Draw Lines and Circles and Get Measurements

Radial Line

The radial lines emanating from the origin of the plot stand for loci of constant R ratio, which is the ratio of the maximum stress to the minimum stress in a cycle.

From: Air current Energy Systems , 2011

Loci applications

Colin H. Simmons , ... Neil Phelps , in Manual of Applied science Drawing (Fifth Edition), 2020

Methods of drawing an ellipse

Two-circle method

Construct two concentric circles equal in bore to the major and pocket-sized axes of the required ellipse. Let these diameters be AB and CD in Fig. 12.1.

Fig. 12.i. Two-circle structure for an ellipse.

Dissever the circles into whatever number of parts; the parts practise not necessarily take to exist equal. The radial lines now cantankerous the inner and outer circles.

Where the radial lines cross the outer circumvolve, draw short lines parallel to the small centrality CD. Where the radial lines cross the inner circle, draw lines parallel to AB to intersect with those fatigued from the outer circle. The points of intersection lie on the ellipse. Draw a smooth connecting bend.

Trammel method

Draw major and pocket-sized axes at right angles, equally shown in Fig. 12.ii.

Fig. 12.ii. Trammel method for ellipse construction.

Take a strip of paper for a trammel and marker on it one-half the major and minor axes, both measured from the aforementioned end. Let the points on the trammel be East, F, and Chiliad.

Position the trammel on the drawing and then that point F ever lies on the major axis AB and point K always lies on the minor axis CD. Mark the point E with each position of the trammel, and connect these points to give the required ellipse.

Note that this method relies on the difference betwixt one-half the lengths of the major and pocket-sized axes, and where these axes are most the same in length, it is difficult to position the trammel with a loftier degree of accurateness. The following alternative method tin be used.

Describe major and small-scale axes as before, but extend them in each direction as shown in Fig. 12.iii.

Fig. 12.iii. Culling trammel method.

Have a strip of newspaper and mark one-half of the major and minor axes in line, and let these points on the trammel be Due east, F, and G.

Position the trammel on the drawing and so that indicate Grand always moves forth the line containing CD; also, position point E along the line containing AB. For each position of the trammel, marking signal F and join these points with a smooth curve to requite the required ellipse.

To draw an ellipse using the two foci

Depict major and minor axes intersecting at point O, as shown in Fig. 12.4. Let these axes exist AB and CD. With a radius equal to half the major centrality AB, draw an arc from center C to intersect AB at points F₁ and F_ii. These two points are the foci. For any ellipse, the sum of the distances PF₁ and PF₂ is a constant, where P is any betoken on the ellipse. The sum of the distances is equal to the length of the major axis.

Fig. 12.4. Ellipse by foci method.

Separate distance OF_i into equal parts. 3 are shown here, and the points are marked G and H.

With center F₁ and radius AG, depict an arc above and below line AB.

With middle F₂ and radius BG, describe an arc to intersect the above arcs.

Repeat these ii steps past firstly taking radius AG from indicate F_ii and radius BG from F_i.

The above procedure should now be repeated using radii AH and BH. Depict a polish bend through these points to requite the ellipse.

It is ofttimes necessary to draw a tangent to a betoken on an ellipse. In Fig. 12.5 P is whatever point on the ellipse, and F₁ and F₂ are the 2 foci. Bisect angle F_one PF_two with line QPR. Erect a perpendicular to line QPR at point P, and this volition be a tangent to the ellipse at indicate P. The methods of cartoon ellipses illustrated to a higher place are all authentic. Gauge ellipses can be constructed as follows.

Fig. 12.v. Tangency to an ellipse.

Approximate method 1: Draw a rectangle with sides equal in length to the major and minor axes of the required ellipse, equally shown in Fig. 12.six.

Fig. 12.vi. Tangency to an ellipse approximate method 1.

Divide the major axis into an equal number of parts; eight parts are shown here. Divide the side of the rectangle into the same equal number of parts. Draw a line from A through point i, and let this line intersect the line joining B to point i   at the side of the rectangle as shown. Repeat for all other points in the aforementioned manner, and the resulting points of intersection will lie on the ellipse.

Judge method ii: Draw a rectangle with sides equal to the lengths of the major and minor axes, as shown in Fig. 12.7.

Fig. 12.7. Tangency to an ellipse estimate method 2.

Bisect EC to give point F. Join AF and Exist to intersect at point G. Join CG. Depict the perpendicular bisectors of lines CG and GA, and these volition intersect the center lines at points H and J.

Using radii CH and JA, the ellipse can be constructed past using 4 arcs of circles.

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Development of Patterns from Sheet Materials

Colin H. Simmons I.Eng, FIED , ... The Late Dennis E. Maguire CEng, MIMechE, Mem ASME, REng.Des, MIED , in Manual of Engineering Drawing (4th Edition), 2012

An instance of radial-line development is given in Fig. xv.8. The dimensions required to make the evolution are the circumference of the base and the slant height of the cone. The chordal distances from the plan view have been used to mark the length of arc required for the pattern; alternatively, for a higher degree of accuracy, the bending can be calculated and then sub-divided. In the front acme, lines O1 and O7 are truthful lengths, and distances OG and OA take been plotted direct onto the design. The lines O2–O6 inclusive are not true lengths, and, where these lines cross the sloping face on the elevation of the conical frustum, horizontal lines have been projected to the side of the cone and marked B, C, D, E, and F. Truthful lengths OF, OE, OD, OC, and OB are then marked on the blueprint. This process is repeated for the other half of the cone. The view on the sloping face will exist an ellipse, and the method of projection has been described in Affiliate 14.

Effigy xv.8.

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POLAR COORDINATES

D.S. DUGDALE B.Sc., Ph.D. , in Elements of Elasticity, 1968

6.6 Force at apex of wedge

Consider a sector bounded by direct radial lines making angles α with the line θ = 0, every bit shown in Fig. 6.3. As the range over which angle θ can vary is limited no stress functions need be excluded on grounds of displacements condign multi-valued. As the sector is assumed to extend to infinity in the r-direction there is no fixed dimension of length to which distances can be referred. At the tip of the wedge, concentrated forces P and Q act in the management of the positive x and y axes while moment M acts in the positive θ-direction.

FIG. 6.3. Concentrated forces at apex of wedge.

A portion of the body is isolated past cartoon an arc of some radius r. Equally it is assumed that the lateral surfaces defined by θ = α and θ = − α are stress-gratuitous this portion of the sector must be in equilibrium under the action of the concentrated forces and the stresses σ_r and σ_rθ acting on the curved purlieus. This is expressed past eqns. (6.26).

As these equilibrium equations must agree good for whatsoever value of r, the beginning two call for stress functions varying as r while the third calls for a part that is constant with respect to r. Past examining the general serial solution (6.11) and the additional functions derived in (half dozen.13) and (6.14), it will be seen that the only functions satisfying these requirements are given by eqn. (half dozen.27). By evaluating stresses using eqns. (half-dozen.four) for polar coordinates, and

(six.26) $\begin{array}{l}P+{\displaystyle \underset{-\alpha }{\overset{\alpha }{\int }}r\left({\sigma }_r\cos \theta -{\sigma }_{r\theta }\sin \theta \right)\text{d}\theta =0,}\hfill \\ Q+{\displaystyle \underset{-\alpha }{\overset{\alpha }{\int }}r\left({\sigma }_r\sin \theta +{\sigma }_{r\theta }\cos \theta \right)\text{d}\theta }=0,\hfill \\ \mathrm{Grand}+{\displaystyle \underset{-\alpha }{\overset{\alpha }{\int }}{r}^2{\sigma }_{r\theta }\text{d}\theta }=0.\hfill \end{array}\}$

(half-dozen.27) $F=Ar\theta \sin \theta +Br\theta \cos \theta +C\theta +D\sin 2\theta .$

inserting these stresses into the equilibrium eqns. (6.26), it is found that

(six.28) $\begin{array}{ll}A\hfill & =-P/2\left(\alpha +\sin 2\alpha \right),\hfill \\ B\hfill & =-Q/2\left(\alpha +\sin 2\alpha \right),\hfill \\ C\hfill & =\mathrm{two}M\cos 2\alpha /\mathrm{ii}\left(\sin 2\alpha -2\alpha \cos 2\alpha \right),\hfill \\ D\hfill & =-M/2\left(\sin 2\alpha -\mathrm{two}\alpha \cos 2\alpha \right).\hfill \end{array}\}$

Information technology can now be verified that σ_θ = σ_rθ = 0 on the lateral surfaces θ = α and θ = − α. This is not fortuitous, as any conceivable distribution of stress on these surfaces would require the resultant forces acting on the curved surface to vary with r, and this possibility was excluded. These solutions are valid for any value of semi-angle α. Very small-scale values suggest a tapered cantilever beam loaded at its gratis cease. Bending stresses may be calculated from simple axle theory and compared with those given by the nowadays analysis. When α = π/2, nosotros have a semi-infinite aeroplane loaded at a particular bespeak on its boundary. For a single normal forcefulness P, the stresses are simply expressed by

${\sigma }_r=-\left(2P/\pi r\right)\cos \theta ,{\sigma }_{\theta }={\sigma }_{r\theta }=0.$

When α = π, the body takes the form of an infinite sheet with a slit extending forth the negative x-centrality. It will be recalled that a solid canvass loaded with force P contains hoop and shear stresses given by (6.25). When a slit is made forth the negative x-axis, it can be verified that these components of stress fall to zero at all points.

It has been seen that when full-bodied forces act, it is necessary to consider equilibrium of the body equally a whole in determining stresses. Ordinarily, a biharmonic stress function gives stresses that ensure equilibrium of any element or aggregate of elements. However, the stress equilibrium equations intermission down in the firsthand vicinity of the point of application of a concentrated force, due to the stresses ceasing to be differentiable. Hence an additional equation must exist obtained for the equilibrium of a small region surrounding such a signal.

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Mixed convection heat transfer in rotating elliptic coolant channels

Olumuyiwa Ajani Lasode , in Applications of Heat, Mass and Fluid Boundary Layers, 2020

9.iv.three Zeroth-lodge solutions

The zeroth-gild equation for the streamfunction is obtained by equating coefficients $R{a}_{\tau }^0$ to cypher. Thus,

(9.47) ${\mathrm{\nabla }}^{\mathrm{iv}}{\psi }_0+\frac{1}{R}.\frac{\partial ({\psi }_0,{\mathrm{\nabla }}^2{\psi }_0)}{\partial (R,\theta )}=0.$

Therefore, the leading term of ψ is zero, that is,

(9.48) ${\Psi }_0=0,$

since at that place can be no menstruum in the (r, Θ)-airplane when $R{a}_{\tau }=0$ , due to the absence of apportionment or secondary menstruation. This corresponds to no heating and no rotation status. The basic equation for the zeroth-order axial velocity is

(9.49) ${\mathrm{\nabla }}^2{W}_0+\frac{1}{R}.\frac{\partial ({\psi }_0,W_0)}{\partial (R,\theta )}+\mathrm{iv}R{\mathrm{eastward}}_m=0.$

Substituting ${\Psi }_0=0$ into the in a higher place equation reduces it to

(nine.fifty) ${\mathrm{\nabla }}^2{\mathrm{West}}_0+\mathrm{four}R{e}_{\mathrm{yard}}=0,$

where

(9.51) ${\mathrm{\nabla }}^2=\frac{{\partial }^2}{\partial R^2}+\frac{1}{R}.\frac{\partial }{\partial R}+\frac{1}{R^2}.\frac{{\partial }^2}{\partial {\theta }^2}.$

The radial management is to be considered since the integration is performed along the radial lines for any angular position. The athwart position is fixed and frozen while integrating forth any detail radius. Then Eq. (9.51) reduces to

(9.52) ${\mathrm{\nabla }}^2=\frac{{\partial }^{\mathrm{ii}}}{\partial R^{\mathrm{ii}}}+\frac{\mathrm{ane}}{R}.\frac{\partial }{\partial R}.$

This can exist written in a more compact form as

(9.53) ${\mathrm{\nabla }}^2=\frac{\mathrm{one}}{R}\frac{\partial }{\partial R}\left(R\frac{\partial }{\partial R}\right).$

Substituting this into Eq. (9.50) and rearranging yields

(nine.54) $\frac{\mathrm{one}}{R}\frac{\partial }{\partial R}\left(R\frac{\partial W_0}{\partial R}\right)=-4R{\mathrm{eastward}}_m.$

Multiplying both sides by R and integrating yields

(ix.55) $R\frac{\partial W_0}{\partial R}=-\mathrm{ii}{R}^{\mathrm{two}}R{\mathrm{eastward}}_{\mathrm{1000}}+A_1.$

Dividing both sides of the above by R and integrating yields

(9.56) ${\mathrm{Due\; west}}_0=-\mathrm{two}{R}^{2}R{e}_m+A_{\mathrm{i}}\ln R+B_1,$

where $A_{\mathrm{i}}$ and $B_1$ are unknown constants of integration. They tin can be adamant from the purlieus conditions which are stated below:

At $R=\sqrt{\xi }$ , $W_0=0$ ;

At $R=0$ , $W_0$ is infinite.

From these equations and for the solution to be unique or valid, $A_1$ must be cipher. Upon substituting these weather into Eq. (9.56), we have

$B_1=R{e}_m\xi$

Putting the values of $A_1$ and $B_1$ into Eq. (9.56) and simplifying gives

(9.57) ${\mathrm{Due\; west}}_0=R{\mathrm{due\; east}}_{\mathrm{grand}}(\xi -R^{\mathrm{two}}).$

The equation for the zeroth-order class of the temperature distribution is retrieved every bit

(nine.58) ${\mathrm{\nabla }}^{\mathrm{two}}{\eta }_0+\frac{Pr}{R}.\frac{\partial ({\psi }_0,{\eta }_0)}{\partial (R,\theta )}+W_0=0.$

Substituting the values of ${\psi }_0\text{and}{W}_0$ into Eq. (ix.58) yields, upon rearranging,

(ix.59) ${\mathrm{\nabla }}^2{\eta }_0=-R{e}_m(\xi -R^2).$

Following the same steps for the zeroth-gild axial velocity distribution yields, after integration,

(nine.sixty) ${\eta }_0=-R{e}_{\mathrm{1000}}(\xi \frac{R^2}{4}-\frac{R^4}{16})+A_{\mathrm{two}}\ln R+B_2,$

where $A_2$ and $B_{\mathrm{two}}$ are constants of integration to be determined past considering the following boundary weather condition:

At $R=\sqrt{\xi }$ , $:{\eta }_0=0$ ,

At $R=0,{\eta }_0\mathrm{is}\mathrm{finite}$ .

Substituting the purlieus atmospheric condition into Eq. (9.60) gives $A_2=0$ , for validity of the resulting equation, and also

$B_2=\frac{\left(3R{\mathrm{due\; east}}_m{\xi }^2\right)}{16}.$

Substituting the values of $A_2$ and $B_2$ into Eq. (9.60) gives

(nine.61) ${\eta }_0=\frac{R{\mathrm{due\; east}}_{\mathrm{grand}}}{16}[3{\xi }^{\mathrm{two}}-4\xi R^2+R^4],$

or

(9.62) ${\eta }_0=\frac{R{e}_m}{16}(\xi -R^2)(3\xi -R^{\mathrm{two}}).$

Proceeding further to the first-order solutions is what we do in the following.

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Time- and Frequency-Domain Controls of Feedback Systems

Nicolae Lobontiu , in Organization Dynamics for Technology Students (Second Edition), 2018

Underdamped 2d-Social club Feedback Systems

Many command system applications involve values of the damping ratio that are less than 1, so they address the underdamped example. For first-order systems, two fourth dimension-domain specifications, the rise fourth dimension and the settling fourth dimension, have been introduced to characterize the response to unit input. For 2d-order systems, in improver to these fourth dimension specifications, two other parameters are of import, namely the peak fourth dimension and the maximum overshoot. They are best defined based on the plot of c(t), and so c(t) volition exist determined explicitly for underdamping. C(s) of Eq. (13.6) is expanded in partial fractions equally:

(13.7) $C\left(s\right)=\frac{1}{s}-\frac{s+\mathrm{\xi }·{\mathrm{\omega }}_{\mathrm{northward}}}{{\left(\mathrm{south}+\mathrm{\xi }·{\mathrm{\omega }}_{\mathrm{north}}\right)}^{\mathrm{two}}+\left(\mathrm{i}-{\mathrm{\xi }}^2\right)·{\mathrm{\omega }}_{\mathrm{north}}^2}-\frac{\mathrm{\xi }}{\sqrt{1-{\mathrm{\xi }}^2}}·\frac{\sqrt{1-{\mathrm{\xi }}^2}·{\mathrm{\omega }}_n}{{\left(s+\mathrm{\xi }·{\mathrm{\omega }}_n\right)}^{\mathrm{two}}+\left(\mathrm{ane}-{\mathrm{\xi }}^2\right)·{\mathrm{\omega }}_{\mathrm{northward}}^{\mathrm{ii}}}$

The inverse Laplace transform of C(s) of Eq. (xiii.7) is:

(13.viii) $c\left(t\right)=1-{\mathrm{eastward}}^{-\mathrm{\xi }·{\mathrm{\omega }}_n·t}·\left[\cos \left({\mathrm{\omega }}_d·t\right)+\frac{\mathrm{\xi }}{\sqrt{1-{\mathrm{\xi }}^2}}·\sin \left({\mathrm{\omega }}_{d}t\right)\right]=\mathrm{ane}-\frac{\mathrm{ane}}{\sqrt{\mathrm{ane}-{\mathrm{\xi }}^{\mathrm{two}}}}·e^{-\mathrm{\xi }·{\mathrm{\omega }}_n·t}·\sin \left({\mathrm{\omega }}_d·t+{\tan }^{-\mathrm{ane}}\frac{\sqrt{1-{\mathrm{\xi }}^{\mathrm{two}}}}{\mathrm{\xi }}\right)$

where ${\mathrm{\omega }}_d=\sqrt{1-{\mathrm{\xi }}^2}·{\mathrm{\omega }}_{\mathrm{north}}$ is the damped frequency, which is introduced in Affiliate 3. Figure xiii.3 contains several plots of c(t) that were obtained for various values of the damping ratio and for ω _n   =   100   rad/s.

Effigy thirteen.3. Response Curves c(t) for 2d-Order Control System With Underdamping and Unit-Step Input.

Information technology can exist seen that smaller amounts of damping generate larger and more sustained oscillations (eastward.g., ξ   =   0.05), whereas larger damping ratios produce vibrations that are more rapidly decaying to the input level (every bit the instance is with ξ   =   0.7). In Effigy 13.3, envelope curves are drawn that are tangent to the crests of i of the curves corresponding to ξ   =   0.05 and are defined by the equations: $f_{\mathrm{fifty}}\left(t\right)=1-{\mathrm{east}}^{-\mathrm{\xi }·{\mathrm{\omega }}_n·t}$ and $f_u\left(t\right)=1+e^{-\mathrm{\xi }·{\mathrm{\omega }}_n·t}$ , where the subscript "l" stands for lower and the subscript "u" denotes the upper curve. Both are decaying curves whose limits are 1 when time grows to infinity. A like exponentially decaying curve is the unit-footstep time response of a first-gild command system—Eq. (13.iii), where the exponential is $f\left(t\right)={\mathrm{eastward}}^{-t/\mathrm{\tau }}$ , with τ being the time abiding. If we compare f(t) with the exponential parts of f _l (t) and f _u (t), it follows that the time constant of an underdamped second-order control system under unit-step input is:

(13.ix) $\tau =\frac{1}{\mathrm{\xi }·{\mathrm{\omega }}_n}$

Large values of the damping ratio and the natural frequency effect in small time constants and therefore in fast system responses—this is seen in Figure thirteen.three and the response curves corresponding to ξ   =   0.2 and ξ   =   0.vii.

The 4 parameters that ascertain the fourth dimension response of second-order control systems under unit-footstep input are: the top time, maximum overshoot, ascension fourth dimension, and settling time, and they are divers based on Figure xiii.4.

Figure 13.four. Typical Response Curve c(t) for Second-Order Control System With Underdamping and Unit-Pace Input.

Because the response reaches the input and exceeds it, the rise time τ _r can be divers as the time the control system needs to accomplish the unit input for the first time, and in our case, the equation to decide it is c(τ _r )   =   1. By using this condition with c(t) of Eq. (13.8) yields:

(xiii.10) ${\mathrm{east}}^{-\mathrm{\xi }·{\mathrm{\omega }}_n·{\mathrm{\tau }}_r}·\left[\cos \left({\mathrm{\omega }}_d·{\mathrm{\tau }}_r\right)+\frac{\mathrm{\xi }}{\sqrt{1-{\mathrm{\xi }}^2}}·\sin \left({\mathrm{\omega }}_d·{\mathrm{\tau }}_r\right)\right]=0\mathrm{or}{\mathrm{\tau }}_r=\frac{{\tan }^{-\mathrm{i}}\left(\frac{\sqrt{1-{\mathrm{\xi }}^{\mathrm{two}}}}{\mathrm{\xi }}\right)}{\sqrt{\mathrm{one}-{\mathrm{\xi }}^{\mathrm{two}}}·{\mathrm{\omega }}_n}$

The ascension fourth dimension results from annulling the bracket of Eq. (xiii.10) and when the minus sign is neglected.

The settling time τ _southward is the time afterward which the response reaches and remains within ii% of the input. Mathematically, this requirement is: $\left|c\left({\mathrm{\tau }}_s\right)-\mathrm{ane}\right|\le 0.02$ ; in conjunction with Eq. (13.8), this results in:

(thirteen.11) $\frac{1}{\sqrt{\mathrm{one}-{\mathrm{\xi }}^2}}·e^{-\mathrm{\xi }·{\mathrm{\omega }}_{\mathrm{north}}·{\mathrm{\tau }}_s}·\left|\sin \left({\mathrm{\omega }}_d·{\mathrm{\tau }}_s+{\tan }^{-\mathrm{i}}\frac{\sqrt{1-{\mathrm{\xi }}^2}}{\mathrm{\xi }}\right)\right|\le \frac{\mathrm{ane}}{\sqrt{\mathrm{ane}-{\mathrm{\xi }}^{\mathrm{ii}}}}·{\mathrm{east}}^{-\mathrm{\xi }·{\mathrm{\omega }}_n·{\mathrm{\tau }}_s}\le 0.02$

which, at limit, means solving the equation:

(13.12) $\frac{1}{\sqrt{\mathrm{ane}-{\mathrm{\xi }}^2}}{e}^{-\mathrm{\xi }·{\mathrm{\omega }}_n·{\mathrm{\tau }}_s}=0.02\mathrm{or}{\mathrm{\tau }}_{\mathrm{south}}=-\frac{\ln \left(0.02\sqrt{1-{\mathrm{\xi }}^{\mathrm{ii}}}\right)}{\mathrm{\xi }·{\mathrm{\omega }}_n}$

An accurate approximation of the settling time given in Eq. (13.12) is

(13.13) ${\mathrm{\tau }}_{\mathrm{due\; south}}=\frac{4}{\mathrm{\xi }·{\mathrm{\omega }}_{\mathrm{northward}}}=\mathrm{iv}\mathrm{\tau }$

where τ is the time constant of Eq. (13.9).

The typical response curve that is plotted in Figure 13.4 displays a superlative value, which defines the maximum overshoot (c _max  −   ane). The time respective to this maximum response is known equally peak time, and information technology can be computed by solving the equation: ${dc\left(t\right)/dt|}_{t={\mathrm{\tau }}_p}=0$ . This condition is combined with the offset form of Eq. (thirteen.8) and the result is:

(13.14) $\frac{{\mathrm{\omega }}_n}{\sqrt{1-{\mathrm{\xi }}^{\mathrm{ii}}}}·e^{-\mathrm{\xi }·{\mathrm{\omega }}_{\mathrm{northward}}·{\mathrm{\tau }}_p}\sin \left({\mathrm{\omega }}_d·{\mathrm{\tau }}_p\right)=0\mathrm{or}{\mathrm{\omega }}_d·{\mathrm{\tau }}_p=\mathrm{\pi }\mathrm{or}{\mathrm{\tau }}_p=\frac{\mathrm{\pi }}{{\mathrm{\omega }}_d}=\frac{\mathrm{\pi }}{\sqrt{1-{\mathrm{\xi }}^2}·{\mathrm{\omega }}_{\mathrm{due\; north}}}$

For an input different from unity, the maximum (peak) overshoot is divers as:

(13.fifteen) $O_p=\frac{c_{\max }-c\left(\infty \right)}{c\left(\infty \right)}$

where c(∞) is the steady-country response (that is equal to the input in the case nosotros clarify). For the physical situation where the input is a unit of measurement footstep and where c(∞)   →   1, the maximum overshoot becomes:

(13.16) $O_p=c_{\max }-\mathrm{one}=e^{-\frac{\mathrm{\pi }·\mathrm{\xi }}{\sqrt{1-{\mathrm{\xi }}^{\mathrm{two}}}}}$

which resulted past substituting the peak time of Eq. (xiii.14) into c(t) of the get-go-form Eq. (13.viii). While the tiptop time depends on both the natural frequency and the damping ratio, the maximum overshoot is only dependent on damping ratio.

The time specifications defined herein can exist connected by ways of the damping ratio and the natural frequency to the pole position of a 2nd-lodge, underdamped unity-feedback control system with unit of measurement-step input. Figure 13.5 shows the position of one of the 2 cohabit closed-loop poles respective to underdamping.

Figure 13.5. Airtight-Loop Complex Aeroplane Pole Position for Underdamped Unity-Feedback Control Organization With Unit of measurement-Step Input.

The underdamped pole can exist expressed in complex number grade as p  =   σ _d   +   ω _d ·j where σ _d is the existent office and ω _d is the imaginary part, as shown in Figure xiii.v; the pole magnitude and its direction are:

(13.17) $\left|p\right|=OP=\sqrt{{\mathrm{\sigma }}_d^2+{\mathrm{\omega }}_d^2}={\mathrm{\omega }}_n;\left|\cos \mathrm{\alpha }\right|=\frac{{\mathrm{\sigma }}_d}{{\mathrm{\omega }}_n}=\mathrm{\xi }$

This effect indicates that if the pole moves around a round trajectory (centered at O ), the natural frequency of the system remains unchanged. It also shows that when the pole moves radially along a line of constant inclination, the damping ratio is unaltered. Clockwise rotation of the radial line OP (which increases the angle α) reduces the damping ratio and conversely, counterclockwise rotation increases the damping result.

Because σ _d   =   ξ·ω _north , the settling time of Eq. (13.13) can be written equally ${\mathrm{\tau }}_{\mathrm{southward}}=4/{\mathrm{\sigma }}_d$ , which shows that the settling time volition remain unmodified every bit long equally the pole moves up or downwards on a specified vertical line in the circuitous airplane. Moving the vertical line to the left in Effigy 13.5 increases the abscissa σ _d and thus the settling fourth dimension decreases. Eq. (13.14) demonstrates that the peak fourth dimension is inversely proportional to the damped frequency ω _d , which shows that when the pole moves on a specified horizontal line (to the left or to the right in Figure 13.five), the superlative time remains abiding. To reduce the peak time, a horizontal line has to be displaced upward in the complex plane, which increases the coordinate ω _d .

If a value of ω _northward   =   1   rad/s is used in Eq. (13.10), the rise time becomes a role of simply the damping ratio—this function is plotted in Figure 13.6(a).

Figure 13.6. (a) Rise Time Versus Damping Ratio; (b) Maximum Overshoot Versus Damping Ratio.

It can be seen that the rise time τ _r decreases when the damping ratio increases. Also, when a radial line OP rotates counterclockwise (which reduces the angle α—see Figure 13.5), the damping ratio becomes larger. The rotation of the pole while preserving the radius keeps the natural frequency unchanged (the natural frequency is in the denominator of τ _r ), only, overall, the ascension fourth dimension is reduced. There is a similar trend in the maximum overshoot O _p , which decreases when the damping increases, equally shown in Figure xiii.vi(b). Again, counterclockwise rotation of the radial line OP in Effigy 13.five generates an increase in damping, which reduces the maximum overshoot.

Example 13.1

The closed-loop transfer function of a unit-sensitivity feedback control system is ${\mathrm{1000}}_{C\mathrm{50}}\left(s\right)=\frac{2500}{{\mathrm{due\; south}}^2+40s+2500}$ , and its time response under unit-step input is being analyzed. We demand reductions in the maximum overshoot of 15% and in the rise time of 20%. What changes in the damping ratio and natural frequency need to exist operated for that? What will be the modifications in the settling time and tiptop fourth dimension? How will the closed-loop poles move in the complex plane as a effect of these modifications?

Solution

The airtight-loop transfer office of this problem shows that ω _n   =   fifty   rad/southward, and therefore 2ξ·ω _n   =   xl, which results in a damping ratio of ξ   =   0.4. The original rise time is calculated by means of Eq. (13.x) as τ _r   =   0.0253   s, and the original maximum overshoot is adamant by Eq. (13.sixteen) every bit O _p   =   25.38%. According to the requirements of the problem, the modified rise fourth dimension and maximum overshoot are:

(xiii.xviii) $\{\begin{array}{l}{\mathrm{\tau }}_r^{\ast }={\mathrm{\tau }}_r-0.2·{\mathrm{\tau }}_r=0.0202\\ O_p^{\ast }=O_p-\mathrm{0.xv}·O_p=21.575\%\end{array}$

The modified (new) damping ratio is obtained past expressing it in terms of the maximum overshoot of Eq. (13.sixteen) equally:

(thirteen.19) ${\mathrm{\xi }}^{\ast }=-\frac{\ln \left(\frac{O_p^{\ast }}{100}\right)}{\sqrt{{\mathrm{\pi }}^{\mathrm{ii}}+{\left[\ln \left(\frac{O_p^{\ast }}{100}\right)\right]}^2}}=0.4387$

The modified natural frequency is calculated from Eq. (thirteen.10) every bit:

(xiii.twenty) ${\mathrm{\omega }}_{\mathrm{due\; north}}^{\ast }=\frac{{\tan }^{-1}\frac{\sqrt{1-{\left({\mathrm{\xi }}^{\ast }\right)}^2}}{{\mathrm{\xi }}^{\ast }}}{\sqrt{1-{\left({\mathrm{\xi }}^{\ast }\right)}^2}·{\mathrm{\tau }}_r^{\ast }}=\mathrm{61.four}\mathrm{rad}/s$

With these values of the damping ratio and natural frequency, the airtight-loop transfer function of this example problem modifies to:

(13.21) $G_{CL}^{\ast }\left(s\right)=\frac{3770}{s^{\mathrm{ii}}+53.87s+3770}$

To plot the pole position in the circuitous aeroplane we employ MATLAB'due south pzmap(sys), which enables plotting the poles and zeroes of an LTI object (such as a transfer office in our case) in the complex plane. The post-obit MATLAB code generates the pole position plot of Figure thirteen.7:

Effigy 13.vii. Closed-Loop Pole Position in the Complex Plane.

>> tf1 = tf(2500,[1,40,2500]);

>> tf2 = tf(3770,[1,53.87,3770]);

>> pzmap(tf1,tf2)

The arrows indicate how the poles moved later on the modifications of the damping ratio and natural frequency. The poles move to the left indicates that the settling fourth dimension decreased, while the poles moving away vertically reduce the tiptop fourth dimension value.

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Ability Transmission, High-Voltage

Southward.A. Annestrand , in Encyclopedia of Physical Science and Technology (Third Edition), 2003

I.A A Brief History

The foundation of modern electrical power transmission was laid in 1882 when Thomas A. Edison'south Pearl Street Station, a dc generator and radial line manual organization used primarily for lighting, was built in New York City. The evolution of ac transmission in the United States began in 1885, when George Westinghouse bought the patents for ac systems developed past 50. Gaulard and J. D. Gibbs of France. Both air-conditioning and dc ability systems, at that fourth dimension, consisted of short radial lines between generators and loads, and served customers in the firsthand vicinity of generation stations.

The first high-voltage air-conditioning transmission line in the United States was built in 1890, traversing 20   km between Willamette Falls at Oregon Urban center and Portland, Oregon. Alternate electric current transmission technology developed speedily (Table I) and many ac lines were soon constructed, but for several years most operated as isolated systems. Every bit manual distances lengthened and demand for electric energy grew, the demand to move larger blocks of power adult, reliability factors became important, and interconnected systems (power grids) began to be built. Interconnected systems provide significant economic advantages. Fewer generators are required as reserve capacity for top need, which reduces construction costs for the utilities. Similarly, fewer generators are needed in spinning reserve to handle sudden, unexpected increases in load, which further reduces investment costs. Ability grids likewise give utilities generation options, permitting the utilise of the least expensive sources of power available to the grid at any time. Ability systems continue to grow, and typical regional ability grids today encompass tens of large generation stations, hundreds of substations, and thousands of kilometers of transmission lines. The development of extensive regional grids and interties in the 1950s and 1960s resulted in greater needs for coordination of pattern criteria, protective relay schemes, and power flow control and has led to the development of computerized supervisory control and data acquisition (SCADA) systems.

Tabular array I. Historical Trends in High-Voltage Power Transmission

Arrangement voltage (kV)
Nominal	Maximum	Year introduced	Typical transmission capacity (MW)	Typical right-of-way width (yard)
Alternating current
115	121	1915	l–200	15–25
230	242	1921	200–500	thirty–40
345	362	1952	400–1500	35–xl
500	550	1964	1000–2500	35–45
765	800	1965	2000–5000	xl–55
1100	1200	Tested 1970s	3000–10000	l–75
Direct current
50		1954	fifty–100	25–30
200	(±100)	1961	200–500	30–35
500	(±250)	1965	750–1500	30–35
800	(±400)	1970	1500–2000	35–xl
1000	(±500)	1984	2000–3000	35–forty
1200	(±600)	1985	3000–6000	40–55

The get-go commercial awarding of high-voltage dc transmission was that developed by R. Thury in France most the turn of the century. This system consisted of a number of dc generators continued in series at the source to obtain the desired high voltage. Ionic converters were adult subsequently, and a thirty-kV sit-in project was installed in New York State in the 1930s. The first modern commercial high-voltage dc transmission system using mercury arc valves was built in 1954, connecting the Island of Gotland and the mainland of Sweden with an underwater cablevision. It has since been followed by many other dc transmission systems, lately using thyristor technology. The projects include overhead lines and underground cables, as well as submarine cables, to utilize fully the capacity of dc to reduce the price of long-altitude transmission, avert reactive power problems associated with long ac cables, and serve as asynchronous ties between ac networks.

Today, commercial power systems at voltages of upward to 800   kV air-conditioning and ±600   kV dc are in functioning worldwide. Prototype ac systems at the 1200- to 1800-kV level have been congenital and tested. Power transfer capabilities have increased to several thousand megawatts per line, and economies of scale have led to increased ratings of substation equipment. Extra-loftier-voltage (EHV) transformer banks with ratings of 1500   MVA and above are common. Substations have go more meaty, as metal-clad buses and SF_{half dozen} gas insulation are more widely used. Automatic control of power generation and power flow are essential to the constructive operation of interconnected systems. Computers and microprocessors are widely used for these applications.

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Construction failures due to improper materials, manufacturing, and pattern

Goutam Mukhopadhyay , in Handbook of Materials Failure Assay, 2018

four.two.two Investigation

The fracture surface (Fig. 4.nineteenA) shows shiny granular appearance indicating brittle manner of failure. The chevron marks or radial lines were also clearly visible on the fracture surface pointing out the initiation of fracture. Fig. 4.nineteenB shows the chevron marks pointing towards one of the weld marks of the discrete cardinal/pivot as a location of cleft initiation. The material analysis (Tabular array iv.10) of the tie rod confirmed to En24 grade [eight] of hardened and tempered steel. Microstructural test (Fig. 4.xxA ) revealed a weld zone associated with the HAZ. It was observed that the base metal-HAZ and HAZ-weld interfaces were abrupt instead of being gradual. Information technology was an upshot of adoption of improper welding procedure or absence of pre- and postweld oestrus treatment. In the HAZ, the microstructure (Fig. 4.20B) revealed untempered martensite. This is because of college hardenability of En24 steel, where untempered martensite was formed on the surface during welding of the primal, every bit there was no pre- and postweld heat treatment. The base of operations metal showed tempered martensite structure (Fig. 4.20C), which is typical to En24 in quenched and tempered conditions. The fusion zone of the weld showed typical dendritic construction (Fig. 4.20A).

Figure 4.19. (A) Fracture surface of the tie rod, (B) closer view of the fracture surface showing initiation of brittle fracture from the weld at the periphery.

Table iv.10. Chemical Analysis (wt.%) of Necktie Rod

Sample	C	Mn	S	P	Si	Ti	Cr	Ni	Mo	V	Nb
Tie rod	0.400	0.570	0.026	0.069	0.273	0.002	0.960	1.310	0.180	0.011	0.006
Spec. En 24 [8]	0.38/ 0.45	0.45/ 0.seventy	0.035 max	0.035 max	0.10/ 0.35	—	1.00/ ane.40	i.30/ 1.seventy	0.20/ 0.35	—	—

Figure 4.twenty. (A) Microstructure at the periphery showing fusion zone and HAZ due to weld, (B) HAZ shows presence of untempered martensite, and (C) base metal shows tempered martensite matrix.

Hardness profile (Fig. iv.21) was measured, which showed that the hardness of HAZ region was significantly higher than that of weld and base metal. This made HAZ very brittle, which was highly undesirable as the functioning involved jerk loads. The keys were not the integral role of the rod; they were superficially welded on it. no proper pre- and postweld heat treatment led to the formation of a layer of untempered martensite.

Effigy 4.21. Hardness profile showing increase in hardness at the HAZ because of the germination of untempered martensite during welding on the En24 pin material.

Welding should not have been done in this case and the central should have been an integral part of the pattern. If the welding needed to be done at all, in that location should have been proper pre- and postweld estrus treatment.

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Study of Catamenia and Mixing in a Generic GT Combustor using LES

B. Wegner , ... J. Janicka , in Engineering Turbulence Modelling and Experiments six, 2005

Quantitative Description

The samples taken from both the coarse and fine grid simulations were time averaged and profiles of statistical quantities were extracted on the radial lines shown in figure 1. In the post-obit, nosotros compare these profiles with experimental results. The average axial and radial velocities are plotted in figure 4 while the average tangential velocity and the mean mixture fraction are shown in effigy v. A very good overall agreement of the simulation results from both grids with the LDV velocity and PLIF concentration measurements can be observed. Especially the results close to the swirler go out at 10  =   i mm lucifer very well with the experiments. Here, it can be seen that there is dorsum-flow into the swirler non only in a local, merely in a hateful sense due to the separation of the flow from the key fuel injector cone. Surely, this interaction betwixt the nozzle and chamber could not exist described by neglecting the swirler geometry. The only major discrepancy between the LES and the experiments is in the region of the fuel jet which in the simulation is too slow. The reason for this is, that the temperature of the fuel was initially assumed to be equal to the ambient temperature of 298 1000 and the co-ordinate fuel density and mass catamenia rate were prepare in the simulations. It was suspected, that the fuel stream is significantly heated up by the preheated air due to rut conduction in the nozzle and inlet geometry. Thermocouple measurements confirmed this hypothesis showing a existent fuel temperature at the nozzle exit of approximately 398 K. Hence, the velocity based on the mass flow and density is also low in the LES. As a consequence, the penetration of the fuel jet into the CRZ is non predicted deep plenty, just the overall structure and shape of the CRZ is predicted to a satisfactory caste of accuracy.

Figure 4. Radial profiles of the hateful axial (left) and radial (right) velocity component. Black symbols: LDV data; solid line: LES, coarse filigree; dashed line: LES, fine grid.

Figure 5. Radial profiles of the mean tangential velocity component (left) and the mean mixture fraction (right). Blackness symbols: LDV/PLIF data; solid line: LES, fine grid; dashed line: LES, fine grid.

Too plotted in effigy 5 is the hateful mixture fraction which matches well the results from the PLIF measurements. In accordance with the velocity results, the mixing of fuel and air is somewhat to fast.

As well the mean velocity components, second moments, namely the velocity fluctuations were too compared. As tin exist seen in effigy six, the predicted resolved fluctuation of the axial and tangential velocity components in broad ranges show a remarkable degree of agreement with the measured profiles. Both the position and pinnacle of the peaks compare very well. In full general, as expected, the fine grid results compare better to the measured data than the results obtained on the coarse grid. Even the fluctuation peaks stemming from the purlieus shear layers in the swirler are predicted correctly by the LES although (as mentioned in a higher place) no wall model was used. This is an indicator that the grid resolution in the wall nearly region was sufficiently fine. By evaluating the y ⁺ distance of the wall next grid point which is y ⁺  <   5 in nigh cells, it could be confirmed that indeed at that place is at least one signal in the laminar sub-layer fifty-fifty on the coarse grid.

Figure 6. Radial profiles of the resolved axial velocity fluctuation (left) and resolved tangential velocity fluctuation (right). Black symbols: LDV data; dashed line: LES, fine grid.

It should be noted that compared to the mean velocity, the magnitude of the fluctuations is very big. This is some other indicator for the presence of highly energetic coherent structures in the catamenia. To further analyse this, the time series recorded throughout the simulations were used to compute power spectra. Typical spectra obtained at a betoken located in the lower role of the CRZ 1 mm higher up the nozzle exit are shown in figure 7. Both the spectra for the velocity likewise as for mixture fraction exhibit strong dominant peaks at frequencies f _ane  ≈   1090 Hz  =   and f ₂  ≈   2390 Hz. Power spectra obtained from the LDV information evidence similar peaks at slightly higher frequencies of 1370 Hz and 2900 Hz, respectively. With a factor between the higher and lower frequencies of 2.19 in the simulation and 2.12 in the experiments, it seems that the higher frequency is non exactly a harmonic of the lower one. Obviously, the frequencies belong to the precessing vortex core, conspicuously showing an influence of the PVC on the scalar field.

Figure 7. Normalized ability spectral density of the axial velocity component (left) and mixture fraction (right).

In order to larn more about how the mixing is influenced past the PVC, phase averaging was used. To this finish, the time series used for the spectral analysis were filtered with a aught phase-lag band pass filter effectually the higher of the two dominant frequencies. The resulting filtered time series was then used as a trigger signal for phase averaging. Figure 8 shows a comparison of the phase boilerplate and time average of mixture fraction in several cantankerous sections. Every bit can be seen, in that location are keen differences betwixt the results obtained by the two averaging procedures. The time averaging results in sightly tilted and squeezed, but otherwise symmetric contours, which are in accordance with the experimental findings. This is probably due to the fact that the combustion chamber is not rotationally symmetric because of the quartz windows necessary for optical access. In contrast, the phase averaging reveals a highly asymmetric distribution of the mixture fraction contours. This is due to the vortices forming the PVC, which by entraining air impose their double helical structure on the spreading fuel jet. Since the sense of rotation of the PVC vortices is such that it counteracts the direction of the fuel jet, the PVC imposes very high shear on the mixing layer of air and fuel. Thus, using the model for the filtered scalar dissipation charge per unit $\tilde{\mathrm{\chi }}$ by Cook et al. (1997).

Effigy 8. Top: Fourth dimension averaged contours of mixture fraction; Bottom: Phase averaged contours of mixture fraction. Left to right: Cuts in planes z  =   0, y  =   0 and x  =   10 mm.

(5) $\tilde{\mathrm{\chi }}=2\left(\mathit{D}+{\mathit{D}}_{\mathit{t}}\right)\frac{\partial \tilde{\mathit{f}}}{\partial {\mathit{x}}_{\mathit{i}}}\frac{\partial \tilde{\mathit{f}}}{\partial {\mathit{x}}_{\mathit{i}}}$

1 obtains values exceeding 2000 s ^{−   1} in the shear layer effectually the nozzle exit. From a flamelet concept point of view, this would surely lead to a quenching of the flame near the nozzle. In fact, the reacting example experiments show that the flame is lifted. It seems therefore, that the PVC has a significant influence on the mixing process and potentially also on combustion.

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Sheet Metal Forming

Erik Tempelman , ... Bruno Ninaber van Eyben , in Manufacturing and Design, 2014

4.8 Application of the FLD: Deep Drawing

Ane of the simplest of all double-curved products is a cylindrical chalice. The concept of the FLD tin can be instructively practical to how such a product tin can be manufactured. Figure 4.12 shows, in three steps, how a beaker can be formed out of a cylindrical blank. The beaker has diameter D_b and depth H_b ; the blank initially has diameter D_p . The double-curved zones are found at the transitions from flange to sidewall and from sidewall to bottom. The detail of how we produce such a shape will be discussed later in this section, but for at present we simply assume that we printing a punch die through the bare, with the outer shape of the die matching the inner shape of the chalice.

Figure 4.12. Three stages of deep cartoon of a circular beaker.

Consider the blank to exist divided into iv areas labeled A, B, C, and D, with representative points marked along a radial line (see Effigy 4.12). At the start of the forming process, the circular area A lies in the middle of the blank, whereas areas B, C, and D prevarication equally concentric rings out toward the border. During forming, these 4 areas experience different combinations of strains:

A.

Throughout the process, this area is elongated in the radial management ε ₁ as well as in the tangential (circumferential) direction ε ₂—a combination of biaxial tensile strains known equally "stretching."

B.

This surface area start sees elongation ε _i and ε ₂ (in the radial and circumferential directions), as A; but later on, information technology gets pulled "over the edge" of the base of operations and changes to a combination of elongation ε ₁ (along the length of the beaker) plus zero strain ε ₂ (effectually the circumference).

C.

This area get-go sees elongation ε _i (in the radial management) plus compression ε ₂ (in the circumferential direction), referred to as the "deep drawing strains." Later, it gets pulled "over the edge" at the meridian, and changes to a combination of elongation ε ₁ (forth the depth of the beaker) plus zero strain ε ₂ (around the circumference); the farther away that the band lies from the border of the chalice, the later the transition in strain path.

D.

Throughout the process, this surface area undergoes deep drawing strains.

Practice 4.28

In an FLD, sketch the trajectories that the four areas A to D travel through, each ane beginning at the origin. Which area will reach the FLC (wherever that curve may be) the soonest, and what does that mean for the manufacture of this beaker?

Practise 4.29

Make up one's mind how much a small segment of area D must exist compressed if it is to end up at the edge of the beaker, starting at the edge of the blank. Hint: its arc length before and later deformation is proportional to its "onetime" and "new" diameter, and you may assume that the thickness remains constant. If the tangential radial compression is ε ₂, then how large is the radial strain ε _ane?

Exercise iv.30

Assuming now that the bottom does not deform at all, determine all three strain components ε ₁, ε ₂, and ε _iii in the wall of the beaker. Hint: ΔV  =   0.

You will accept noticed that you need to make several clever assumptions and simplifications to answer questions like the previous 2. For starters, if the product is rotationally symmetrical, and then you lot can presume that all strains must show the same symmetry (provided that the material is isotropic—a useful supposition, but one that unfortunately is not quite true in normal practice). Also, consider that because the beaker'south circumference is constant, the strains along its circumference must exist equal to zero. However, this is not the case in the flange, where these tangential strains are negative. Finally, consider that the volume cannot change; for the wall of the beaker this ways that the depth H_b depends on the strain across the thickness ε ₃.

Exercise four.31

Cull realistic values for the dimensions of the beaker and blank, and then select one alloy from Figure 4.11. Can this detail chalice be produced from this metal? Explain.

Note that the material in area C experiences some other deformation when it slides from the flange over the edge of the beaker into the sidewall: when this happens, it gets aptitude out-of-plane besides as deformed inside its plane. Seen in cross-section, we get a combination of bending with stretching, with more than strain in the radial direction on the exterior of the radius, and less strain on the inside. Of course, this also happens in area B at the transition from bottom to sidewall.

Exercise four.32

Calculate the extra strains due to this out-of-aeroplane angle, if the angle radius is given equally five times the blank thickness. Give your answer every bit a true strain. Exactly where (over the thickness of the cloth) do these extra strains occur? Are they meaning?

Let united states now consider briefly how such a beaker can exist made. In exercise, blanks tin be compressed merely very footling earlier they begin to buckle. For this reason, we need a "blank holder"—that is, a heavy steel ring placed concentrically around the dial die and pressing downwards onto the blank (see also Section 4.10). A lubricant ensures that the blank can slide inward from under the blank holder and deform. (Incidentally, this likewise keeps the material from getting thicker, which is why for Exercise 4.30 nosotros could presume that ε ₃ in the flange is zero.) Yet, all deformation in the flange is driven past the drawing force in the beaker wall, generated by the punch die pushing on the lesser. If the flange is besides large, the wall of the beaker volition fail, and for this reason, very deep (or thin) products are produced in multiple forming steps, with a separate die for each step, getting successively smaller in diameter (in other words, this is done working from the outside inward).

Exercise 4.33

Take a closer await at your double-curved product sample(s) from Exercise iv.i. Which areas will accept experienced deep drawing, and which ones stretching? Which areas have seen much deformation, and which only petty? Any new thoughts on the fabric that has been used?

A final annotation on a applied difficulty with deep cartoon, particularly for thin-walled products such as beverage cans. We have assumed isotropic behavior—the aforementioned yield response in all directions. Merely blanks for circular products are stamped out of rolled sheet, and because of the rolling process the sheets show anisotropy—dissimilar strengths forth and across the canvass. A common source of issues in deep drawing rolled canvas is that the outer edge of the bare goes wavy during deep drawing—a phenomenon known as "earing". The underlying reason is that the sideslip planes in the grains are not randomly arranged, but rolling has forced the crystals to get oriented in more than specific patterns with respect to the axes of the canvass. The statistical spread of orientations can be measured and mapped—it is called crystallographic texture. Surprisingly, annealing heat treatments to fully recrystallize rolled canvas do not eliminate texture, but a unlike anisotropy emerges. Texture, and the resulting anisotropy, depend in complex ways on the alloy limerick, on the rolling and annealing atmospheric condition, and on all sorts of microstructural detail inherited from the original casting and other previous processes. It may not look it, just in metallurgical terms the thin-walled beverage can is i of the most heavily researched and sophisticated products.

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Orbits in Three Dimensions

Howard D. Curtis , in Orbital Mechanics for Applied science Students (Second Edition), 2010

4.eight Basis Tracks

The projection of a satellite's orbit onto the earth'southward surface is chosen its ground rails . At a given instant one can imagine a radial line fatigued outward from the center of the earth to the satellite. Where this line pierces the earth's spherical surface is a indicate on the ground track. Nosotros locate this point by giving its latitude and longitude relative to the earth. Every bit the satellite moves around the earth, the trace of these points is its ground track.

Considering the satellite reaches a maximum and minimum latitude ("aamplitude") during each orbit while passing over the equator twice, on a Mercator projection the ground rail of a satellite in low world orbit often resembles a sine bend. If the globe did not rotate, there would be just ane sinusoid-like track, traced repeatedly as the satellite orbits the world. However, the earth rotates eastward beneath the satellite orbit at 15.04° per hr, and then the ground rail advances westward at that charge per unit. Figure 4.23 shows about ii and a half orbits of a satellite, with the get-go and end of this portion of the ground track labeled. The distance between two successive crossings of the equator is measured to be 23.2°, which is the amount of earth rotation in one orbit of the spacecraft. Therefore, the basis track reveals that the period of the satellite is

Figure 4.23. Ground rails of a satellite.

$T\text{=}\frac{23.2\text{ degrees}}{\text{15}\text{.04}\text{degrees}/\text{hr}}\text{=}\mathrm{i.54}\text{ hr}\text{=}\mathrm{92.vi}\text{ min}$

This is a typical low earth orbital menses.

Given a satellite's position vector r, we tin utilise Algorithm four.1 to notice its right rise and declination relative to the geocentric equatorial XYZ frame, which is stock-still in infinite. The world rotates at an athwart velocity ω_East relative to this organisation. Allow united states attach an x′y′z′ Cartesian coordinate organization to the earth with its origin located at the globe's heart, as illustrated in Figure 1.18. The x′y′ axes prevarication in the equatorial aeroplane and the z′ centrality points north. (In Figure ane.18 the ten′ axis is directed towards the prime height, which passes through Greenwich, England.) The XYZ and x′y′z′ differ only by the angle θ between the stationary X axis and the rotating x′ axis. If the X and 10′ axes line upward at time t ₀, then at any fourth dimension t thereafter the angle θ will be given by ω_E (t–t ₀). The transformation from XYZ to x′y′z′ is represented past the elementary rotation matrix (call back Equation 4.34),

(4.55) $\begin{array}{ccc}\text{[}{R}_3\text{(}\theta \text{)]}\text{=}\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0\\ \text{-}\sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]& & \theta \text{=}{\omega }_{\mathrm{Eastward}}\text{(}t\text{-}{t}_0\text{}\text{)}\end{array}$

Thus, if the components of the position vector r in the inertial XYZ frame are given by {r} ₁₀ , its components {r} _x′ in the rotating, earth-fixed ten′y′z′ frame are:

(iv.56) ${\text{{}r\text{}}}_{{\mathrm{ten}}^{\prime }}\text{=}\left[R_3\text{(}\theta \text{)}\right]{\text{{}r\text{}}}_{\mathrm{Ten}}$

Knowing {r} _x ′, we utilise Algorithm 4.one to determine the right ascension (longitude due east of x′) and declination (latitude) in the earth-fixed system. These points are usually plotted on a rectangular Mercator projection of the earth'southward surface, as in Effigy four.23.

Algorithm 4.6

Given the initial orbital elements (h, e, a, T, i, ω ₀, Ω ₀, and θ ₀) of a satellite relative to the geocentric equatorial frame, compute the right ascension α and declination δ relative to the rotating globe after a time interval Δt. This algorithm is implemented in MATLAB every bit the script ground_track.thousand in Appendix D.23.

ane.

Compute $\dot{\Omega }$ and $\dot{\omega }$ from Equations 4.52 and 4.53.

two.

Calculate the initial time t ₀ (time since perigee passage):

a.

Observe the eccentric bibelot E ₀ from Equation iii.13b.

b.

Find the mean anomaly K ₀ from Equation 3.14.

c.

Notice t ₀ from Equation 3.15.

3.

At time t = t ₀ + Δt, summate α and δ.

a.

Calculate the true anomaly:

i.

Find 1000 from Equation 3.8

two.

Discover Due east from Equation iii.14 using Algorithm three.1.

iii.

Find θ from Equation 3.13a.

b.

Update Ω and ω:

$\Omega \text{=}{\Omega }_0\text{+}\dot{\Omega }\Delta t\text{undefined}\omega \text{=}{\omega }_0\text{+}\dot{\omega }\Delta t$

c.

Discover {r} _x using Algorithm 4.5.

d.

Observe {r}x′ using Equations four.55 and 4.56.

e.

Apply Algorithm 4.1 to obtain compute α and δ from {r} _x′ .

4.

Repeat Step three for additional times ( $t\text{=}{t}_0\text{+}2\Delta t$ , $t\text{=}{t}_0\text{+}3\Delta t$ , etc.).

Example 4.12

An globe satellite has the following orbital parameters:

$r_p\text{=}670\text{0}\text{km}$	Perigee
$r_a\text{=}10,000\text{km}$	Apogee
${\theta }_0\text{=}230\text{°}$	True bibelot
${\Omega }_0\text{=}270\text{°}$	Right rising of the ascending node
$i_0\text{=}60\text{°}$	Inclination
${\omega }_0\text{=}45\text{°}$	Argument of perigee

Calculate the correct rise (longitude east of x′) and declination (latitude) relative to the rotating earth 45 minutes later.

Solution

First, we compute the semimajor axis a, eccentricity e, the angular momentum h, the semimajor axis a, and the period T. For the semimajor axis nosotros call up that

$a\text{=}\frac{r_p\text{+}{r}_a}{2}\text{=}\frac{6700\text{+}\mathrm{x},000}{2}\text{=}8350\text{km}$

From Equation ii.84 we become

$e\text{=}\frac{r_a\text{-}{r}_p}{r_a\text{+}{r}_p}\text{=}\frac{10,000\text{-}6700}{\mathrm{ten},000\text{+}6700}\text{=}0.19760$

Equation 2.fifty yields

$h\text{=}\sqrt{\mu r_p\text{(}\mathrm{one}\text{+}e\text{)}}\text{=}\sqrt{398,600\cdot 6700\cdot \text{(}1\text{+}0.19760\text{)}}\text{=}56,554{\text{ km}}^{\mathrm{two}}\text{/s}$

Finally, we obtain the catamenia from Equation two.83:

$T\text{=}\frac{2\pi }{\sqrt{\mu }}{a}^{3/2}\text{=}\frac{2\pi }{\sqrt{398,600}}{8350}^{\mathrm{iii}/\mathrm{ii}}\text{=}7593.5\text{south}$

At present we tin can proceed with Algorithm 4.6.

Step 1.

$\dot{\Omega }\text{=-}\left[\frac{\mathrm{iii}}{2}\frac{\sqrt{\mu }{J}_2{R}_{\text{earth}}^2}{\left(1\text{-}{\mathrm{due\; east}}^2\right)a^{7/2}}\right]\cos i\text{=-}\left[\frac{\mathrm{iii}}{2}\frac{\sqrt{398,600}\cdot 0.0010836\cdot {6378}^2}{\text{(}1\text{-}{0.19760}^2\text{)}{8350}^{7/\mathrm{two}}}\right]\cos \mathrm{lx}\text{°=-}2.3394\text{×}{\mathrm{x}}^{\text{-}\mathrm{vii}}\text{°/}\text{s}$

$\dot{\omega }\text{=}\dot{\Omega }\frac{5/2{\sin }^{2}i\text{-}2}{\cos i}\text{=-}2.3394\text{×}{\mathrm{ten}}^{\text{-}\mathrm{v}}\left(\frac{5/\mathrm{two}{\sin }^2\mathrm{sixty}\text{°-}2}{\cos 60\text{°}}\right)\text{=}\mathrm{v.8484}\text{×}{10}^{\text{-}\mathrm{vi}}\text{°/}\text{s}$

Step two.

a.

$E\text{=}2{\tan }^{\text{-}1}\left(\tan \frac{\theta }{\mathrm{two}}\sqrt{\frac{1\text{-}\mathrm{due\; east}}{1\text{+}e}}\right)\text{=}\mathrm{two}{\tan }^{\text{-}1}\left(\tan \frac{230\text{°}}{2}\sqrt{\frac{1\text{-}0.19760}{\mathrm{one}\text{+}0.19760}}\right)\text{=-}2.1059\text{rad}$

b.

$\mathrm{Chiliad}\text{=}E\text{-}e\sin \mathrm{Due\; east}\text{=-}2.1059\text{-}0.19760\sin \text{(}\text{}\text{-}\mathrm{ii.1059}\text{)}\text{=-}1.9360\text{rad}$

c.

$t_0\text{=}\frac{\mathrm{Yard}}{2\pi }T\text{=}\frac{\text{-}\mathrm{ane.9360}}{\mathrm{two}\pi }\cdot \mathrm{7593.v}\text{=-}2339.7\text{ southward}\text{undefined}\text{(2339}\text{.7 s}unti\mathrm{50}\text{ perigee)}$

Step three. $t\text{=}{t}_0\text{+}45\text{ min}\text{=-}2339.7\text{+}45\cdot 60\text{=}360.33\text{undefined}\mathrm{south}\text{undefined}(360.33\text{ s}after\text{undefined}p\mathrm{east}rige\mathrm{eastward}\text{)}$

a.

$\mathrm{Chiliad}\text{=}2\pi \frac{t}{T}\text{=}2\pi \frac{360.33}{7593.5}\text{=}0.29815\text{rad}$

$\mathrm{Eastward}\text{-}0.19760\sin \mathrm{Eastward}\text{=}0.29815\stackrel{\text{Algorithm 3}\text{.ane}}{\overbrace{\Rightarrow }}E\text{=}0.36952\text{rad}$

$\theta \text{=}\mathrm{ii}{\tan }^{\text{-}1}\left(\tan \frac{\mathrm{Eastward}}{2}\sqrt{\frac{1\text{+}e}{1\text{-}e}}\right)\text{=}2{\tan }^{\text{-}\mathrm{one}}\left(\tan \frac{0.36952}{2}\sqrt{\frac{1\text{+}0.19760}{\mathrm{one}\text{-}0.19760}}\right)\text{=}25.723\text{°}$

b.

$\Omega \text{=}{\Omega }_0\text{+}\dot{\Omega }\Delta t\text{=}270\text{°+}\text{(}\text{}\text{-}2.3394\text{×}{\mathrm{ten}}^{\text{-}\mathrm{five}}\text{undefined}\text{°/}\text{s)(}2700\text{ s)}\text{=}269.94\text{°}$

$\omega \text{=}\omega 0_{}\text{+}\dot{\omega }\Delta t\text{=}45\text{°+}\text{(}5.8484\text{×}{10}^{-6}\text{undefined}\text{°/}\text{s)(}2700\text{ s)}\text{=}45.016\text{°}$

c.

${\text{{}r\text{}}}_X\stackrel{\text{Algorithm 4}\text{.5}}{\overbrace{\text{=}}}\left\{\begin{array}{c}3212.6\\ \text{-}2250.5\\ 5568.6\end{array}\right\}\text{ (km)}$

d.

$\theta \text{=}{\omega }_E\Delta t\text{=}\frac{360\text{°}\left(1\text{+}\frac{1}{365.26}\right)}{24\cdot 3600\text{s}}\cdot 2700\text{s}\text{=}\mathrm{eleven.281}\text{°}$

$\text{[}{R}_3\text{(}\theta \text{)]}\text{=}\left[\begin{array}{ccc}\cos 11.281\text{°}& \sin 11.281\text{°}& 0\\ \text{-}\sin \mathrm{eleven.281}\text{°}& \cos \mathrm{eleven.281}\text{°}& 0\\ 0& 0& 1\end{array}\right]\text{=}\left[\begin{array}{ccc}\text{undefined}0.98068& 0.19562& 0\\ \text{-}0.19562& 0.98068& 0\\ 0& 0& 1\end{array}\right]$

${\text{{}r\text{}}}_{x^{\prime }}\text{=}\text{[}{R}_3\text{(}\theta \text{)]e {}r{\text{}}}_{\mathrm{10}}\text{=}\left[\begin{array}{ccc}0.98068& 0.19562& 0\\ \text{-}0.19562& 0.98068& 0\\ 0& 0& \mathrm{one}\end{array}\right]\left\{\begin{array}{c}3212.6\\ \text{-}2250.5\\ \mathrm{5568.half\; dozen}\end{array}\right\}\text{=}\left\{\begin{array}{c}2710.3\\ \text{-}2835.4\\ 5568.6\end{array}\right\}\text{ (km)}$

e.

The script ground_track.m in Appendix D.23 can be used to plot ground tracks. For the data of Example iv.12 the ground track for 3.25 periods appears in Figure iv.24. The footing track for ane orbit of a Molniya satellite is featured more elegantly in Figure 4.25.

Figure 4.24. Footing rails for 3.25 orbits of the satellite in Example 4.6.

Figure iv.25. Ground track for two orbits of a Molniya satellite with a 12-hour period. Tick marks are 1 hour autonomously.

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URL:

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Source: https://www.sciencedirect.com/topics/engineering/radial-line