 10515 So with motor inaction deriving from the peripheral effort,
 EMI36.7
 there is opposed a final action equal to 79 per cent of said motor action.



   But, given that the torque of wheels just described gave in practical tests an efficiency of 70 per cent, it is evident that 30 per cent of the effect of the peripheral force is absorbed by the resistors. passive which translate into a catchy action due to the whole set, while the 70
 EMI36.8
 The remaining percent is balanced by the 0.79 braking faction which has been determined below, with an excess of 9 percent as the safety margin.

 <Desc / Clms Page number 37>

 



   And here it is demonstrated how we verify the phenomenon of irreversibility between parallel axes.



   However, in order to reassure the Patent Examiner, it can be specified that the yield of 70 percent was noted by the Professors of the Higher School of Engineers of NAPLES, the measurements having been made in two different experimental laboratories and in using different devices and means.



   In addition, the same Examiner should be reminded that, in the patents filed several years ago by the applicant, it was also claimed irreversibility between parallel axes; but it was obtained by a process that was entirely insufficient for the practical application of the invention, since in fact no practical applications could ever be realized. For the genesis of the invention, it should therefore be noted that, several years ago, we had only vaguely seen a principle, and that it then took several years of study and experience to arrive at the con- a satisfactory clusion which has been amply demonstrated above.



   It is also obvious that if we want to transmit a power greater than that which can be transmitted by means of a single pinion, it will suffice to cause several pinions (two or three, depending on what we want) to mesh with the driven wheel. doubling or tripling the power to be transmitted), without modifying any of the components of the driven wheel.



   Naturally, these pinions will be connected to each other and to the motor shaft by means of one of the various known systems (gears, chains, trapezoidal belts) as has been shown by way of example in FIGS. 21, 22 and 23.



   In addition, it is obvious that one can easily increase the speed ratio given by a single pinion meshing with the wheel, by acting on the transmissions shown in the aforementioned drawings, fig. 21, 22 and 23.


    

Claims (1)

REV END 1 CAT ION S l - System of mono- or bi-helical, reversible and irreversible gear wheels, meshing in both cases between parallel axes, characterized in that the diameters of the driven wheels are appreciably smaller than they should be in relation to the respective ratios of speeds, the contact between the teeth waiting over almost the entire surface between the meshing surfaces. 2 - System of mono- or bi-helical gear wheels, reversible and irreversible, meshing in both cases between parallel axes, characterized in that the diameters of the driven wheels are appreciably less than they should be compared to the reports respective speeds, the contact surface between the teeth, and the profile of the teeth themselves, as well as the measurements of the angles at the top and the measurements of the diameters being such that they allow to achieve the irreversibility.