A Gear Train Is Shown in Figure 1 on the Next Page UPDATED

A Gear Train Is Shown in Figure 1 on the Next Page

Low power manual plastic gear trains: which parameters affect the subjective acoustic quality?

R. Faventi , ... M. Torrente Rodriguez , in International Gear Conference 2014: 26th–28th August 2014, Lyon, 2014

Abstract

Gear trains are often used in consumer goods, for case in the brush-bar of vacuum cleaners, and tin can generate problematic racket. The user's impression of the noise is of highest importance equally poor sound quality can reflect badly on the product. Therefore primal parameters affecting gear noise are investigated and assessed using psychoacoustic metrics. For all the gear trains tested the optimum acoustic response was attained whilst operating at their nominal heart altitude. Further, the spur gear trains were more sensitive to changes in heart distances than the helical gear set. Lastly, information technology was found that an increasing contact ratio gave a better acoustic response.

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Investigation of the effect of manufacturing errors on dynamic characteristics of herringbone planetary gear trains

F. Ren , D. Qin , in International Gear Conference 2014: 26th–28th Baronial 2014, Lyon, 2014

1 INTRODUCTION

Planetary gear railroad train (PGT) has been widely used in the transmissions of helicopters, automobiles, wind turbine, shipping engines, etc. The principal advantages of this manual are its high efficiency, its compactness, its large transmission ratio and its large power-to-weight ratio, etc. Planetary gear trains consist of either spur, single-helical or herringbone gears. Compared with single-helical gears, the herringbone gears take significant advantages of college load carrying capacity, lager total contact ratio and lower axial force. Therefore, the herringbone planetary gear train (HPGT) is too applied to heavy machinery such every bit coal cutter, aerospace engine, etc. Since the vibration and noise significantly influent the fatigue life and reliability of the gear transmission system, in order to achieve quiet and reliable transmissions, information technology is of great necessity to construct a dynamic model for the HPGT for the dynamic performance analysis.

Dynamic modeling and dynamic analysis of spur or single-helical PGT take been carried out past many researchers. J. Lin and Parker (1-2) set up a translation-torsional dynamic model for a single-stage spur planetary gear to study the free vibration. Kahraman (three) developed a single-stage helical PGT dynamic model to investigate the dynamic beliefs of a four-planet PGT organisation. Notwithstanding, the published literature on HPGT is very limited. Sondkar et al. (4) developed a linear, time-invariant dynamic model of a unmarried-phase double–helical planetary gear prepare to investigate the free and forced vibration characteristics. The model formulation adopted 6(N   +   3) DOF model of Kahraman (iii) to represent the left and right side of the double helical planetary gear separately, and the formulations for the Euler axle elements were employed to combine the left and right sides of the gear set. Nevertheless, the presented formulations of a double–helical planetary gear prepare with groove appeared too circuitous due to besides many DOFs. Bu et al. (v) developed a dynamic model for HPGT with journal bearings to simply investigate its modal properties, only they regard herringbone gears as the absence of centric force while modeling and only considered the unmarried-side for the sun and each planet while respectively establishing the equations of motion of the dominicus and each planet. In addition, there are inevitably manufacturing errors in a actual PGT, while the in a higher place literatures don't consider manufacturing errors. Chaari et al. (6), taking manufacturing errors (eccentricity and profile fault) into consideration, discussed the influence of manufacturing errors on the dynamic performance of spur PGT. Kahraman (seven), because manufacturing and assembly errors, constructed a nonlinear time-varying dynamic model of a spur PGT, and investigated effects of manufacturing and assembly variations on load sharing. However, the studies on the gratuitous and forced vibration for the PGT are mostly focused on the spur or unmarried-helical PGT, and studies of the effect of manufacturing errors on HPGT haven't been establish yet.

In this newspaper, in consideration of the structure characteristics of the herringbone gear, by applying the lumped-parameter method, a new and generalized dynamic model for the HPGT is presented to study the forced vibration characteristics. The developed model is capable of existence utilized to investigate the vibration performances of the HPGT with different types of manufacturing errors and arbitrary number of planets. The forced vibrations of an example simulation for an HPGT with 3 planets are numerically computed with the model presented, and these are analyzed respectively in time domain and frequency domain. The effects of manufacturing errors such as the sun gear eccentric mistake East _s on dynamic characteristics are discussed.

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Engineering applications of supply-demand-based optimization

Weiguo Zhao , ... Zhenxing Zhang , in New Optimization Algorithms and their Applications, 2021

7.5 Gear train design

Gear train design aims to minimize the fault betwixt the obtained gear ratio and a required gear ratio of one/6.931. The constraints are merely limits on pattern variables (side constraints). Design variables to be optimized are in a discrete form considering each gear has to have an integral number of teeth. The pattern variables T _a , T _b , T _d , and T _f represent the teeth number of four gears, respectively. The lower and upper bounds of integer design variables are 12 and 60, respectively (Sadollah et al., 2013). The gear railroad train design is shown in Fig. seven.seven. The optimization equation for this problem is given in Appendix B.ix.

Fig. 7.7. Gear train blueprint trouble.

Tabular array seven.7 shows the comparison of the best solution for SDO and other algorithms such MFO, MBA, and PSO in terms of the value of design variables and function value. It tin can be seen that, although the mistake between the obtained gear ratio and a required gear ratio provided by AEO is the same as that provided by MBA and MFO, the optimal design variables obtained are different. Therefore, the results evidence that AEO provides   a   good   culling to solve this trouble. Fig. seven.8 shows objective function value, pattern variables versus iterations provided past the SDO for gear railroad train blueprint problem.

Table 7.vii. Values of objective functions, design variables for gear railroad train design problem.

Algorithms	Optimal variables				Role value
	T a	T b	T d	T f
MFO	43	19	16	49	two.7009E   −   12
SDO	49	xvi	19	43	two.7009E   −   12
MBA	43	sixteen	19	49	2.7009E   −   12
PSO	52	15	29	58	2.35764E   −   9

Fig. 7.viii. Blueprint variables versus iterations, value of objective function.

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Gears

N. Gokarneshan , ... C.B. Senthil Kumar , in Mechanics and Calculations of Textile Machinery, 2013

2.iv Unproblematic gears

A uncomplicated gear train uses two gears, which may be of unlike sizes. If one of these gears is attached to a motor or a crank then information technology is called the driver gear. The gear that is turned by the commuter gear is called the driven gear. When a simple gear train has three meshed gears, the intermediate gear between the driver and driven gear is called an idler gear. An idler gear does not affect the gear ratio (velocity ratio) between the driver gear and the driven gear. It is mainly used to bridge the gap betwixt the driver and driven wheels which are a altitude apart and may as well be used to become aforementioned management of rotation of driven as that of commuter.

2.1. Simple gear train.

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Fundamentals of motion in textile machinery

Northward. Gokarneshan , ... C.B. Senthil Kumar , in Mechanics and Calculations of Cloth Machinery, 2013

14.ix.viii Epicyclic gear train

In an epicyclic gear train, the axes of the shafts, over which the gears are mounted, motion relative to a fixed axis. A simple epicyclic gear train is shown in Fig. xiv.15. L where a gear A and the arm C have a common axis at O_i about which they tin can rotate. The gear B meshes with gear A and has its axis on the arm O _two, almost which the gear B tin can rotate. If the arm is stock-still, the gear train is simple and gear A can drive gear B or vice versa, only if gear A is stock-still and the arm is rotated about the axis of gear A (i.e. O₁) then the gear B is forced to rotate upon and around gear A. Such a motion is chosen epicyclic and the gear trains arranged in such a mode that one or more of their members move upon and around another member are known every bit epicyclic gear trains (epi means upon and cyclic means around). The epicyclic gear trains may be uncomplicated or compound.

14.fifteen. Epicyclic gear railroad train.

The epicyclic gear trains are useful for transmitting high velocity ratios with gears of moderate size in a comparatively lesser space. They are used as differential gears in the speed frames, for driving bobbins and in the Roper positive let off in looms.

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A brief overview of design of conformal gearings: land of the art

Stephen P. Radzevich , in High-Conformal Gearing (Second Edition), 2020

2.4.27 Ability gear train with a mixed-type meshing (Zhuravlev, Thou.A., 1986)

A novel pattern of ability gear train with a mixed-type of meshing was developed by Dr. 1000.A. Zhuravlev in 1986 (filed in December 22, 1986) [46].

A power gear railroad train with a mixed type of meshing relates to mechanical engineering and tin be used when designing gear trains for transmitting a high ability. An increment the ability capacity is the primary goal of the gear train under consideration. Elimination of the geometrical stress riser at the middle of the tooth top is the manner how the goal is attained.

Meshing of ii gears in the ability gear railroad train with a mixed blazon of meshing is illustrated in Fig. 2.44.

Figure 2.44. Power gear train with a mixed type of meshing.

Source: Adapted from: Pat. 1,663,283 (USSR), Power Gear Train with a Mix-Blazon of Meshing, G.A. Zhuravlev, Int. Cl. F16h 55/08, Filed: 22.12.1986, Patented: 15.07.1991.

A power gear train with a mixed type of meshing is comprised of two gears ane and two. The portions 3-4 and 5-6 of the tooth flanks of a gear include the transient involute portions four-5, in the vicinity of the pitch point. The portions seven-8 and ix-10 of the molar flanks consist of the transient portions 8-9. The portions iv-5 and 8-9 of the tooth contour are conjugate with the addendum, and the other ane—with the dedendum of the gear molar.

A curvature of the tooth flank portions 5-6 and 9-10 at the addendum and/or dedendum portions 3-four and 7-8 are of the aforementioned sign as the cohabit transient portions iv-5 and 8-9 correspondingly.

Afterward run-in the portions of the tooth addendum (shown in dashed line) deviates from the nominal tooth profile.

When the gear railroad train is operating, the portions of the tooth profile in between points 3, 4, 5, and six interact with the tooth profile in between points seven, 8, 9, and 10 (encounter the dashed lines).

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Power manual and sizing

Richard Crowder , in Electric Drives and Electromechanical Systems (Second Edition), 2020

3.one.1 Conventional gears

In a gear train two or more than gears are fabricated to mesh with each other to transmit power from one shaft to some other. The pattern of the train used in a specific application depends upon the gear ratio required and the relative position of the axes of shafts. A number of approaches for configuring the gear train are possible, (i) simple gear train, (ii) compound gear railroad train, (iii) epicyclic gear train. In the first 2 types of gear trains, the axes of the shafts over which the gears fixed relative to each other, in the case of epicyclic gear trains, the axes of the shafts on which the gears are mounted may motion relative to a fixed axis, equally discussed in Section iii.1.ii.

Every bit discussed in Section 2.1.3 in a simple gear train their volition be a alter in the angular velocity and torque between an input and output shaft; the key speed relationship is given past,

(3.1) $\begin{array}{c}n=\pm \frac{{\omega }_i}{{\omega }_o}=\pm \frac{{\mathrm{North}}_o}{N_i}\end{array}$

where $N_i$ and ${\omega }_i$ are the number of teeth on, and the angular velocity of, the input gear, and $N_o$ and ${\omega }_o$ are the number of teeth on, and the angular velocity of, the output gear. Fig. iii.1A shows a simple gear train, where,

Fig. 3.1. Examples of the dependency of management and velocity of the output shaft on the type of gearing. (A) A elementary gear train consisting of two spur gears. (B) Uncomplicated gear train, with the addition of an idler gear.

(3.2) $\begin{array}{c}\mathrm{northward}=-\frac{N_{\mathrm{ii}}}{N_1}\end{array}$

If an idler gear is included, the gear ratio can exist calculated in an identical way, hence for an external gear train, Fig. iii.1B,

(3.3) $\begin{array}{c}n=\frac{{\omega }_i}{{\omega }_o}=\left(-\frac{N_{\mathrm{ii}}}{N_{\mathrm{i}}}\right)\left(-\frac{N_3}{{\mathrm{Due\; north}}_{\mathrm{two}}}\right)=\frac{{\mathrm{Northward}}_{\mathrm{three}}}{{\mathrm{Due\; north}}_1}\end{array}$

While the inclusion of idler gears is useful by assuasive the distance between the input and output shaft to be determined past the designer, which in a two-gear wheel railroad train is solely dependent on the size of the gears, while non changing the overall gear ratio, Eq. (3.2). In many cases the distance between the driver and the driven shafts must exist bridged over by intermediate gears, and if the gear ratio is required cannot be met with 2 gears, then compound gears on intermediate shafts can be used, as shown in (Fig. 3.two). In a compound gear, each intermediate shaft has two gears, rotating at the aforementioned speed.

Fig. 3.2. A gear train using compound gears; gear 1 is on the input shaft, gears 2 and 3 are compound gears on shaft A; gears 4 and 5 are compound gears on shaft B and the gear 6 is mounted on the output shaft.

If the input shaft speed is ${\omega }_1$ , the speeds of the subsequent shafts are given by;

(3.4) $\begin{array}{c}{\omega }_A=\frac{N_1}{N_{\mathrm{two}}}{\omega }_i\\ {\omega }_B=\frac{N_3}{{\mathrm{Due\; north}}_4}{\omega }_A\\ {\omega }_o=\frac{N_{\mathrm{five}}}{N_6}{\omega }_B\end{array}$

Hence,

(three.5) $\begin{array}{c}n=\frac{{\omega }_o}{{\omega }_i}=\frac{{\mathrm{Northward}}_2}{N_1}\frac{{\mathrm{Northward}}_4}{{\mathrm{Northward}}_3}\frac{{\mathrm{North}}_6}{N_5}\end{array}$

which can exist generalised to,

(3.half-dozen) $\begin{array}{c}n=\frac{\text{Product}\text{of}\text{the}\text{teeth}\text{numbers}\text{on}\text{the}\text{driven}\text{gears}}{\text{Product}\text{of}\text{the}\text{teeth}\text{numbers}\text{on}\text{the}\text{driver}\text{gears}}\end{array}$

The advantage of a compound railroad train over a simple gear train, is that a much larger speed reduction from the first shaft to the last shaft tin can be obtained with smaller gears. If a elementary gear train is used to give a large speed reduction, the concluding gear may accept a significant diameter compared to the input gear.

In practice the bodily gear railroad train can consist of either spur, or helical gear wheels. A spur gear (encounter Fig. 3.3A) is usually employed within conventional gear trains and has the reward of producing minimal axial forces which reduce bug connected with motility of the gear's begetting. Helical gears (see Fig. iii.3B) are widely used in robotic systems since they requite a higher contact ratio than spur gears for the same size; the penalty is an axial gear load. The limiting factors in gear transmission are the stiffness of the gear teeth, which can exist maximised past selecting the largest-diameter gear bicycle which is practical for the application, and backlash or lost motion between private gears. The net result of these problems is a loss in accuracy through the gear railroad train, which tin have an agin issue on the overall accuracy of a controlled axis.

Fig. three.3. Conventional gears. (A) Spur gear train. (B) A helical gear.

In the gear trains so far discussed, the input shaft is parallel to the output shaft, if a change of direction is required a worm or bevel gears tin can be used. Fig. 3.4A shows a worm gear. The gear ratio can be determined by because the pb of the worm, where the pb is the distance the worm moves forrad in i revolution, hence,

Fig. 3.iv. The use of a worm and worm gear allowing the input and output shaft to exist displaced by 90   degrees. (A) A worm gear arrangement. (B) Representation of a worm gear'south lead showing the relationship betwixt the diameter of the piece of work, its atomic number 82 and the atomic number 82 angle (λ).

(3.vii) $\begin{array}{c}L={\mathrm{North}}_1{p}_a\end{array}$

where $p_a$ is the axial pitch and ${\mathrm{North}}_1$ the number of teeth on the worm. If the axial pitch equals the lead, there is only one molar on the worm. If a molar of the worm is effectively unwrapped, Fig. 3.4(b), the following relationship is given,

(three.8) $\begin{array}{c}\tan \lambda =\frac{L}{\pi d_w}\end{array}$

where $\lambda$ is the lead bending and $d_1$ is the diameter of the worm. For successful performance the pitch of the worm gear should be the same every bit the helical output gear, giving a gear ratio of,

(3.9) $\begin{array}{c}\frac{{\omega }_i}{{\omega }_o}=\frac{N_{\mathrm{i}}}{N_2}\end{array}$

1 point that should exist noted is that, a worm gear organisation is reversible depending on the gear's friction, the limiting value is given by $\lambda <{\tan }^{-1}\mu$ , where μ is the coefficient of friction.

In whatsoever gear railroad train clearance is provided betwixt the mating gear teeth, to ensure that the gears mesh without binding and to provide space for a flick of lubricating oil between the teeth. Notwithstanding, this clearance will cause lost move between the input and output of the gear train, resulting in positional errors. To eliminate backlash several approaches can be employed;

•

Specially manufactured precision gear trains that minimise error and bearing play.

•

Other types of gearboxes can exist used, for example the harmonic gearboxes have backfire in the lodge of arc-minutes depending on the power transmission requirement. An culling is the cycloid gearbox which has no teeth, only rolling surfaces, giving cipher backlash.

•

The design of the gear can be modified to reduce backlash. A normally used method is to use Dissever gearing, where the gearwheel is effectively split into ii. One-half is fixed to a shaft while the other one-half is displaced by a molar, the spring results in the lost motion existence significantly reduced, providing the load limitations are not exceeded. This approach is shown in Fig. iv.15, when two resolvers are continued to increase the resolution of a measurement organization.

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Spur gears

Dan B. Marghitu , Mihai Dupac , in Auto Component Analysis with MATLAB, 2019

six.1 Introduction

The standards for gear are specified according to American Gear Manufacturers Clan (AGMA). The spur gears are used to shift the motility between parallel shafts. The gears are durable and accept very good efficiency (up to 98%). Ii gears in contact are shown in Fig. 6.1. For the gears in contact, the common normal to the surfaces at the bespeak of contact intersects the line of gear centers at the same signal P, called the pitch bespeak, equally seen in Fig. half-dozen.one. This is divers every bit the law of conjugate gear-molar action and satisfies the constant condition for the angular speed ratios. The involute curve complies with the conjugate gear-tooth action basic law. The involute of a circle (base circle) of radius $r_b$ is given by the Cartesian parametric equations

Effigy half dozen.i. Two gears in contact. From Dan B. Marghitu, Kinematic Chains and Machine Elements Pattern, Elsevier, 2005.

$\begin{array}{ccc}\hfill & \hfill \hfill & x=(\cos \theta +\theta \sin \theta )r_b,\hfill \\ \hfill & \hfill \hfill & y=(\sin \theta -\theta \cos \theta )r_b,\hfill \end{array}$

where θ is the angle in radians, or in MATLAB^®

syms rb theta

rb = 0.one;

theta = 0:pi/100:ii*pi;

%   circle of radius rb

xc = rb*cos(theta);

yc = rb*sin(theta);

% the parametric equation of the anfractuous

xi = (cos(theta) + theta.*sin(theta))*rb;

yi = (sin(theta) - theta.*cos(theta))*rb;

figure

plot(xc,yc,'-r', xi,yi,'-b','linewidth',2)

filigree on

axis equal

Fig. vi.2 shows the involute of a base circle with radius $r_b=\mathrm{0.i}$ . The pitch circles are the imaginary tangent circles of the gears in contact. The contact point for the pitch circles is the pitch point P. For two continued gears, the smaller of the ii is the pinion and the larger is the gear, while the athwart speed ratio is divers as

Figure six.ii. Involute of a base circumvolve of radius r _b  = 0.1.

(6.one) $i=\frac{{\omega }_p}{{\omega }_{\mathrm{thou}}}=-\frac{d_g}{d_p}\text{or}{i}_{12}=\frac{{\omega }_{\mathrm{ane}}}{{\omega }_{\mathrm{ii}}}=-\frac{d_{\mathrm{ii}}}{d_1},$

where d is the pitch diameter, ω represents the angular speed expressed in rad/s, and, when used, symbol northward represents the athwart speed in rpm. The minus sign in Eq. (6.i) shows that the two gears are rotating in opposite directions. The center distance between the gears defined as

(vi.2) $c=\frac{d_{\mathrm{thousand}}+d_p}{2}=r_{\mathrm{thou}}+r_p,$

where the radius of the pitch circle is calculated as $r=d/\mathrm{ii}$ . For the gears shown in Fig. vi.2, line tt is the common tangent to the pitch circles, and line nn is the common normal line to the surfaces at the contact signal. The pressure angle, ϕ, is the angle of the line nn with the line tt and has the standard value of ${20}^{\circ }$ for English language and SI units. The pressure level angle $\varphi ={25}^{\circ }$ is a standard value in the US also. The outer circle of a gear is the annex circumvolve and its radius is

$r_a=a+r,\text{or}{r}_{ap}=a+r_p\text{and}{r}_{a\mathrm{thousand}}=a+r_g,$

where a is the addendum shown in Fig. 6.3.

Effigy 6.3. Gear teeth.

The external teeth profiles are reduced from the pitch circle, a length b named dedendum. The circular pitch, p, is measured in mm (SI units) or in inches (English language units) and is given by

(vi.3) $p=\frac{\pi d}{N}\text{or}p=\frac{\pi d_p}{N_p}=\frac{\pi d_{\mathrm{thou}}}{N_m},$

where $N_p$ and $N_g$ represent the number of teeth of the pinion and gear, respectively.

For the English units the diametral pitch, $P_d$ , is

(6.4) $P_d=\frac{\mathrm{Due\; north}}{d}\text{or}{P}_d=\frac{{\mathrm{North}}_p}{d_p}=\frac{N_g}{d_g}.$

If p is in inches and $P_d$ is in teeth per inch then $p{P}_d=\pi$ .

The module, m, is used only with SI in mm and is given by

(6.5) $k=\frac{d}{N}\text{or}m=\frac{d_p}{N_p}=\frac{d_{\mathrm{grand}}}{N_g}.$

If p is in mm and m is in mm so $p/\mathrm{thou}=\pi$ .

For the full-depth involute teeth with $\varphi ={20}^{\circ }$ , the addendum standard value and the dedendum minimum value in SI units are $a=m$ and $b=\mathrm{i.25}m$ , respectively.

For English language units the annex and the minimum dedendum are [7] equally follows:

• total-depth involute with $\varphi ={14.5}^{\circ }$ : $a=\frac{1}{P_d}$ and $b=\frac{1.157}{P_d}$ ;

• full-depth anfractuous with $\varphi ={20}^{\circ }$ : $a=\frac{1}{P_d}$ and $b=\frac{1.157}{P_d}$ ;

• stub involute with $\varphi ={20}^{\circ }$ : $a=\frac{8}{10P_d}$ and $b=\frac{\mathrm{i}}{P_d}$ .

The gear interference is defined every bit the contact of parts of molar profiles that are not conjugate. This can be the example when segments of teeth that are not involute will interfere during gear meshing.

The maximum possible addendum circle radius without interference for external gearing is

(half dozen.six) $\begin{array}{cc}\hfill & \hfill r_{a(ma\mathrm{ten})}=\sqrt{r_b^{\mathrm{two}}+c^2{\sin }^2\varphi }\text{or}\hfill \\ \hfill & \hfill r_{a(max)p}=\sqrt{r_{bp}^{\mathrm{two}}+c^2{\sin }^2\varphi }\text{and}{r}_{a(\mathrm{grand}a\mathrm{ten})g}=\sqrt{r_{b\mathrm{one\; thousand}}^2+c^2{\sin }^2\varphi },\hfill \end{array}$

where $r_{bg}=r_g\cos \varphi$ and $r_{bp}=r_p\cos \varphi$ are the base circle radii of the gear and pinion, respectively. The pitch base, $p_b$ , is given by

(6.7) $p_b=\frac{\pi d_b}{N}\text{or}{p}_b=\frac{\pi d_{bp}}{N_p}=\frac{\pi d_{bk}}{{\mathrm{Due\; north}}_k}.$

A link of two or more contacting gears forms a gear railroad train. Fig. 6.4 represents a gear railroad train formed by three gears connected in series. The ratio of the gear train is

Effigy 6.4. Simple gear railroad train.

(six.8) $i_{13}=\frac{{\omega }_{\mathrm{one}}}{{\omega }_{\mathrm{three}}}=\frac{{\omega }_{\mathrm{i}}}{{\omega }_2}\frac{{\omega }_2}{{\omega }_3}=(-\frac{{\mathrm{North}}_{\mathrm{two}}}{N_1})(-\frac{N_3}{{\mathrm{North}}_2})=\frac{N_3}{{\mathrm{Due\; north}}_1}.$

The eye gear 2 is an idler and does not bear on the overall ratio.

The gears of a gear train rotate almost their ain axes. An epicyclic (or planetary) gear railroad train has gear axes that rotate also with respect to a reference frame. When a circle (planet gear) rolls without slipping on the circumference of another circumvolve (directing circle or sun gear), any point of the circumvolve describes an epicycloid. The parametric equation of the epicycloids is given by

$\begin{array}{ccc}\hfill & \hfill \hfill & \mathrm{ten}=(b+a)\cos \theta -b\cos \frac{(b+a)\theta }{b},\hfill \\ \hfill & \hfill \hfill & y=(b+a)\sin \theta -b\sin \frac{(b+a)\theta }{b}.\hfill \end{array}$

The MATLAB program for the epicycloid shown in Fig. 6.5 is given by:

Figure 6.5. Epicycloid bend.

b = 0.i;

a = 0.1;

theta = 0:pi/100:2*pi;

% circle of radius a (sun)

xa = a*cos(theta);

ya = a*sin(theta);

% parametric equation of epicycloid

xi = (b + a)*cos(theta) - b*cos((b + a)*theta./b);

yi = (b + a)*sin(theta) - b*sin((b + a)*theta./b);

effigy

plot(xa,ya,'-r', xi,yi,'-b','linewidth',2)

filigree on

axis equal

Example 6.1

A planetary gear train is shown in Fig. 6.6. The system consists of a directing/dominicus gear 1 with an angular speed of 75 rpm, two generating/planet gears two and 2′ with $N_2=\mathrm{xx}$ teeth, and a fixed ring gear 4 with ${\mathrm{Northward}}_{\mathrm{iv}}=70$ teeth. The module of the gears is $m=2$ mm. If arm iii drives a machine, determine its angular speed.

Figure 6.6. Planetary gear train.

Solution

The planet gear 2′ is a passive element. This planet gear tin can be eliminated, and the number of degrees of freedom of the machinery remains the same. To calculate the number of degrees of freedom for the planetary gear train, the passive element must be eliminated as shown in Fig. 6.seven. The number of degrees of freedom of the epicyclic gear train is calculated by

Figure 6.seven. Planetary gear train without the passive planet gear.

$G=3n-2c_5-c_4,$

where n, $c_{\mathrm{five}}$ , and $c_{\mathrm{iv}}$ represent the number of moving links, and the number of one and two degree of freedom joints, respectively. There are three moving links (gear 1, gear 2, and arm 3) continued by joints, $n=3$ . There is a single one degree of freedom joint (revolute joint) between the gear 1 and fixed frame 0 located at O. There is a revolute articulation betwixt gear 2 and arm 3 located at $O_2$ . There is a gear joint (two caste of freedom joint) between gear 1 and gear 2 at A, and between gear 2 and gear iv at B. The ring gear 4 is fixed to frame 0. There is a revolute joint between the fixed frame 0 and planet arm 3 at O. The number of degrees of freedom of the epicyclic gear train is $\mathrm{Thousand}=3\mathrm{northward}-2c_{\mathrm{iv}}-c_4=\mathrm{iii}(3)-\mathrm{ii}(3)-2=\mathrm{i}$ .

The gears will mesh at their pitch circles. The pitch circle radii of the planet and ring gear are $r_{\mathrm{ii}}=m{\mathrm{Northward}}_2/2=2(\mathrm{xx})/2=\mathrm{twenty}\text{mm}$ and $r_4=\mathrm{thousand}{N}_{\mathrm{four}}/\mathrm{ii}=2(\mathrm{lxx})/2=70\text{mm}$ . The circle radius of the lord's day gear is $r_{\mathrm{ane}}=r_4-2r_2=30\text{mm}$ .

The position vectors ${\mathbf{r}}_A$ , ${\mathbf{r}}_{O_2}$ , and ${\mathbf{r}}_B$ are

$\begin{array}{ccc}\hfill & \hfill \hfill & {\mathbf{r}}_A=x_A\mathrm{ı}+y_A\mathbf{ȷ}=r_1\mathbf{ȷ},\hfill \\ \hfill & \hfill \hfill & {\mathbf{r}}_{O_2}={\mathrm{10}}_{O_2}\mathrm{ı}+y_{O_2}\mathbf{ȷ}=(r_{\mathrm{ane}}+r_{\mathrm{ii}})\mathbf{ȷ},\hfill \\ \hfill & \hfill \hfill & {\mathbf{r}}_B=x_B\mathrm{ı}+y_B\mathbf{ȷ}=r_4\mathbf{ȷ}.\hfill \end{array}$

The MATLAB commands for position and graphic analysis are:

m = 2*10^-3; % (m)

N2 = twenty; % (teeth)

N4 = 70; % (teeth)

% pitch radii

r2 = yard*N2/two; % (m)

r4 = m*N4/2; % (thousand)

r1 = r4 - 2*r2;

% origin of the reference frame

% center of gear 1: O1=O

rO1_ = [0 0 0];

% center of gear 4: O4

rO4_ = rO1_;

% center of gear 2: O2

rO2_ = [0 r1+r2 0];

rA_ = [0 r1 0];

rB_ = [0 r4 0];

figure(1)

af = .08; axis([-af af -af af])

axis equal

axis square

grid on

hold on

xlabel('x(m)'), ylabel('y(m)'), zlabel('z(thou)')

t = linspace(0,ii*pi);

% plot gear 1 of radius r1 at O1

plot(rO1_(1)+r1*cos(t),rO1_(2)+r1*sin(t),...

'color','b','LineWidth',3)

% plot gear 2 of radius r2 at O2

plot(rO2_(1)+r2*cos(t),rO2_(2)+r2*sin(t),...

'color','r','LineWidth',three)

plot(rO4_(1)+r4*cos(t),rO4_(two)+r4*sin(t),...

'color','k','LineWidth',3)

text(rO1_(one),rO1_(2)+0.004,' O','fontsize',12)

text(rO2_(1),rO2_(2),' O_2','fontsize',12)

text(rA_(i),rA_(two)+0.004,' A','fontsize',12)

text(rB_(1),rB_(two)+0.004,' B','fontsize',12)

plot(rO1_(i),rO1_(2),'o','colour','k')

plot(rO2_(1),rO2_(ii),'o','colour','yard')

plot(rA_(1),rA_(ii),'o','color','r')

plot(rB_(1),rB_(2),'o','color','r')

line([rO1_(1),rO2_(one)],[rO1_(2),rO2_(2)],...

'LineStyle','-','LineWidth',4)

line([-af,af],[0,0],...

'LineStyle','-.','LineWidth',i.5)

line([0,0],[-af,af],...

'LineStyle','-.','LineWidth',1.five)

The velocity of point $A_1$ on the sun gear 1 is

${\mathbf{v}}_{A_{\mathrm{one}}}={\mathbf{v}}_O+{\mathit{\omega }}_{\mathrm{ane}}\times {\mathbf{r}}_A=\mathbf{0}+\left|\begin{array}{ccc}\hfill \mathrm{ı}\hfill & \hfill \mathbf{ȷ}\hfill & \hfill \mathbf{k}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill {\omega }_1\hfill \\ \hfill 0\hfill & \hfill r_{\mathrm{ane}}\hfill & \hfill 0\hfill \end{array}\right|=-{\omega }_{\mathrm{ane}}{r}_{\mathrm{ane}}\mathrm{ı}.$

The velocity of point $A_{\mathrm{ii}}$ on the planet gear two is

${\mathbf{5}}_{A_2}={\mathbf{5}}_{B_{\mathrm{two}}}+{\mathit{\omega }}_2\times {\mathbf{r}}_{BA}=\mathbf{0}+\left|\begin{array}{ccc}\hfill \mathrm{ı}\hfill & \hfill \mathbf{ȷ}\hfill & \hfill \mathbf{1000}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill {\omega }_2\hfill \\ \hfill 0\hfill & \hfill -2r_2\hfill & \hfill 0\hfill \end{array}\right|=2{\omega }_{\mathrm{ii}}{r}_2\mathrm{ı}.$

The pitch line velocities at A are equal, ${\mathbf{v}}_{A_{\mathrm{i}}}={\mathbf{5}}_{A_2}$ . The planet arm angular velocity is calculated every bit

${\omega }_{\mathrm{ii}}=-{\omega }_{\mathrm{i}}{r}_1/(2r_2).$

The velocity of centre $O_{\mathrm{ii}}$ on link 2 is

${\mathbf{5}}_{O_{\mathrm{ii}}}={\mathbf{v}}_{B_2}+{\mathit{\omega }}_2\times {\mathbf{r}}_{B{O}_{\mathrm{two}}}=\mathbf{0}+\left|\begin{array}{ccc}\hfill \mathrm{ı}\hfill & \hfill \mathbf{ȷ}\hfill & \hfill \mathbf{k}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill {\omega }_2\hfill \\ \hfill 0\hfill & \hfill -r_{\mathrm{ii}}\hfill & \hfill 0\hfill \end{array}\right|={\omega }_{\mathrm{two}}{r}_2\mathrm{ı}=-\mathrm{0.v}{\omega }_{\mathrm{i}}{r}_1\mathrm{ı}.$

The velocity of $O_{\mathrm{two}}$ on link 3 is

${\mathbf{five}}_{O_2}={\mathbf{v}}_O+{\mathit{\omega }}_3\times {\mathbf{r}}_{O_2}=\mathbf{0}+\left|\begin{array}{ccc}\hfill \mathrm{ı}\hfill & \hfill \mathbf{ȷ}\hfill & \hfill \mathbf{k}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill {\omega }_3\hfill \\ \hfill 0\hfill & \hfill r_{\mathrm{ane}}+r_2\hfill & \hfill 0\hfill \end{array}\right|=-{\omega }_{\mathrm{three}}(r_1+r_2)\mathrm{ı}.$

From the previous equations the angular velocity of arm 3 is

${\omega }_3=0.5{\omega }_1{r}_{\mathrm{i}}/(r_1+r_2)=18.850\text{rad/s}.$

The MATLAB code for the angular velocity of the planet arm is:

n1 = 75;   % (rpm)

omega1 = n1*pi/30;   % (rad/s)

omega1_ = [0 0 omega1];

vO_ = [0 0 0];

vA1_ = vO_ + cross(omega1_,rA_);

syms omega2

vB2_ = [0 0 0];

omega2_ = [0 0 omega2];

vA2_ =   vB2_ + cross(omega2_,rA_-rB_);

eqvA_ = vA1_-vA2_;

eqvA   = eqvA_(i);

omega2   = eval(solve(eqvA));

omega2_ = [0 0 omega2];

syms omega3

omega3_ = [0 0 omega3];

vO22_ = vB2_ + cross(omega2_,rO2_-rB_);

vO23_ = vO_ + cross(omega3_,rO2_);

eqvO2_ = vO22_-vO23_;

eqvO2 = eqvO2_(1);

omega3   = eval(solve(eqvO2));

omega3_ = [0 0 omega3];

The results are:

r1 =   0.030 (m)

r2 =   0.020 (grand)

r4 =   0.070 (m)

omega1= 7.854(rad/s)=75.000(rpm)

omega2=-five.890(rad/southward)=-56.250(rpm)

omega3= 2.356(rad/due south)=22.500(rpm)

Example six.2

The gears of the planetary mechanism shown in Fig. six.8 mesh at their pitch circles. The pitch radius of gear 1 (sun) is $r_{\mathrm{i}}=\mathrm{iv}$ in and the pitch radius of gear ii (planet) is $r_2=4$ in. Planet gear ii′ with the pitch radius of $r_{{\mathrm{ii}}^{\prime }}=\mathrm{eight}$ in is fixed – as planet gear two – on the same shaft. At a given instant the sunday gear i has angular velocity ${\omega }_{\mathrm{ane}}=-3$ rad/south (clockwise) and angular acceleration ${\alpha }_{\mathrm{i}}=-7$ rad/s² (clockwise) and the outer ring gear 3 has athwart velocity ${\omega }_3=\mathrm{ii}$ rad/s (counterclockwise) and angular acceleration ${\alpha }_3=5$ rad/sⁱⁱ (counterclockwise). Notice the athwart acceleration of planet gear ii at this instant.

Figure six.8. Planetary gear train.

Solution

The input data for the planetary gear train are introduced in MATLAB with the commands:

syms r1 r2 r2p

syms omega1 omega2 omega3 alpha1 alpha2 alpha3

% simbolical values

lists = {r1, r2, r2p, omega1, omega3, alpha1, alpha3};

% numerical values

listn = {four, iv, 8, -3, two, -7, v};

and the position vectors of points A, B, and C are

rA_ = [0 -r1 0];

rB_ = [0 -r1-r2 0];

rC_ = [0 -r1-r2-r2p 0];

The velocity at A is

${\mathbf{v}}_A={\mathbf{v}}_O+{\mathit{\omega }}_{\mathrm{i}}\times {\mathbf{r}}_A={\mathit{\omega }}_{\mathrm{one}}\times {\mathbf{r}}_A.$

The velocity of C on the planet gear 2′ is calculated equally

${\mathbf{v}}_{C_2}={\mathbf{five}}_A+{\mathit{\omega }}_2\times {\mathbf{r}}_{AC}.$

The velocity of C on the band gear 3 is

${\mathbf{five}}_{C_3}={\mathbf{v}}_O+{\mathit{\omega }}_{\mathrm{three}}\times {\mathbf{r}}_3.$

The angular velocity ${\mathit{\omega }}_2$ is calculated from ${\mathbf{five}}_{C_{\mathrm{ii}}}={\mathbf{v}}_{C_3}$ , or in MATLAB:

vO_ = [0 0 0];

vA_ = vO_ + cross(omega1_,rA_);

% velocity of C on planet gear 2p

vC2_ = vA_ + cantankerous(omega2_,rC_-rA_);

% velocity of C on ring gear iii

vC3_ = vO_ + cross(omega3_,rC_);

eqvC_ = vC2_-vC3_;

eqvC   = eqvC_(i);

omega2 = solve(eqvC,omega2);

omega2n = eval(subs(omega2,lists, listn));

% omega2 = -(omega1*r1 - omega3*(r1 + r2 + r2p))/(r2 + r2p)

% omega2 =   iii.667 (rad/s)

The athwart acceleration ${\mathit{\omega }}_2$ is calculated from ${\mathbf{a}}_{C_2}={\mathbf{a}}_{C_3}$ , or in MATLAB:

alpha1_ = [0 0 alpha1];

alpha2_ = [0 0 alpha2];

alpha3_ = [0 0 alpha3];

aO_ = [0 0 0];

aA_ = aO_ + cross(alpha1_,rA_)...

+cantankerous(omega1_,cross(omega1_,rA_));

aC2_ = aA_ + cross(alpha2_,rC_-rA_)...

+cantankerous(omega2_,cantankerous(omega2_,rC_-rA_));

aC3_ = aO_ + cross(alpha3_,rC_)...

+cantankerous(omega3_,cross(omega3_,rC_));

eqaC_ = aC2_-aC3_;

eqaC   = eqaC_(1);

alpha2 = solve(eqaC,alpha2);

alpha2n = eval(subs(alpha2,lists, listn));

% alpha2 = -(alpha1*r1 - alpha3*(r1 + r2 + r2p))/(r2 + r2p)

% alpha2 =   9.000 (rad/s^2)

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Solar thermal free energy

S.C. Bhatia , in Advanced Renewable Free energy Systems, 2014

Agile tracker

Agile trackers utilise motors and gear trains to direct the tracker every bit commanded by a controller responding to the solar management. In guild to control and manage the movement of these massive structures special slewing drives are designed and rigorously tested. Active two-axis trackers are also used to orient heliostats-moveable mirrors that reflect sunlight toward the absorber of a central power station. As each mirror in a large field volition accept an individual orientation these are controlled programmatically through a central estimator system, which also allows the system to be close down when necessary. Light-sensing trackers typically have 2 photosensors, such every bit photodiodes, configured differentially and so that they output a null when receiving the same light flux. Mechanically, they should exist omnidirectional (i.e. flat) and are aimed 90° autonomously.

This will cause the steepest part of their cosine transfer functions to balance at the steepest part, which translates into maximum sensitivity. Since the motors eat energy, one wants to use them just as necessary. So instead of a continuous motion, the heliostat is moved in discrete steps. Also, if the light is below some threshold at that place would non be plenty power generated to warrant reorientation. This is likewise true when there is not plenty difference in light level from one direction to another, such as when clouds are passing overhead. Consideration must be made to keep the tracker from wasting energy during cloudy periods.

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Manual gearboxes and overdrives

Heinz Heisler MSc., BSc., F.I.Thou.I., M.Southward.O.Eastward., M.I.R.T.E., M.C.I.T., One thousand.I.L.T. , in Avant-garde Vehicle Engineering science (Second Edition), 2002

3.7.1 Epicyclic overdrive gearing

Epicyclic gear train overdrives are and so arranged that the input shaft drives the pinion carrier while the output shaft is driven by the annular gear ring (Figs 3.27 and 3.28). The gear railroad train may be either of simple (unmarried stage) or compound (double phase) design and the derived formula for each arrangement is equally follows:

Fig. three.27. Elementary epicycle overdrive gear train

Fig. 3.28. Chemical compound epicycle gear train

Simple gear train (Fig. 3.27)

$\begin{array}{l}O5erdrive\mathrm{one\; thousand}earratio=\frac{A}{S+S}\\ also\text{\hspace{0.17em}A=S+2P}\\ wher\mathrm{east}A=\mathrm{northward}ugberofannulusrin\mathrm{one\; thousand}g\mathrm{east}arte\mathrm{due\; east}th\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Due south=number}of\mathrm{south}un\mathrm{yard}eart\mathrm{due\; east}\mathrm{due\; east}th\\ p_{\mathrm{southward}}=nu\mathrm{one\; thousand}b\mathrm{east}rofs\mathrm{chiliad}a\mathrm{50}lplanetg\mathrm{eastward}arteeth\\ p_L=\mathrm{northward}u\mathrm{thou}berof\mathrm{50}\arg \mathrm{eastward}plan\mathrm{due\; east}tgear\end{array}$

Compound gear train (Fig. three.28)

$\begin{array}{l}Overdri\mathrm{five}e\mathrm{thousand}\mathrm{east}arratio=\frac{p_{\mathrm{50}}(p_L+p_{\mathrm{Due\; south}}+\mathrm{Southward})}{p_{\mathrm{Fifty}}(p_L+p_{\mathrm{Due\; south}}+S)+P_{S}S}\\ a\mathrm{fifty}\mathrm{south}o\text{\hspace{0.17em}A=}{p}_L+p_S+S\\ \mathrm{due\; west}hereA=\mathrm{north}umberofannulusrin\mathrm{thou}g\mathrm{due\; east}arte\mathrm{eastward}th\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Southward=number}of\mathrm{south}ungearteeth\\ p_s=\mathrm{north}umberofsma\mathrm{fifty}\mathrm{fifty}p\mathrm{fifty}anetgearteeth\\ p_{\mathrm{Fifty}}=\mathrm{due\; north}umberof\mathrm{fifty}\arg \mathrm{eastward}planet\mathrm{thou}ear\end{array}$

The amount of overdrive (undergearing) used for cars, vans, coaches and commercial vehicles varies from as footling every bit fifteen% to as much as 45%. This corresponds to undergearing ratios of between 0.87:1 and 0.69: ane respectively. Typical overdrive ratios which accept been frequently used are 0.82:i (22%), 0.78:i (28%) and 0.75:1 (37%).

Example one An overdrive simple epicyclic gear train has sun and annulus gears with 21 and 75 teeth respectively. If the input speed from the engine drives the planet carrier at 3000 rev/min, determine

a)

the overdrive gear ratio,

b)

the number of planet gear teeth,

c)

the annulus ring and output shaft speed,

d)

the percentage of overdrive.

$a)Ov\mathrm{east}rdri5\mathrm{east}\mathrm{yard}earratio=\frac{A}{A+\mathrm{Due\; south}}=\frac{75}{75+21}=\frac{75}{96}=0.78125$

$\begin{array}{l}b)\text{\hspace{0.17em}A}=\mathrm{South}+\mathrm{two}P\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Therefor\hspace{0.17em}p=}\frac{A-S}{2}=\frac{75-21}{2}=\frac{54}{2}=27t\mathrm{eastward}eth\end{array}$

$c)\text{\hspace{0.17em}Output}\mathrm{south}pe\mathrm{east}d=\frac{3000}{0.78125}=3840rev/\min$

$d)\text{\hspace{0.17em}Persentage}ofo\mathrm{five}erdrive=\left(\frac{3840-3000}{3000}\right)100=\frac{840\times 100}{3000}=28\%$

Example ii A chemical compound epicyclic gear train overdrive has sun, small planet and big planet gears with 21, 15 and 24 teeth respectively. Determine the following if the engine drives the input planet carrier at 4000 rev/min.

a)

The overdrive gear ratio,

b)

the number of annulus ring gear teeth,

c)

the annulus band and output shaft speed,

d)

the percentage of overdrive.

$a)\text{\hspace{0.17em}Overdrive}\mathrm{1000}earratio=\frac{p_L(P_{\mathrm{Fifty}}+P_S+S)}{P_{\mathrm{Fifty}}(P_L+P_{\mathrm{South}}+S)+P_{S}S}=\frac{24(24+15+21)}{24(24+15+21)+(15\times 21)}=\frac{24\times \mathrm{lx}}{(24\times 60)+315}=\frac{1400}{1755}=0.82$

$\begin{array}{l}b){\text{\hspace{0.17em}A=P}}_{\mathrm{Fifty}}+P_{\mathrm{Due\; south}}+\mathrm{Southward}\\ =21+24+15\\ =60\text{\hspace{0.17em}teeth}\end{array}$

$c)\text{\hspace{0.17em}Output}\mathrm{due\; south}pe\mathrm{eastward}d=\frac{4000}{0.82}=4878rev/\min$

$\begin{array}{l}d)Persentageofo\mathrm{five}\mathrm{eastward}rdri5e=\frac{4878-4000}{4000}\times 100\\ =\frac{878\times 100}{4000}\\ =21.95\%\end{array}$

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