{"id":1032,"date":"2021-09-09T13:34:13","date_gmt":"2021-09-09T13:34:13","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=1032"},"modified":"2021-10-08T14:02:25","modified_gmt":"2021-10-08T14:02:25","slug":"4-2-2","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch4\/4-2-2\/","title":{"rendered":"4-2-2"},"content":{"rendered":"
\n
\n\t

4.2<\/b> Mekanizmalarda H\u0131z ve \u0130vme Analizi -2<\/h1>\n

\u00d6rnek:<\/strong><\/p>\n

\"\"<\/p>\n

Cismin A ve B noktalar\u0131n\u0131n h\u0131zlar\u0131 \u015fekilde g\u00f6sterildi\u011fi gibidir. S noktas\u0131n\u0131n h\u0131z vekt\u00f6r\u00fcn\u00fc bulun.<\/p>\n

v<\/strong>A<\/sub>\u00a0ve\u00a0v<\/strong>B<\/sub>\u00a0h\u0131z vekt\u00f6rleri ov<\/sub> ba\u015flang\u0131\u00e7 noktas\u0131ndan \u00e7izildi\u011finde u\u00e7lar\u0131 a ve b noktalar\u0131 olarak i\u015faretlenecektir. H\u0131z poligonunda olu\u015facak olan abs \u00fc\u00e7geni ile cisim \u00fczerinde mevcut ABS \u00fc\u00e7genlerinin benzer yapacak \u015fekilde olu\u015fturulmas\u0131 gerekmektedir. Benzer \u00fc\u00e7genlerin a\u00e7\u0131lar\u0131 ayn\u0131d\u0131r. Bu nedenle b noktas\u0131ndan xb do\u011frusunu \u2220xba = \u2220SBA olacak \u015fekilde ve a noktas\u0131ndan ua do\u011frusunu \u2220uab = \u2220SAB olacak \u015fekilde \u00e7izdi\u011fimizde ua ve xb\u00a0do\u011frular\u0131n\u0131n kesim noktas\u0131 s dir ve ov<\/sub>\u00a0s<\/strong> vekt\u00f6r\u00fc S noktas\u0131n\u0131n v<\/strong>S<\/sub>\u00a0h\u0131z vekt\u00f6r\u00fcd\u00fcr (\u015eekil b).<\/p>\n

\u00d6rnek:<\/strong><\/p>\n

\"\"<\/p>\n

\u015eekilde g\u00f6sterilen bir d\u00f6rt-\u00e7ubuk mekanizmas\u0131nda |A0<\/sub>B0<\/sub>| = a1<\/sub>, |A0<\/sub>A| = a2<\/sub>, |AB| = a3<\/sub>, |AC| = b3<\/sub>, |B0<\/sub>B| = a4<\/sub>. H\u0131z ve ivme analizini yapal\u0131m. Devre kapal\u0131l\u0131k denklemi ve e\u015fleni\u011fi:<\/p>\n

a2<\/sub>ei\u03b812<\/sub><\/sup> + a3<\/sub>ei\u03b813<\/sub><\/sup> = a1<\/sub> + a4<\/sub>ei\u03b814<\/sub><\/sup><\/p>\n

a2<\/sub>e\u2212i\u03b812<\/sub><\/sup> + a3<\/sub>e\u2212<\/sup>i\u03b813<\/sub><\/sup> = a1<\/sub> + a4<\/sub>e-i\u03b814<\/sub><\/sup><\/p>\n

Devre kapal\u0131l\u0131k denkleminin birinci t\u00fcrevi al\u0131nd\u0131\u011f\u0131nda elde edilen h\u0131z devre denklemi:<\/p>\n

ia2<\/sub>\u03c912<\/sub>ei\u03b812<\/sub><\/sup> + ia3<\/sub>\u03c913<\/sub>ei\u03b813<\/sub><\/sup> = ia4<\/sub>\u03c914<\/sub>ei\u03b814<\/sub><\/sup><\/p>\n

\u2212ia2<\/sub>\u03c912<\/sub>e\u2212<\/sup>i\u03b812<\/sub><\/sup> \u2212 ia3<\/sub>\u03c913<\/sub>e\u2212<\/sup>i\u03b813<\/sub><\/sup> = \u2212ia4<\/sub>\u03c914<\/sub>e\u2212<\/sup>i\u03b814<\/sub><\/sup><\/p>\n

elde edilecektir. Dikkat edilir ise, h\u0131z devre denklemi vekt\u00f6rel olarak:<\/p>\n

v<\/strong>A<\/sub> + v<\/strong>B\/A<\/sub> = v<\/strong>B<\/sub><\/p>\n

denklemidir. Karma\u015f\u0131k say\u0131larla yaz\u0131lm\u0131\u015f olan her bir h\u0131z teriminin tan\u0131m\u0131 genellikle kolayca yap\u0131labilir ve mutlaka ba\u011f\u0131l h\u0131z kavram\u0131 ile elde edilecek vekt\u00f6r e\u015fitlik denklemi ile ayn\u0131 sonu\u00e7 vermesi gereklidir. Bir ba\u015fka anlat\u0131mla, karma\u015f\u0131k say\u0131lar ile yaz\u0131lan h\u0131z devre denklemleri, vekt\u00f6rel h\u0131z denklemlerinin farkl\u0131 bir yaz\u0131l\u0131\u015f \u015feklidir. Vekt\u00f6rel h\u0131z denklemleri analitik veya geometrik y\u00f6ntemle \u00e7\u00f6z\u00fclebilir. Analitik olarak karma\u015f\u0131k say\u0131lar ile yaz\u0131lm\u0131\u015f olan h\u0131z devre denklemlerinin bilinmeyen h\u0131z denklemlerine g\u00f6re \u00e7\u00f6z\u00fcm\u00fc genellikle daha basit sonu\u00e7 vermektedir. Denklemler verilen giri\u015f h\u0131z\u0131 \u03c912<\/sub> ile bilinmeyen \u03c913<\/sub> ve \u03c914<\/sub> a\u00e7\u0131sal h\u0131z de\u011fi\u015fkenlerine g\u00f6re lineer olup e\u011fer konum analizi \u00f6nceden yap\u0131lm\u0131\u015f ve verilen \u03b812<\/sub> a\u00e7\u0131s\u0131na g\u00f6re \u03b813<\/sub> ile \u03b814<\/sub> a\u00e7\u0131lar\u0131n\u0131n o konumda ald\u0131\u011f\u0131 de\u011ferler bulunmu\u015f ise \u03c913<\/sub> ve \u03c914<\/sub> h\u0131z de\u011fi\u015fkenlerine Cramer kural\u0131 kullan\u0131larak \u00e7\u00f6z\u00fcm yap\u0131labilir:<\/p>\n

\\displaystyle {{\\text{\u03c9}}_{{13}}}=\\frac{{\\left| {\\begin{array}{cc} {-{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} & {-{{\\text{a}}_{4}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{14}}}}}}} \\\\ {-{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} & {-{{\\text{a}}_{4}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{14}}}}}}} \\end{array}} \\right|}}{{\\left| {\\begin{array}{cc} {{{\\text{a}}_{3}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} & {-{{\\text{a}}_{4}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{14}}}}}}} \\\\ {{{\\text{a}}_{3}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} & {-{{\\text{a}}_{4}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{14}}}}}}} \\end{array}} \\right|}}=\\frac{{{{\\text{a}}_{2}}{{\\text{a}}_{4}}\\left( {{{\\text{e}}^{{\\text{i}\\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{14}}}} \\right)}}}-{{\\text{e}}^{{-\\text{i}\\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{14}}}} \\right)}}}} \\right)}}{{{{\\text{a}}_{3}}{{\\text{a}}_{4}}\\left( {-{{\\text{e}}^{{-\\text{i}\\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}}+{{\\text{e}}^{{\\text{i}\\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}}} \\right)}}{{\\text{\u03c9}}_{{12}}}=\\frac{{{{\\text{a}}_{2}}}}{{{{\\text{a}}_{3}}}}\\frac{{\\sin \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{14}}}} \\right)}}{{\\sin \\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}{{\\text{\u03c9}}_{{12}}} <\/span><\/p>\n

Benzer bir \u015fekilde:<\/p>\n

\\displaystyle {{\\text{\u03c9}}_{{14}}}=\\frac{{\\left| {\\begin{array}{cc} {{{\\text{a}}_{3}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} & {-{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} \\\\ {{{\\text{a}}_{3}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} & {-{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{12}}}}}}} \\end{array}} \\right|}}{{\\left| {\\begin{array}{cc} {{{\\text{a}}_{3}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} & {-{{\\text{a}}_{4}}{{\\text{e}}^{{\\text{i}{{\\text{\u03b8}}_{{14}}}}}}} \\\\ {{{\\text{a}}_{3}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{13}}}}}}} & {-{{\\text{a}}_{4}}{{\\text{e}}^{{-\\text{i}{{\\text{\u03b8}}_{{14}}}}}}} \\end{array}} \\right|}}=\\frac{{{{\\text{a}}_{3}}{{\\text{a}}_{2}}\\left( {-{{\\text{e}}^{{-\\text{i}\\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}}+{{\\text{e}}^{{\\text{i}\\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}}} \\right)}}{{{{\\text{a}}_{3}}{{\\text{a}}_{4}}\\left( {-{{\\text{e}}^{{-\\text{i}\\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}}+{{\\text{e}}^{{-\\text{i}\\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}}} \\right)}}{{\\text{\u03c9}}_{{12}}}=\\frac{{{{\\text{a}}_{2}}}}{{{{\\text{a}}_{4}}}}\\frac{{\\sin \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}{{\\sin \\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}{{\\text{\u03c9}}_{{12}}} <\/span><\/p>\n

C noktas\u0131n\u0131n h\u0131z\u0131 isteniyor ise, C nin konum vekt\u00f6r\u00fc:<\/p>\n

r<\/strong>C<\/sub>\u00a0= a2<\/sub>ei\u03b812<\/sub><\/sup> + b3<\/sub>ei(\u03b813 <\/sub>+\u00a0<\/sub>\u03b2)<\/sup><\/p>\n

r<\/strong>C<\/sub>\u00a0konum vekt\u00f6r\u00fcn\u00fcn zamana g\u00f6re t\u00fcrevi, C noktas\u0131n\u0131n h\u0131z vekt\u00f6r\u00fcn\u00fc tan\u0131mlar:<\/p>\n

v<\/strong>C<\/sub> = \\displaystyle {\\dot{\\text{x}}} <\/span>C<\/sub> + i \\displaystyle {\\dot{\\text{y}}} <\/span>C<\/sub> = ia2<\/sub>\u03c912<\/sub>ei\u03b812<\/sub><\/sup> + ib3<\/sub>\u03c913<\/sub>ei(\u03b813 <\/sub>+\u00a0<\/sub>\u03b2)<\/sup><\/p>\n

Bu denklem<\/p>\n

v<\/strong>C<\/sub> = v<\/strong>A<\/sub> + v<\/strong>C\/A<\/sub><\/p>\n

vekt\u00f6rel h\u0131z denklemidir. x ve y bile\u015fenleri:<\/p>\n

\\displaystyle {\\dot{\\text{x}}} <\/span>C<\/sub> = \u2212ia2<\/sub>\u03c912<\/sub>sin\u03b812<\/sub> \u2212 ib3<\/sub>\u03c913<\/sub>sin(\u03b813 <\/sub>+\u00a0<\/sub>\u03b2)<\/p>\n

\\displaystyle {\\dot{\\text{y}}} <\/span>C<\/sub> = a2<\/sub>\u03c912<\/sub>cos\u03b812<\/sub> + b3<\/sub>\u03c913<\/sub>cos(\u03b813 <\/sub>+\u00a0<\/sub>\u03b2)<\/p>\n

d\u0131r. E\u011fer devre kapal\u0131l\u0131k denklemi ve h\u0131z devre denklemleri \u00f6nceden \u00e7\u00f6z\u00fclm\u00fc\u015f ise, yukar\u0131da verilmi\u015f olan denklemlerin sa\u011f taraf\u0131nda kalan terimlerin t\u00fcm\u00fc bilinmektedir. H\u0131z\u0131n x ve y bile\u015fenleri \u00e7\u00f6z\u00fcld\u00fckten sonra istenildi\u011finde kutupsal koordinat sistemi kullan\u0131larak h\u0131z bir \u015fiddet ve y\u00f6n a\u00e7\u0131s\u0131 ile g\u00f6sterilebilir. Bu durumda C noktas\u0131n\u0131n h\u0131z vekt\u00f6r\u00fc karma\u015f\u0131k say\u0131 ile<\/p>\n

v<\/strong>C<\/sub> = vC<\/sub>ei\u03b7<\/sup><\/p>\n

d\u0131r. Bu denklemde vC<\/sub> = \\displaystyle \\sqrt{{{{{\\dot{\\text{x}}}}_{\\text{C}}}^{2}+{{{\\dot{\\text{y}}}}_{\\text{C}}}^{2}}} <\/span> h\u0131z \u015fiddeti ve \u03b7 = tan-1<\/sup>( \\displaystyle {\\dot{\\text{y}}} <\/span>C<\/sub>\/ \\displaystyle {\\dot{\\text{x}}} <\/span>C<\/sub>) h\u0131z vekt\u00f6r\u00fcn\u00fcn pozitif x eksenine g\u00f6re yapt\u0131\u011f\u0131 a\u00e7\u0131d\u0131r.<\/p>\n

\u0130vme analizi i\u00e7in karma\u015f\u0131k say\u0131 ile yaz\u0131lm\u0131\u015f olan h\u0131z devre denkleminin t\u00fcrevi al\u0131narak ivme devre denklemi elde edilecektir:<\/p>\n

ia2<\/sub>\u03b112<\/sub>ei\u03b812<\/sub><\/sup> \u2212 a2<\/sub>\u03c912<\/sub>2<\/sup>ei\u03b812<\/sub><\/sup> + ia3<\/sub>\u03b113<\/sub>ei\u03b813<\/sub><\/sup> \u2212 a3<\/sub>\u03c913<\/sub>2<\/sup>ei\u03b813<\/sub><\/sup> = ia4<\/sub>\u03b114<\/sub>ei\u03b814<\/sub><\/sup> \u2212 a4<\/sub>\u03c914<\/sub>2<\/sup>ei\u03b814<\/sub><\/sup><\/p>\n

\u2212ia2<\/sub>\u03b112<\/sub>e\u2212<\/sup>i\u03b812<\/sub><\/sup> \u2212 a2<\/sub>\u03c912<\/sub>2<\/sup>e\u2212<\/sup>i\u03b812<\/sub><\/sup> \u2212 ia3<\/sub>\u03b113<\/sub>e\u2212<\/sup>i\u03b813<\/sub><\/sup> \u2212 a3<\/sub>\u03c913<\/sub>2<\/sup>e\u2212<\/sup>i\u03b813<\/sub><\/sup> = \u2212ia4<\/sub>\u03b114<\/sub>e\u2212<\/sup>i\u03b814<\/sub><\/sup> \u2212 a4<\/sub>\u03c914<\/sub>2<\/sup>e\u2212<\/sup>i\u03b814<\/sub><\/sup><\/p>\n

Dikkat edilir ise yaz\u0131lm\u0131\u015f olan h\u0131z devre denklemi vekt\u00f6rel olarak:<\/p>\n

at<\/sup><\/strong>A<\/sub> + an<\/sup><\/strong>A<\/sub>\u00a0+ at<\/sup><\/strong>B\/A<\/sub>\u00a0+ an<\/sup><\/strong>B\/A<\/sub>\u00a0= at<\/sup><\/strong>B<\/sub>\u00a0+ an<\/sup><\/strong>B<\/sub><\/p>\n

denkleminden farkl\u0131 de\u011fildir. \u0130vme devre denkleminde bilinen terimleri denklemin sa\u011f taraf\u0131nda k\u00fcmeler isek:<\/p>\n

\u00a0 ia3<\/sub>\u03b113<\/sub>ei\u03b813<\/sub><\/sup> \u2212 ia4<\/sub>\u03b114<\/sub>ei\u03b814<\/sub><\/sup>\u00a0= \u2212ia2<\/sub>\u03b112<\/sub>ei\u03b812<\/sub><\/sup> + a2<\/sub>\u03c912<\/sub>2<\/sup>ei\u03b812<\/sub><\/sup> + a3<\/sub>\u03c913<\/sub>2<\/sup>ei\u03b813<\/sub><\/sup>\u00a0\u2212 a4<\/sub>\u03c914<\/sub>2<\/sup>ei\u03b814<\/sub><\/sup><\/p>\n

\u2212ia3<\/sub>\u03b113<\/sub>e\u2212<\/sup>i\u03b813<\/sub><\/sup> + ia4<\/sub>\u03b114<\/sub>e\u2212<\/sup>i\u03b814<\/sub><\/sup> = ia2<\/sub>\u03b112<\/sub>e\u2212<\/sup>i\u03b812<\/sub><\/sup> + a2<\/sub>\u03c912<\/sub>2<\/sup>e\u2212<\/sup>i\u03b812<\/sub><\/sup> + a3<\/sub>\u03c913<\/sub>2<\/sup>e\u2212<\/sup>i\u03b813<\/sub><\/sup>\u00a0\u2212 a4<\/sub>\u03c914<\/sub>2<\/sup>e\u2212<\/sup>i\u03b814<\/sub><\/sup><\/p>\n

Denklemi bilinmeyen ivme de\u011fi\u015fkenleri \u03b113<\/sub> ve \u03b114<\/sub>\u00a0i\u00e7in \u00e7\u00f6zd\u00fc\u011f\u00fcm\u00fczde:<\/p>\n

\u03b113<\/sub> = \\displaystyle \\frac{{{{\\text{a}}_{2}}{{\\text{\u03b1}}_{{12}}}\\sin \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{14}}}} \\right)+{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}^{2}\\cos \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{14}}}} \\right)-{{\\text{a}}_{4}}{{\\text{\u03c9}}_{{12}}}^{2}+{{\\text{a}}_{3}}{{\\text{\u03c9}}_{{13}}}^{2}\\cos \\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}}{{{{\\text{a}}_{3}}\\sin \\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}} <\/span><\/p>\n

ve<\/p>\n

\u03b114<\/sub> = \\displaystyle \\frac{{{{\\text{a}}_{2}}{{\\text{\u03b1}}_{{12}}}\\sin \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)+{{\\text{a}}_{2}}{{\\text{\u03c9}}_{{12}}}^{2}\\cos \\left( {{{\\text{\u03b8}}_{{12}}}-{{\\text{\u03b8}}_{{13}}}} \\right)-{{\\text{a}}_{3}}{{\\text{\u03c9}}_{{12}}}^{2}\\cos \\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)+{{\\text{a}}_{3}}{{\\text{\u03c9}}_{{13}}}^{2}}}{{{{\\text{a}}_{4}}\\sin \\left( {{{\\text{\u03b8}}_{{14}}}-{{\\text{\u03b8}}_{{13}}}} \\right)}} <\/span><\/p>\n

C noktas\u0131n\u0131n ivmesi ise C noktas\u0131n\u0131n h\u0131z vekt\u00f6r\u00fcn\u00fcn zamana g\u00f6re t\u00fcrevi al\u0131narak elde edilir.<\/p>\n

a<\/strong>C<\/sub> = ia2<\/sub>\u03b112<\/sub>ei\u03b812<\/sub><\/sup> \u2212 a2<\/sub>\u03c912<\/sub>2<\/sup>ei\u03b812<\/sub><\/sup>\u00a0+ ib3<\/sub>\u03b113<\/sub>ei(\u03b813 <\/sub>+\u00a0<\/sub>\u03b2)<\/sup>\u00a0\u2212 b3<\/sub>\u03c913<\/sub>2<\/sup>ei(\u03b813 <\/sub>+\u00a0<\/sub>\u03b2)<\/sup><\/p>\n

Dikkat edilir ise, devre kapal\u0131l\u0131k denklemi, h\u0131z ve ivme devre denklemleri \u00e7\u00f6z\u00fclm\u00fc\u015f ise, sa\u011f tarafta bulunan terimler bilinen de\u011ferlerdir. Bu denklem vekt\u00f6rel olarak<\/p>\n

a<\/strong>C<\/sub>\u00a0= an<\/sup><\/strong>A<\/sub>\u00a0+ at<\/sup><\/strong>A<\/sub>\u00a0+ an<\/sup><\/strong>C\/A<\/sub>\u00a0+ at<\/sup><\/strong>C\/A<\/sub><\/p>\n

denklemi ile ayn\u0131d\u0131r.\u00a0G\u00f6r\u00fcld\u00fc\u011f\u00fc gibi devre kapal\u0131l\u0131k denklemi ve t\u00fcrevleri olan h\u0131z ve ivme denklemleri \u00e7\u00f6z\u00fcld\u00fckten sonra mekanizmada bulunan her hangi bir uzvun \u00fczerinde bir noktan\u0131n konumu, h\u0131z\u0131 ve ivmesi kolayl\u0131kla belirlenebilir.<\/p>\n

\"\"<\/p>\n

Yukar\u0131da ki \u015fekilde d\u00f6rt-\u00e7ubuk mekanizmas\u0131 i\u00e7in h\u0131z ve ivme analizinin grafik \u00e7\u00f6z\u00fcm\u00fc g\u00f6sterilmi\u015ftir.<\/p>\n

H\u0131z ve ivme analizini kademe kademe incelemek i\u00e7in a\u015fa\u011f\u0131da verilen flash animasyonunu kullanabilirsiniz.<\/p>\n