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dla studentów studiów inżynierskich WSH RADOMpuma creepers
http://infzao2009.phorum.pl/viewtopic.php?p=11903#11903
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Autor: <a href='http://infzao2009.phorum.pl/profile.php?mode=viewprofile&u=656'>Jeremy Gissing</a><br /><br />
Wysłany: Sob Sty 04, 2020 04:36<br /><br />
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ÿþwhere A(q) is the n X n kineticenergymatrix; a l ï»¿<a href="http://www.hitchonmusic.com/puma-c-1/puma-creepers-c-1_2/" target="_blank" class="postlink">puma creepers</a> l = 11 12 coa2(82) 13 cos(82)co8(82 63) (3) B(q) is the n x n(n-1)/2matrix of Coriolistorques; z4 cos2(e2 03) C(q) is the n x n matrix of centrifugaltorques; Calculations required: 3 multiplications, 3 additions. g(q) is then-vector of gravitytorques; q is thne-vector of accelerations; where ZI.= d i m 3 dZm3 2 d2d3m3 dgm2 J3yy J3== is the generalizedjoint force vector. r J2== Jzyv Jlzr Jizz; etc.<br />
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And reduction of these expressions with four relations that hold on these partial derivatives. J -- ,p',jj (5) 4. Formation of the needed partial derivatives, expansion of where (qk * it) is the <a href="http://www.hitchonmusic.com/puma-c-1/puma-rihanna-c-1_3/" target="_blank" class="postlink">puma rihanna</a> j t h velocity product in the [q4]vector, and the Coriolis and centrifugal matrix elements in terms of the derivatives, and simplification by combining is the Christoffel symbol. inertia constants as in 2. The number of unique <a href="http://www.hitchonmusic.com/puma-c-1/puma-fenty-c-1_4/" target="_blank" class="postlink">puma fenty</a> non-zero Christoffel symbols required The first step was carried out with a LISP program.<br />
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Which symbolically generates the dynamic model of an equations thathold on the derivatives of the kinetic energy matrixarticulatedmechanism.EMDEGemploys Kane's dynamic for- elements.The first two equationsaregeneral;the last two aremulation [Kane 19681, and produced a result comparable in form specific to the PUMA 560. The equations are:and size to that of ARM [Murry and Neuman 1984). Three sirn-plifying assumptions were made for this <a href="http://www.hitchonmusic.com/puma-c-1/puma-slides-c-1_5/" target="_blank" class="postlink">puma slides</a> analysis.<br />
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Of gravity and the terms of the inertia dyadic.Thewrist,link where I I = M-wg2 ** r12three and link two of a PUMA 560 arm were detached in order Mgto measure these parameters. The mass of each component was r is theinertiaabouthe axis of rotation;determined with a beam balance; the cenotfergravity was located is the weight of the link;by balancing each link on a knife edge, once orthogonal to each w is thdeistance fromeacshuspensionaxis;<br />
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The tolerance values assigned to calculated parameters were de-With this arrangement a rotational pendulum is created about termined by RMS combination of the tolerance assigned to eachanaxisparallel toand halfwaybetween the suspension wires.The link's center of gravity must lie on this axis.The inertiasuspension method of measuring the rotational inertia requires <a href="http://www.hitchonmusic.com/puma-c-1/puma-fenty-slides-c-1_6/" target="_blank" class="postlink">puma fenty slides</a> dyadic and center of gravity parameters of link 3 were measuredknowledge of parameters that are easily measured.<br />
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40is possible to start fundamentalmodeoscillationwithout visi- Link 3 4.80blyexcitingany of the other modes. The relationshipbetween Link 4* 0.82measured properties and rotational inertia is: Link 5* 0.34 Link 6* 0.09 * This method was suggested by Prof. David Powell. Link 3 wiCthomplete LVrist 6.04 Detached Wrist 2.24 * Values derived from external dimensions; f 2 5 % . The positions of the centers of gravity are reported in Table 5. The dimensions rz! ry and <a href="http://www.hitchonmusic.com/puma-fenty-slides-p-84.html" target="_blank" class="postlink"><img src="http://www.hitchonmusic.com/images/large/puma-316gbu.jpg" border="0" /></a> rz refer to the x, y and z coordinates 513http://infzao2009.phorum.pl/viewtopic.php?p=11903#11903Jeremy GissingSob Sty 04, 2020 04:36http://infzao2009.phorum.pl/viewtopic.php?p=11903#11903