{"id":1299,"date":"2021-09-09T22:18:20","date_gmt":"2021-09-09T22:18:20","guid":{"rendered":"https:\/\/blog.metu.edu.tr\/eresmech\/?page_id=1299"},"modified":"2021-09-27T20:46:57","modified_gmt":"2021-09-27T20:46:57","slug":"6-3","status":"publish","type":"page","link":"https:\/\/blog.metu.edu.tr\/eresmech\/mekanizma-teknigi\/ch6\/6-3\/","title":{"rendered":"6-3"},"content":{"rendered":"
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6.3 Konik Di\u015fli Kullanan Di\u015fli Kutular\u0131<\/h1>\n

Hareket iletimi paralel olmayan miller aras\u0131nda yap\u0131lacak ise, genellikle konik di\u015fliler kullan\u0131l\u0131r. Kinematik a\u00e7\u0131dan konik di\u015fliler birbiri \u00fczerinde yuvarlanan iki koni\u011fe e\u015fde\u011ferdir. Konik di\u015fliler basit veya planet di\u015fli sistemlerde kullan\u0131labilirler. Kinematik analizleri, iki fark d\u0131\u015f\u0131nda d\u00fcz di\u015fliler ile olu\u015fturulan di\u015fli sistemlerle ayn\u0131d\u0131r. Birinci \u00f6nemli fark burada anlat\u0131lacak olan y\u00f6ntem ekseni sabit olmayan planet di\u015flilerin a\u00e7\u0131sal h\u0131zlar\u0131n\u0131n hesaplanmas\u0131nda kullan\u0131lamaz, \u00e7\u00fcnk\u00fc uzayda paralel olmayan eksenler etraf\u0131nda d\u00f6nemler toplanamaz. \u0130kinci fark ise, d\u00fcz di\u015flilerde kullan\u0131lan a\u00e7\u0131sal y\u00f6n belirleme y\u00f6ntemi konik di\u015fliler i\u00e7in uygulanamaz.<\/p>\n

\"\"<\/p>\n

\u015eekilde birbirleri ile di\u015fli \u00e7ift olu\u015fturan iki konik di\u015fli g\u00f6r\u00fclmektedir. Eksenler aras\u0131nda a\u00e7\u0131 \u03b8 d\u0131r. P noktas\u0131nda sadece yuvarlanma oldu\u011fundan, iki uzuv \u00fczerinde h\u0131zlar e\u015fit olacakt\u0131r:<\/p>\n

\"\"<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0 \u03b8 = \u03b1 + \u03b2<\/p>\n

ve<\/p>\n

\\displaystyle \\frac{{{{\\text{\u03c9}}_{{13}}}}}{{{{\\text{\u03c9}}_{{12}}}}}=\\frac{{{\\text{r}_{2}}}}{{{\\text{r}_{3}}}}=\\frac{{{\\text{T}_{2}}}}{{{\\text{T}_{3}}}}=\\frac{{{{\\text{r}}_{2}}\/\\text{OP}}}{{{{\\text{r}}_{3}}\/\\text{OP}}}=\\frac{{\\sin \\text{\u03b1}}}{{\\sin \\text{\u03b2}}}={{\\text{R}}_{{23}}}<\/span><\/p>\n

d\u00fcr.\u00a0Uzuvlar\u0131n d\u00f6nme y\u00f6n\u00fcn\u00fc belirlemek i\u00e7in yine P noktas\u0131nda iki koninin h\u0131z\u0131n\u0131n ayn\u0131 olmas\u0131 gerekti\u011fini not edelim. E\u011fer 2 uzvu O noktas\u0131ndan bakt\u0131\u011f\u0131m\u0131zda saat yelkovan\u0131 y\u00f6n\u00fcnde d\u00f6n\u00fcyor ise, 3 uzvu O noktas\u0131ndan bakt\u0131\u011f\u0131m\u0131zda saat yelkovan\u0131na ters y\u00f6nde d\u00f6necektir. Bunu belirlemek i\u00e7in d\u00f6nen bir ok d\u00fc\u015f\u00fcnelim 2 uzvu saat yelkovan\u0131 y\u00f6n\u00fcnde d\u00f6nd\u00fc\u011f\u00fcnden bu ok \u015fekilde g\u00f6r\u00fclen kesit d\u00fczlemine P noktas\u0131ndan (koninin \u00fcst k\u0131sm\u0131ndan) girecek, koninin alt k\u0131sm\u0131ndan ise d\u0131\u015far\u0131 \u00e7\u0131kacakt\u0131r. Bir oka arkadan bakt\u0131\u011f\u0131m\u0131zda “+” \u00f6nden bakt\u0131\u011f\u0131m\u0131zda ise “\u2022<\/b>” g\u00f6r\u00fcr\u00fcz. \u00d6yle ise 2 uzvu i\u00e7in ok \u015fekilde g\u00f6sterildi\u011fi gibi yukar\u0131da + a\u015fa\u011f\u0131da ise\u00a0\u2022<\/b>\u00a0g\u00f6r\u00fclecektir. 3 uzvunda ise koninin P taraf\u0131nda “+” yani okun i\u00e7eriye do\u011fru oldu\u011fu bilindi\u011fine g\u00f6re, di\u011fer tarafta “\u2022<\/b>” yani okun ka\u011f\u0131d\u0131 delerek \u00e7\u0131kmas\u0131 gerekir. O noktas\u0131ndan bakt\u0131\u011f\u0131m\u0131zda ise 3 uzvu saat yelkovan\u0131 y\u00f6n\u00fcne ters d\u00f6n\u00fcyor g\u00f6r\u00fclecektir. Bu y\u00f6ntem ile uzuvlar\u0131n d\u00f6nme y\u00f6nleri belirlenebilir.<\/p>\n

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

\u015eekilde g\u00f6sterilen birle\u015fik basit di\u015fli zincirinde 4 uzvunun 2 uzvuna g\u00f6re a\u00e7\u0131sal h\u0131z oran\u0131n\u0131 (R24<\/sub>\u00a0= N24<\/sub>) bulunuz.<\/p>\n

\"\"<\/p>\n

\\displaystyle {{\\text{R}}_{{23}}}=\\frac{{80}}{{20}}=\\frac{{{\\text{\u03c9}_{{13}}}}}{{{\\text{\u03c9}_{{12}}}}}<\/span>\u00a0 \u00a0ve\u00a0 \u00a0 \\displaystyle {{\\text{R}}_{{34}}}=\\frac{{18}}{{60}}=\\frac{{{\\text{\u03c9}_{{14}}}}}{{{\\text{\u03c9}_{{13}}}}}<\/span><\/p>\n

veya<\/p>\n

\\displaystyle {{\\text{R}}_{{24}}}=\\frac{{{\\text{\u03c9}_{{14}}}}}{{{\\text{\u03c9}_{{12}}}}}=\\frac{{20\\cdot 18}}{{80\\cdot 60}}=\\frac{3}{40}=\\frac{{\\text{Tahrik eden di\u015fli di\u015f say\u0131lar\u0131 \u00e7arp\u0131m\u0131}}}{{\\text{Tahrik edilen di\u015fli di\u015f say\u0131lar\u0131 \u00e7arp\u0131m\u0131}}}<\/span><\/p>\n

0larak bulunur. Ancak bu hesaplamada d\u00fcz di\u015flilerde oldu\u011fu gibi y\u00f6n g\u00f6z \u00f6n\u00fcne al\u0131nmam\u0131\u015ft\u0131r. 2 uzvunun her hangi bir y\u00f6nde d\u00f6nd\u00fc\u011f\u00fcn\u00fc kabul edelim. \u00d6rne\u011fin sa\u011fdan bakt\u0131\u011f\u0131m\u0131zda 2 uzvu saat yelkovan\u0131 y\u00f6n\u00fcnde d\u00f6n\u00fcyor olsun. Bu d\u00f6nme y\u00f6n\u00fcn\u00fc g\u00f6steren ok di\u015flinin \u00fcst taraf\u0131ndan ka\u011f\u0131ttan d\u0131\u015far\u0131 \u00e7\u0131kacak, alt taraf\u0131ndan ise i\u00e7eri do\u011fru girecektir. Buna uygun olarak ilk i\u015faretimizi koydu\u011fumuzda, 3 uzvunun d\u00f6nme y\u00f6n\u00fc, P1<\/sub>\u00a0noktas\u0131nda okun i\u00e7eri do\u011fru girmesi, P2<\/sub>\u00a0noktas\u0131nda ise okun d\u0131\u015far\u0131 do\u011fru \u00e7\u0131kmas\u0131 ile neticelenir. 4 uzvu ise bu durumda P2<\/sub>\u00a0noktas\u0131nda ok d\u0131\u015far\u0131 \u00e7\u0131k\u0131yor ise, koninin \u00fcst taraf\u0131nda ok i\u00e7eri do\u011fru girecektir. \u00d6yle ise, 2 ve 4 uzuvlar\u0131 ayn\u0131 y\u00f6nde d\u00f6nmektedirler. \u015eekilde g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi (+) ve (\u2022<\/b>) sembollerini s\u0131ra ile bu kurala uygun yerle\u015ftirdi\u011fimizde uzuvlar\u0131n y\u00f6nleri belirlenmi\u015f olur. \u00d6yle ise:<\/p>\n

\\displaystyle {{\\text{R}}_{{24}}}=\\frac{{{\\text{\u03c9}_{{14}}}}}{{{\\text{\u03c9}_{{12}}}}}=\\frac{{20\\cdot 18}}{{80\\cdot 60}}=+\\frac{3}{40}=+0.075<\/span><\/p>\n

E\u011fer konik di\u015fliler planet di\u015flilerde kullan\u0131l\u0131yor ise, sabit eksenli di\u015fliler aras\u0131nda di\u015fli oran\u0131n\u0131, h\u0131zlar\u0131n kola g\u00f6re ba\u011f\u0131l h\u0131zlar\u0131na e\u015fitliyerek bulmam\u0131z m\u00fcmk\u00fcnd\u00fcr. Ancak planet di\u015flinin h\u0131z\u0131n\u0131 bu y\u00f6ntemle bulmak do\u011fru de\u011fildir.<\/p>\n

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

\"\"<\/p>\n

\u015eekilde g\u00f6sterilen di\u015fli sisteminde 2 ve 4 uzuvlar\u0131 aras\u0131nda a\u00e7\u0131sal h\u0131z oranlar\u0131n\u0131 ve bu a\u00e7\u0131sal h\u0131zlar\u0131n birbirlerine g\u00f6re y\u00f6n\u00fcn\u00fc bulmak istiyoruz. Dikkat edilir ise 3 uzvu planet di\u015flidir ve 2 uzvuna d\u00f6ner mafsalla ba\u011fl\u0131 oldu\u011fundan 2 uzvu koldur. 2 uzvunu sabit tuttu\u011fumuzda 1, 3 ve 4 uzuvlar\u0131 aras\u0131nda elde edilen basit di\u015fli sistem i\u00e7in di\u015fli oran\u0131:<\/p>\n

\\displaystyle {{\\text{R}}_{{14}}}=\\frac{{40\\cdot 42}}{{20\\cdot 18}}=\\frac{14}{3}=\\frac{{{{\\text{n}}_{{14}}}-{{\\text{n}}_{{12}}}}}{{{{\\text{n}}_{{11}}}-{{\\text{n}}_{{12}}}}} <\/span><\/p>\n

Bu basit di\u015fli sistem i\u00e7in (2 uzvu sabit varsay\u0131ld\u0131\u011f\u0131nda) d\u00f6nme y\u00f6nleri d\u00f6nme okunun y\u00f6n\u00fcne uygun olacak \u015fekilde i\u015faretlenerek belirlendi\u011finde, 1 ve 4 uzuvlar\u0131n\u0131n ayn\u0131 y\u00f6nde d\u00f6nd\u00fckleri g\u00f6r\u00fclmektedir. n11<\/sub>\u00a0= 0 oldu\u011fundan n14<\/sub>\u00a0i\u00e7in:<\/p>\n

n14<\/sub> = (1 \u2212 14\/3)n12<\/sub><\/p>\n

veya<\/p>\n

N24<\/sub>\u00a0= n14<\/sub>\/n12<\/sub>\u00a0=\u00a0\u221211\/3<\/p>\n

elde edilir. Dikkat edilir ise, sonu\u00e7ta 2 ve 4 uzuvlar\u0131 birbirlerine ters y\u00f6nde d\u00f6nmektedir.<\/p>\n

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

\"\"<\/p>\n

\u015eekilde “Humpage’in konik di\u015fli diferansiyeli” olarak adland\u0131r\u0131lan sistem g\u00f6r\u00fclmektedir. N24<\/sub>\u00a0h\u0131z oran\u0131n\u0131 bulmam\u0131z istenmektedir.<\/p>\n

2, 3 ve 1 uzvundan olu\u015fan zincir incelendi\u011finde, 3 uzvu planet di\u015fli ve 5 uzvu koldur. \u00d6yle ise:<\/p>\n

\\displaystyle {{\\text{R}}_{12}}=\\frac{{{{\\text{n}}_{12}}-{{\\text{n}}_{15}}}}{{{{\\text{n}}_{11}}-{{\\text{n}}_{15}}}}=-\\frac{{76\\cdot 56}}{{56\\cdot 20}}=-\\frac{19}{5} <\/span><\/p>\n

n11<\/sub> = 0 oldu\u011fundan n15<\/sub> = (5\/24)n12<\/sub><\/p>\n

1, 3 ve 4 uzvundan olu\u015fan zincirde 3 uzvu planet ve 5 uzvu onu ta\u015f\u0131yan koldur. \u00d6yle ise:<\/p>\n

\\displaystyle {{\\text{R}}_{14}}=\\frac{{{{\\text{n}}_{14}}-{{\\text{n}}_{15}}}}{{{{\\text{n}}_{11}}-{{\\text{n}}_{15}}}}=+\\frac{{76\\cdot 24}}{{56\\cdot 35}}=+\\frac{228}{245} <\/span><\/p>\n

yaz\u0131labilir. n11<\/sub> = 0 oldu\u011fundan n14<\/sub> = (1 \u2212 228\/245)n15<\/sub> = (17\/245)n15<\/sub> = (17\/1176)n12<\/sub> veya N24<\/sub> = n14<\/sub>\/n12<\/sub> = 0.0145\u00a0olacakt\u0131r.<\/p>\n

Bir diferansiyel nas\u0131l \u00e7al\u0131\u015f\u0131r?<\/strong><\/em><\/p>\n

T\u00fcm arabalarda viraj al\u0131rken d\u0131\u015farda bulunan teker daha fazla mesafe kat edece\u011finden daha i\u00e7erde bulunan tekere g\u00f6re daha h\u0131zl\u0131 d\u00f6necektir. Tahrik edilen tekerler ayn\u0131 motor taraf\u0131ndan tahrik edildi\u011finden e\u011fer tek serbestlik dereceli bir mekanizma kullan\u0131l\u0131r ise, her iki teker ayn\u0131 miktar d\u00f6nd\u00fcr\u00fclece\u011finden dolay\u0131, ara\u00e7 kayar. Bunu \u00f6nlemek i\u00e7in tekerlekler diferansiyel olarak adland\u0131r\u0131lan iki serbestlik dereceli bir konik planet di\u015fli sistemi ile tahrik edilirler.<\/p>\n

\"\"<\/p>\n

Viraj s\u0131ras\u0131nda teker d\u00f6nmesi<\/p>\n

\"\"<\/p>\n

Di\u015fli kutusundan \u00e7\u0131kan milin \u00e7evirdi\u011fi giri\u015f di\u015flisi planet di\u015fli sistemde kolu \u00e7evirmektedir. Kola d\u00f6ner mafsal ile planet di\u015fli ba\u011fl\u0131d\u0131r (genellikle 3 veya 4 planet di\u015fli bulunur). Sa\u011f ve sol teker di\u015flileri sabit ve \u00e7ak\u0131\u015fan bir eksende d\u00f6nmektedir. Sistem g\u00f6r\u00fcld\u00fc\u011f\u00fc \u015fekilde iki serbestlik derecesine sahiptir.<\/p>\n

E\u011fer iki tekere gelen y\u00fck e\u015fit ise t\u00fcm sistem, kolun d\u00f6nme a\u00e7\u0131s\u0131 h\u0131z\u0131nda birlikte d\u00f6ner.<\/p>\n

E\u011fer tekerlerden birine etki eden y\u00fck di\u011fer tekere gelen y\u00fcke g\u00f6re \u00e7ok fazla ise, bir teker sabit durur ve di\u011fer teker iki misli h\u0131z ile d\u00f6ner. Virajlarda i\u00e7 teker d\u0131\u015f tekere g\u00f6re daha fazla s\u00fcrt\u00fcnme kuvveti ile kar\u015f\u0131laca\u011f\u0131ndan i\u00e7 teker d\u0131\u015f tekere g\u00f6re daha az d\u00f6necektir.<\/p>\n

Sistem iki serbestlik dereceli oldu\u011fundan uygulamada tekerleklerin d\u00f6nme miktar\u0131 etki eden d\u0131\u015f kuvvetlere ve sistemin dinami\u011fine ba\u011fl\u0131 olacakt\u0131r.<\/p>\n

Benzer diferansiyel sistemi bir engele \u00e7arpt\u0131\u011f\u0131nda y\u00f6n de\u011fi\u015ftiren oyuncak arabalarda g\u00f6r\u00fcrs\u00fcn\u00fcz.<\/p>\n

G\u00fcneyi G\u00f6steren Sava\u015fc\u0131 (<\/strong><\/i><\/span>Milattan \u00f6nce 2600 y\u0131llar\u0131nda):<\/em><\/strong><\/p>\n

A\u015fa\u011f\u0131da g\u00f6r\u00fclen araba \u00c7inliler taraf\u0131ndan Gobi \u00e7\u00f6l\u00fcn\u00fc ge\u00e7erken kullan\u0131lan \u00f6k\u00fczler taraf\u0131ndan \u00e7ekilen bir araban\u0131n temsili resmidir. Di\u015fliler g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi zaman\u0131nda tahta pimlerden yap\u0131lm\u0131\u015ft\u0131r. Ancak o devirdede planet di\u015fli sistemi ve diferansiyel kavram\u0131n\u0131 bildikleri g\u00f6r\u00fclmektedir. Araba d\u00fcz giderken sava\u015fc\u0131n\u0131n y\u00f6n\u00fc de\u011fi\u015fmemekte, ara\u00e7 sa\u011fa veya sola d\u00f6nd\u00fc\u011f\u00fcnde, ayn\u0131 a\u00e7\u0131da sava\u015fc\u0131 arabaya g\u00f6re ters y\u00f6nde d\u00f6nmekte ve yere g\u00f6re hi\u00e7 bir a\u00e7\u0131sal d\u00f6nme yapmam\u0131\u015f olmaktad\u0131r. Bu \u015fekilde sava\u015fc\u0131 yola ba\u015flarken g\u00f6sterdi\u011fi y\u00f6n\u00fc devaml\u0131 g\u00f6stermektedir.<\/p>\n

\"\"<\/i><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>\n\n\n

\"\"<\/a>\"\"<\/a>\"\"<\/a>\"\"<\/a>\"\" <\/p>\n","protected":false},"excerpt":{"rendered":"

6.3 Konik Di\u015fli Kullanan Di\u015fli Kutular\u0131 Hareket iletimi paralel olmayan miller aras\u0131nda yap\u0131lacak ise, genellikle konik di\u015fliler kullan\u0131l\u0131r. Kinematik a\u00e7\u0131dan konik di\u015fliler birbiri \u00fczerinde yuvarlanan iki koni\u011fe e\u015fde\u011ferdir. Konik di\u015fliler basit veya planet di\u015fli sistemlerde kullan\u0131labilirler. Kinematik analizleri, iki fark …<\/p>\n

6-3<\/span> Devam\u0131n\u0131 Oku »<\/a><\/p>\n","protected":false},"author":7747,"featured_media":0,"parent":1292,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"class_list":["post-1299","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1299","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/users\/7747"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/comments?post=1299"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1299\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/pages\/1292"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/eresmech\/wp-json\/wp\/v2\/media?parent=1299"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}