{"id":2284,"date":"2026-06-23T07:12:36","date_gmt":"2026-06-23T07:12:36","guid":{"rendered":"https:\/\/helicalcutgears.top\/?p=2284"},"modified":"2026-06-23T07:12:36","modified_gmt":"2026-06-23T07:12:36","slug":"helical-gear-ratio-centre-distance-calculation","status":"publish","type":"post","link":"https:\/\/helicalcutgears.top\/tr\/helical-gear-ratio-centre-distance-calculation\/","title":{"rendered":"Helisel Di\u015fli Oran\u0131, Merkez Mesafesi ve Di\u015fli Sistemi Hesaplamalar\u0131 \u2014 Eksiksiz \u00c7\u00f6z\u00fcml\u00fc K\u0131lavuz"},"content":{"rendered":"
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Helisel Di\u015fli Oran\u0131, Merkez Mesafesi ve Di\u015fli Sistemi Hesaplamalar\u0131 \u2014 Eksiksiz \u00c7\u00f6z\u00fcml\u00fc K\u0131lavuz<\/h1>\n

Helisel di\u015fli tahrik sistemlerinde di\u015fli oranlar\u0131n\u0131, merkez mesafelerini ve di\u015f say\u0131lar\u0131n\u0131 hesaplamak, tek tek basit olan ancak daha sonra bir boyut de\u011fi\u015ftirildi\u011finde hatalara neden olabilecek \u015fekilde etkile\u015fime giren bir dizi form\u00fcl i\u00e7erir. Bu k\u0131lavuz, eksiksiz \u00fc\u00e7 a\u015famal\u0131 bir di\u015fli kutusu \u00f6rne\u011fiyle her form\u00fcl\u00fc kapsar ve ters hesaplamay\u0131 da i\u00e7erir; yani gerekli orandan ve sabit g\u00f6vde merkez mesafesinden di\u015f say\u0131s\u0131n\u0131 bulmay\u0131.<\/p>\n

Di\u015fli Sistemi Tasar\u0131m\u0131 Deste\u011fi Talep Et \u2192<\/a><\/p>\n<\/div>\n<\/div>\n

Helisel Di\u015fli Hesaplamalar\u0131 i\u00e7in Temel Form\u00fcl Seti<\/h2>\n

T\u00fcm helisel di\u015fli<\/strong> Hesaplamalar d\u00f6rt temel ili\u015fkiden t\u00fcretilir. Bu d\u00f6rt ili\u015fkiyi ezberlemek ve aralar\u0131ndaki ba\u011flant\u0131lar\u0131 anlamak, \u00e7o\u011fu pratik tasar\u0131m durumunda tablolara olan ihtiyac\u0131 ortadan kald\u0131r\u0131r:<\/p>\n

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D\u00f6rt Temel Helisel Di\u015fli Form\u00fcl\u00fc<\/p>\n

1. Hat \u00e7ap\u0131: d = Mn \u00d7 z \/ cos \u03b2<\/p>\n

2. Merkez mesafesi: C = Mn \u00d7 (z\u2081 + z\u2082) \/ (2 \u00d7 cos \u03b2)<\/p>\n

3. H\u0131z (di\u015fli) oran\u0131: i = z\u2082 \/ z\u2081 = n\u2081 \/ n\u2082 = T\u2082 \/ T\u2081<\/p>\n

4. \u00c7\u0131k\u0131\u015f torku: T\u2082 = 9550 \u00d7 P \u00d7 \u03b7 \/ n\u2082 [N\u00b7m, kW, RPM]<\/p>\n

Temel kar\u015f\u0131l\u0131kl\u0131 ba\u011f\u0131ml\u0131l\u0131k:<\/strong> \u00c7\u00fcnk\u00fc d = Mn \u00d7 z \/ cos \u03b2 (d\u00fcz di\u015flilerdeki gibi Mn \u00d7 z de\u011fil), ayn\u0131 Mn ve z'yi korurken \u03b2'y\u0131 de\u011fi\u015ftirmek hem d'yi hem de C'yi de\u011fi\u015ftirir. helisel di\u015fli<\/strong> Belirli bir merkez mesafesi C etraf\u0131nda tasarlanan di\u015fli kutusu g\u00f6vdesi, d\u00fcz di\u015fli tasar\u0131m\u0131n\u0131 helisel di\u015fliye d\u00f6n\u00fc\u015ft\u00fcr\u00fcrken s\u0131k kar\u015f\u0131la\u015f\u0131lan bir hata kayna\u011f\u0131 olan cos \u03b2 terimini hesaba katmal\u0131d\u0131r.<\/div>\n<\/div>\n

H\u0131z Oran\u0131 Hesaplamas\u0131 \u2014 Tek Kademeli ve \u00c7ok Kademeli<\/h2>\n

Tek Kademeli Helisel Di\u015fli Oran\u0131<\/h3>\n

Tek bir ki\u015fi i\u00e7in helisel di\u015fli<\/strong> \u00e7ift \u200b\u200b(pinyon tahrik di\u015flisi):<\/p>\n

i = z\u2082 \/ z\u2081 (her zaman tahrik edilen di\u015flinin di\u015f say\u0131s\u0131, tahrik eden di\u015fliye b\u00f6l\u00fcn\u00fcr)<\/p>\n

\u00c7\u0131k\u0131\u015f h\u0131z\u0131: n\u2082 = n\u2081 \/ i. \u00c7\u0131k\u0131\u015f torku (kay\u0131plar ihmal edilerek): T\u2082 = T\u2081 \u00d7 i. A\u011f verimlili\u011fi \u03b7 dahil (tipik olarak topraklama i\u00e7in 0,985\u20130,995) helisel di\u015fli<\/strong> sahne):<\/p>\n

T\u2082 = T\u2081 \u00d7 i \u00d7 \u03b7\u2081<\/p>\n

\u00c7ok Kademeli Helisel Di\u015fli Sistemi<\/h3>\n

K a\u015famas\u0131 i\u00e7in helisel di\u015fli<\/strong> Di\u015fli kutusunda, toplam oran t\u00fcm kademe oranlar\u0131n\u0131n \u00e7arp\u0131m\u0131d\u0131r ve toplam verimlilik t\u00fcm kademe verimliliklerinin \u00e7arp\u0131m\u0131d\u0131r:<\/p>\n

i_toplam = i\u2081 \u00d7 i\u2082 \u00d7 i\u2083 \u00d7 \u2026 \u00d7 i_k<\/p>\n

\u03b7_toplam = \u03b7\u2081 \u00d7 \u03b7\u2082 \u00d7 \u03b7\u2083 \u00d7 \u2026 \u00d7 \u03b7_k<\/p>\n

T_\u00e7\u0131kt\u0131 = 9550 \u00d7 P \u00d7 \u03b7_toplam \/ n_\u00e7\u0131kt\u0131 [N\u00b7m]<\/p>\n

Minimum \u015eanz\u0131man Boyutu \u0130\u00e7in Oran\u0131n A\u015famalar Aras\u0131nda Nas\u0131l Da\u011f\u0131t\u0131laca\u011f\u0131<\/h3>\n

3 a\u015famal\u0131 bir i\u015flem i\u00e7in helisel di\u015fli<\/strong> Toplam oran\u0131 i = 40 olan bir di\u015fli kutusunda, oran\u0131n 8:1 \u00d7 5:1 \u00d7 1:1 \u015feklinde da\u011f\u0131t\u0131lmas\u0131 verimsizdir; birinci kademe di\u015flisi (pinyon \u00e7ap\u0131n\u0131n 8 kat\u0131) g\u00f6vde boyutuna hakim olurken, \u00fc\u00e7\u00fcnc\u00fc kademe hi\u00e7bir \u015fey eklemez. Minimum boyut da\u011f\u0131l\u0131m\u0131nda yakla\u015f\u0131k olarak e\u015fit kademe oranlar\u0131 vard\u0131r: 3,4:1 \u00d7 3,4:1 \u00d7 3,4:1 \u2248 39,3 \u2014 hedef 40'a yak\u0131n. Pratik kural: kademe oranlar\u0131 birbirine 1,5-2 kat yak\u0131n olmal\u0131d\u0131r. Geometrik ilerleme (her kademe oran\u0131 = i_toplam^(1\/k)) minimum g\u00f6vde hacmi i\u00e7in matematiksel optimumdur.<\/p>\n

Merkez Mesafesi Hesaplamas\u0131 \u2014 Konutun Kritik Boyutu<\/h2>\n

Merkez mesafesi C, bir motorun giri\u015f mili merkez \u00e7izgisi ile \u00e7\u0131k\u0131\u015f mili merkez \u00e7izgisi aras\u0131ndaki mesafedir. helisel di\u015fli<\/strong> A\u015fama. Muhafaza taraf\u0131ndan belirlenir ve muhafazan\u0131n yeniden i\u015flenmesi olmadan de\u011fi\u015ftirilemez. Merkez mesafesi form\u00fcl\u00fc:<\/p>\n

C = Mn \u00d7 (z\u2081 + z\u2082) \/ (2 \u00d7 cos \u03b2)<\/p>\n

cos \u03b2 terimine dikkat edin: a helisel di\u015fli<\/strong> Ayn\u0131 Mn ve di\u015f say\u0131s\u0131na sahip d\u00fcz di\u015fli \u00e7iftine k\u0131yasla, bir di\u015fli \u00e7iftinin merkez mesafesi daha b\u00fcy\u00fckt\u00fcr. Mn = 5, z\u2081 = 20, z\u2082 = 60 ve \u03b2 = 25\u00b0 i\u00e7in:<\/p>\n

C = 5 \u00d7 (20 + 60) \/ (2 \u00d7 cos 25\u00b0) = 5 \u00d7 80 \/ (2 \u00d7 0,906) = 400 \/ 1,812 = 220,8 mm<\/p>\n

E\u015fde\u011fer d\u00fcz di\u015fli \u00e7iftinin C = 5 \u00d7 80 \/ 2 = 200 mm \u2014 10,4% daha k\u00fc\u00e7\u00fck olmas\u0131 gerekir. Bu, tasar\u0131mc\u0131lar\u0131n sonradan uyarlama yapmaya \u00e7al\u0131\u015f\u0131rken s\u0131k\u00e7a yapt\u0131\u011f\u0131 bir hesaplama hatas\u0131d\u0131r. helisel di\u015fliler<\/strong> Mevcut bir d\u00fcz di\u015fli muhafazas\u0131na tak\u0131ld\u0131\u011f\u0131nda: \u03b2 > 0\u00b0 ise ayn\u0131 Mn ve di\u015f say\u0131lar\u0131 mevcut merkez mesafesine uymayacakt\u0131r.<\/p>\n

Standart d\u0131\u015f\u0131 merkez mesafesi i\u00e7in profil kaymas\u0131 (ek de\u011fi\u015fiklik):<\/strong> E\u011fer g\u00f6vde merkez mesafesi C_housing, yukar\u0131da hesaplanan standart C de\u011ferinden yakla\u015f\u0131k \u00b15%'den daha az farkl\u0131l\u0131k g\u00f6steriyorsa, profil kaymas\u0131 (x\u2081 + x\u2082 \u2260 0), di\u015f say\u0131s\u0131n\u0131 veya mod\u00fcl\u00fc de\u011fi\u015ftirmeden \u00e7al\u0131\u015fma merkez mesafesini g\u00f6vdeye uyacak \u015fekilde ayarlayabilir. Korea Ever-Power, tasar\u0131m inceleme s\u00fcrecinin bir par\u00e7as\u0131 olarak, standart olmayan herhangi bir merkez mesafesi i\u00e7in do\u011fru profil kaymas\u0131 katsay\u0131lar\u0131n\u0131 hesaplar.<\/div>\n

Gerekli Torktan Mod\u00fcl Se\u00e7imi \u2014 Yeni Bir Tasar\u0131ma Ba\u015flama<\/h2>\n

Basitle\u015ftirilmi\u015f Mod\u00fcl Boyutland\u0131rma Form\u00fcl\u00fc<\/h3>\n

Yeni bir helisel di\u015fli<\/strong> Mod\u00fcl\u00fcn hen\u00fcz belirlenmedi\u011fi tasar\u0131mlarda, ISO 6336 di\u015f k\u00f6k\u00fc e\u011filme dayan\u0131m\u0131na dayal\u0131 basitle\u015ftirilmi\u015f bir ba\u015flang\u0131\u00e7 \u200b\u200bform\u00fcl\u00fc, iletilen tork, \u015faft h\u0131zlar\u0131 ve malzeme kalitesinden gerekli mod\u00fcl\u00fc verir:<\/p>\n

Mn_min \u2248 1,5 \u00d7 \u221b(T\u2081 \u00d7 KA \/ (z\u2081 \u00d7 b\/d \u00d7 \u03c3_F_lim \/ SF))^(1\/3) [mm, N\u00b7m]<\/p>\n

Bu basitle\u015ftirilmi\u015f form yaln\u0131zca ilk tahmini vermektedir; tam ISO 6336 hesaplamas\u0131, dinamik y\u00fck fakt\u00f6r\u00fc K_v, y\u00fczey y\u00fck da\u011f\u0131l\u0131m\u0131 K_H\u03b2 ve hem k\u00f6k e\u011filmesi hem de temas yorulmas\u0131n\u0131 kapsamaktad\u0131r. Korea Ever-Power'\u0131n m\u00fchendislik ekibi, t\u00fcm \u00f6zel sipari\u015fler i\u00e7in tam hesaplamay\u0131 ger\u00e7ekle\u015ftirir. Standart uygulamalar i\u00e7in, giri\u015f g\u00fcc\u00fcne g\u00f6re a\u015fa\u011f\u0131daki deneysel mod\u00fcl aral\u0131klar\u0131, 20CrMnTi karb\u00fcrlenmi\u015f i\u00e7in durumlar\u0131n \u00e7o\u011funu kapsamaktad\u0131r. helisel di\u015fliler<\/strong> \u03b2 = 20\u00b0'de, SF = 1,3:<\/p>\n

\n\n\n\n\n\n\n\n\n\n\n
n\u2081 = 1500 RPM'de Giri\u015f G\u00fcc\u00fc (kW)<\/th>\nBa\u015flang\u0131\u00e7 \u200b\u200bMod\u00fcl\u00fc Tahmini (Mn)<\/th>\nTipik Y\u00fcz Geni\u015fli\u011fi b<\/th>\nTipik Pinyon Di\u015f Say\u0131s\u0131 z\u2081<\/th>\n<\/tr>\n<\/thead>\n
5\u201315 kW<\/td>\nM2\u2013M3<\/td>\n20\u201340 mm<\/td>\n18\u201325<\/td>\n<\/tr>\n
15\u201350 kW<\/td>\nM3\u2013M5<\/td>\n30\u201360 mm<\/td>\n20\u201328<\/td>\n<\/tr>\n
50\u2013150 kW<\/td>\nM4\u2013M6<\/td>\n50\u201390 mm<\/td>\n20\u201330<\/td>\n<\/tr>\n
150\u2013500 kW<\/td>\nM5\u2013M8<\/td>\n80\u2013140 mm<\/td>\n22\u201332<\/td>\n<\/tr>\n
500\u20132000 kW<\/td>\nM8\u2013M14<\/td>\n120\u2013240 mm<\/td>\n22\u201335<\/td>\n<\/tr>\n
2000\u201310.000 kW<\/td>\nM12\u2013M22<\/td>\n200\u2013450 mm<\/td>\n20\u201330<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

\u00dc\u00e7 A\u015famal\u0131 \u015eanz\u0131man\u0131n Tam \u00c7al\u0131\u015fma \u00d6rne\u011fi<\/h2>\n
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75 kW, 1500\u219238 RPM g\u00fcc\u00fcnde, 3 kademeli s\u0131ral\u0131 helisel di\u015fli kutusu tasarlay\u0131n:<\/p>\n

Ad\u0131m 1: Toplam oran<\/p>\n

i_total = n_input \/ n_output = 1500 \/ 38 = 39.5<\/p>\n

Ad\u0131m 2: A\u015fama oran\u0131 da\u011f\u0131l\u0131m\u0131 (geometrik)<\/p>\n

i_per_stage = 39.5^(1\/3) = 3.41 per stage<\/p>\n

Kullan\u0131m: A\u015fama 1 = 3,55, A\u015fama 2 = 3,35, A\u015fama 3 = 3,32 \u2192 toplam = 3,55 \u00d7 3,35 \u00d7 3,32 = 39,5 \u2713<\/p>\n

Ad\u0131m 3: \u015eaft h\u0131zlar\u0131<\/p>\n

n\u2081 = 1500 RPM n\u2082 = 1500\/3,55 = 422 RPM n\u2083 = 422\/3,35 = 126 RPM n\u2084 = 126\/3,32 = 38 RPM<\/p>\n

Ad\u0131m 4: \u00c7\u0131k\u0131\u015f torku (\u03b7_total = 0,97^3 \u2248 0,915, 3 a\u015fama i\u00e7in)<\/p>\n

T\u2084 = 9550 \u00d7 75 \u00d7 0,915 \/ 38 = 17.278 N\u00b7m \u2248 17,3 kN\u00b7m<\/p>\n

Ad\u0131m 5: Mod\u00fcl se\u00e7imi (1500 RPM'de A\u015fama 1, P=75 kW)<\/p>\n

Pinyon torku T\u2081 = 9550 \u00d7 75 \/ 1500 = 477 N\u00b7m \u2192 tablodan: Mn = M4\u2013M6 \u2192 M5'i se\u00e7in<\/p>\n

126 RPM'de 3. A\u015fama, P = 75 \u00d7 0,97\u00b2 \u2248 70,6 kW \u2192 Pinyon T\u2083 = 9550 \u00d7 70,6 \/ 126 = 5.348 N\u00b7m \u2192 Mn = M8\u2013M10 \u2192 M8'i se\u00e7in<\/p>\n

Ad\u0131m 6: A\u015fama 1 i\u00e7in di\u015f say\u0131lar\u0131 (i = 3,55, Mn = 5, \u03b2 = 20\u00b0)<\/p>\n

z\u2081 = 24, z\u2082 = 24 \u00d7 3,55 = 85,2 \u2192 z\u2082'yi 85'e yuvarlarsak, ger\u00e7ek i = 85\/24 = 3,542<\/p>\n

D\u00fczeltilmi\u015f zincir: 1500 \u2192 423.1 \u2192 126.3 \u2192 38.1 RPM (hata <0.3%)<\/p>\n

Ad\u0131m 7: Merkez mesafesi A\u015fama 1<\/p>\n

C\u2081 = 5 \u00d7 (24 + 85) \/ (2 \u00d7 cos 20\u00b0) = 5 \u00d7 109 \/ 1.879 = 290.1 \u200b\u200bmm<\/p>\n<\/div>\n

Sabit Merkez Mesafesinden Di\u015f Say\u0131lar\u0131n\u0131n Geriye Do\u011fru Hesaplanmas\u0131<\/h2>\n

Sabit merkez mesafesi C olan mevcut bir di\u015fli kutusundaki di\u015flileri de\u011fi\u015ftirirken, zorluk, se\u00e7ilen Mn ve \u03b2 de\u011ferleri verildi\u011finde, gerekli i oran\u0131n\u0131 veren ve sabit C'ye uyan di\u015f say\u0131s\u0131 z\u2081 ve z\u2082'yi bulmakt\u0131r. \u0130\u015flem \u015fu \u015fekildedir:<\/p>\n

C = Mn \u00d7 (z\u2081 + z\u2082) \/ (2 cos \u03b2'dan): z\u2081 + z\u2082 = 2C \u00d7 cos \u03b2 \/ Mn<\/p>\n

i = z\u2082 \/ z\u2081 denkleminden: z\u2082 = i \u00d7 z\u2081<\/p>\n

Birle\u015ftirme: z\u2081 \u00d7 (1 + i) = 2C cos \u03b2 \/ Mn \u2192 z\u2081 = 2C cos \u03b2 \/ (Mn \u00d7 (1 + i))<\/p>\n

Sonu\u00e7 tam say\u0131ya yuvarlanmal\u0131d\u0131r. Yuvarlanm\u0131\u015f de\u011ferler tam olarak istenen oran\u0131 vermezse, ger\u00e7ek oran i_actual = z\u2082_rounded \/ z\u2081_rounded, i_target'tan biraz farkl\u0131 olur. \u00c7o\u011fu de\u011fi\u015ftirme uygulamas\u0131 i\u00e7in \u00b11\u20132% oran tolerans\u0131 kabul edilebilir. Daha s\u0131k\u0131 oran do\u011frulu\u011fu gerekiyorsa, profil kayd\u0131rma, yuvarlanm\u0131\u015f di\u015f say\u0131lar\u0131n\u0131 tam olarak kar\u015f\u0131layacak \u015fekilde \u00e7al\u0131\u015fma merkezi mesafesini ayarlayabilir.<\/p>\n

Korea Ever-Power \u2014 Standart M\u00fchendislik Hizmetinin Bir Par\u00e7as\u0131 Olarak Di\u015fli Sistemi Hesaplamalar\u0131<\/h2>\n

\"Kore<\/p>\n

Korea Ever-Power, her \u00fcr\u00fcn\u00fcn\u00fcn bir par\u00e7as\u0131 olarak eksiksiz di\u015fli sistemi hesaplama deste\u011fi sunmaktad\u0131r. helisel kesme di\u015flisi<\/strong> Sipari\u015f \u2014 \u00fcretim ba\u015flamadan \u00f6nce di\u015f say\u0131s\u0131, merkez mesafesi uyumlulu\u011fu, di\u015fli oran\u0131 do\u011frulu\u011fu ve \u00e7\u0131k\u0131\u015f torkunun onaylanmas\u0131.<\/p>\n

Korea Ever-Power'\u0131n m\u00fchendislik ekibi, her \u00f6zel sipari\u015f i\u00e7in eksiksiz di\u015fli sistemi hesaplamalar\u0131 ger\u00e7ekle\u015ftirir. helisel kesme di\u015flisi<\/a> Sipari\u015f, di\u015f say\u0131lar\u0131n\u0131n, merkez mesafesi uyumlulu\u011funun, di\u015fli oran\u0131 do\u011frulu\u011funun, tork ve h\u0131zdan mod\u00fcl se\u00e7iminin ve y\u00fcz geni\u015fli\u011fi yeterlili\u011finin onaylanmas\u0131n\u0131 i\u00e7erir. Do\u011frudan bir sipari\u015f olarak. helisel di\u015fli \u00fcreticisi<\/a> 10.000'den fazla ile helisel di\u015fli<\/strong> Korea Ever-Power ekibi, sahip oldu\u011fu deneyim sayesinde, \u00fcretim ba\u015flamadan \u00f6nce hesaplama hatalar\u0131n\u0131 ve teknik \u00f6zellik uyu\u015fmazl\u0131klar\u0131n\u0131 tespit ederek, di\u015fli tahrik sistemleri tedarikinde en yayg\u0131n ve pahal\u0131 hatan\u0131n \u00f6n\u00fcne ge\u00e7iyor: teknik olarak \u00e7izime uygun ancak g\u00f6vdeye uymayan veya belirtilen \u00e7\u0131k\u0131\u015f h\u0131z\u0131na ula\u015famayan di\u015flilerin teslim al\u0131nmas\u0131.<\/p>\n

S\u0131k\u00e7a Sorulan Sorular<\/h2>\n
Mod\u00fcl ve di\u015f say\u0131s\u0131n\u0131 ayn\u0131 tutsam bile, helis a\u00e7\u0131s\u0131n\u0131 de\u011fi\u015ftirmek neden merkez mesafesini de\u011fi\u015ftiriyor?<\/strong><\/p>\n

Merkez uzakl\u0131\u011f\u0131 form\u00fcl\u00fc C = Mn \u00d7 (z\u2081 + z\u2082) \/ (2 \u00d7 cos \u03b2), paydada cos \u03b2 i\u00e7erir. \u03b2'y\u0131 art\u0131rmak cos \u03b2'y\u0131 azalt\u0131r, bu da C'yi art\u0131r\u0131r. \u00d6rne\u011fin, Mn = 5, z\u2081 + z\u2082 = 80 oldu\u011funda: \u03b2 = 0\u00b0'de (\u00e7\u0131k\u0131nt\u0131) C = 200 mm; \u03b2 = 20\u00b0'de C = 212,8 mm; \u03b2 = 25\u00b0'de C = 220,8 mm. Bu, bir helis a\u00e7\u0131s\u0131n\u0131 basit\u00e7e art\u0131ramayaca\u011f\u0131n\u0131z anlam\u0131na gelir. helisel di\u015fli<\/strong> Konut merkez mesafesinin de de\u011fi\u015fmesi gerekip gerekmedi\u011fi veya \u00e7al\u0131\u015fma merkez mesafesini orijinal konut boyutuna geri getirmek i\u00e7in profil kayd\u0131rma (ek de\u011fi\u015fiklik) uygulanmas\u0131 gerekip gerekmedi\u011fi kontrol edilmeden.<\/p>\n<\/div>\n

Helisel di\u015flide alt kesmeyi \u00f6nlemek i\u00e7in minimum pinyon di\u015f say\u0131s\u0131 ka\u00e7t\u0131r?<\/strong><\/p>\n

Standart bir protez i\u00e7in minimum di\u015f say\u0131s\u0131 helisel di\u015fli<\/strong> Alt kesme olmaks\u0131z\u0131n yakla\u015f\u0131k olarak: z_min \u2248 17 \/ cos\u00b3\u03b2 (\u03b1_n = 20\u00b0 i\u00e7in). \u03b2 = 0\u00b0'de (d\u00fcz di\u015fli), z_min \u2248 17. \u03b2 = 20\u00b0'de, z_min \u2248 17 \/ cos\u00b3(20\u00b0) = 17 \/ 0,830 \u2248 20,5 \u2192 21'e yuvarlan\u0131r. \u03b2 = 30\u00b0'de, z_min \u2248 17 \/ 0,650 \u2248 26. Bu, daha y\u00fcksek helis a\u00e7\u0131s\u0131na sahip helisel di\u015flilerin, d\u00fcz di\u015flilere g\u00f6re daha y\u00fcksek minimum di\u015f say\u0131s\u0131na sahip oldu\u011fu anlam\u0131na gelir; bu, pinyon di\u015f say\u0131s\u0131n\u0131n aksi takdirde alt kesme s\u0131n\u0131r\u0131n\u0131n alt\u0131na d\u00fc\u015febilece\u011fi kompakt y\u00fcksek oranl\u0131 tasar\u0131mlar i\u00e7in \u00f6nemlidir.<\/p>\n<\/div>\n

Sadece giri\u015f ve \u00e7\u0131k\u0131\u015f devir say\u0131s\u0131n\u0131 bildi\u011fimde, \u00e7ok kademeli bir \u015fanz\u0131man\u0131n di\u015fli oran\u0131n\u0131 nas\u0131l hesaplar\u0131m?<\/strong><\/p>\n

i_total = n_input \/ n_output. Yukar\u0131daki \u00f6rnekte: i = 1500 \/ 38 = 39,47. Di\u015f say\u0131lar\u0131ndan ayr\u0131 ayr\u0131 kademe oranlar\u0131n\u0131 bulmak i\u00e7in, \u015fanz\u0131mandaki t\u00fcm di\u015flilerin di\u015f say\u0131lar\u0131n\u0131 \u00f6l\u00e7\u00fcn (k\u00fc\u00e7\u00fck di\u015fliler i\u00e7in do\u011frudan say\u0131n; do\u011frudan sayman\u0131n zor oldu\u011fu b\u00fcy\u00fck di\u015fliler i\u00e7in di\u015fli analiz\u00f6r\u00fc kullan\u0131n). Her kademe i\u00e7in kademe oran\u0131 = z_driven \/ z_driver. T\u00fcm kademe oranlar\u0131n\u0131n \u00e7arp\u0131m\u0131, \u00b10,5% i\u00e7inde i_total'e e\u015fit olmal\u0131d\u0131r (di\u015f say\u0131lar\u0131n\u0131n yuvarlanmas\u0131 k\u00fc\u00e7\u00fck bir sapmaya neden olur).<\/p>\n<\/div>\n

Korea Ever-Power, merkez mesafesi \u00f6l\u00e7\u00fcm\u00fcnden ve gerekli orandan yola \u00e7\u0131karak di\u015f say\u0131s\u0131n\u0131 tersine hesaplayabilir mi?<\/strong><\/p>\n

Evet, bu standart bir hizmettir. \u015eunlar\u0131 sa\u011flay\u0131n: merkez mesafesi C (g\u00f6vdeden \u00f6l\u00e7\u00fcl\u00fcr), gerekli di\u015fli oran\u0131 i (giri\u015f\/\u00e7\u0131k\u0131\u015f mili h\u0131z \u00f6l\u00e7\u00fcm\u00fcnden veya etiketten), helis a\u00e7\u0131s\u0131 \u03b2 (di\u015fli analiz\u00f6r\u00fc ile mevcut di\u015fliden \u00f6l\u00e7\u00fcl\u00fcr) ve normal mod\u00fcl Mn (di\u015f say\u0131s\u0131 ve d\u0131\u015f \u00e7ap \u00f6l\u00e7\u00fcm\u00fcnden hesaplan\u0131r). Korea Ever-Power, C'ye uyan ve kabul edilebilir yuvarlama tolerans\u0131 i\u00e7inde i veren tam di\u015f say\u0131s\u0131 kombinasyonunu hesaplar, gerekli profil kaymas\u0131n\u0131 belirtir ve yedek par\u00e7ay\u0131 \u00fcretir. helisel di\u015fli<\/strong> Birbirine uyumlu bir \u00e7ift olarak.<\/p>\n<\/div>\n

\n

Di\u015fli tak\u0131m\u0131 hesaplamas\u0131 deste\u011fine mi ihtiyac\u0131n\u0131z var?<\/h2>\n

Giri\u015f h\u0131z\u0131n\u0131z\u0131, \u00e7\u0131k\u0131\u015f h\u0131z\u0131n\u0131z\u0131, g\u00fcc\u00fcn\u00fcz\u00fc, g\u00f6vde merkez mesafesini ve helis a\u00e7\u0131s\u0131n\u0131 belirtin. Korea Ever-Power, di\u015f say\u0131s\u0131n\u0131 hesaplar, oran do\u011frulu\u011funu do\u011frular, merkez mesafesi uyumlulu\u011funu kontrol eder ve 24 i\u015f saati i\u00e7inde teknik \u00f6zellikleri olu\u015fturur.<\/p>\n

Hesaplama Deste\u011fi Talebi<\/a>
\nHelisel Di\u015fli \u00dcr\u00fcn Yelpazesi<\/a><\/div>\n

Oran hesaplamas\u0131 \u00b7 Merkez mesafesi kontrol\u00fc \u00b7 Di\u015f say\u0131s\u0131 optimizasyonu \u00b7 Profil kaymas\u0131 \u00b7 Uygulama \u00f6rne\u011fi \u00b7 Minimum sipari\u015f miktar\u0131 (MOQ) 1 adet<\/p>\n<\/div>\n

Edit\u00f6r: Cxm<\/p>\n<\/div>","protected":false},"excerpt":{"rendered":"

Helisel Di\u015fli Oran\u0131, Merkez Mesafesi ve Di\u015fli Sistemi Hesaplamalar\u0131 \u2014 Eksiksiz \u00c7\u00f6z\u00fcml\u00fc K\u0131lavuz Helisel di\u015fli tahrik sisteminde di\u015fli oranlar\u0131, merkez mesafeleri ve di\u015f say\u0131lar\u0131n\u0131n hesaplanmas\u0131, tek tek basit olan ancak daha sonra bir boyut de\u011fi\u015ftirildi\u011finde hatalara neden olan etkile\u015fimli form\u00fcller k\u00fcmesini i\u00e7erir. Bu k\u0131lavuz, her form\u00fcl\u00fc ayr\u0131nt\u0131l\u0131 olarak ele almaktad\u0131r [\u2026]<\/p>","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_et_pb_use_builder":"","_et_pb_old_content":"","_et_gb_content_width":"","footnotes":""},"categories":[3082],"tags":[],"class_list":["post-2284","post","type-post","status-publish","format-standard","hentry","category-helical-gears"],"_links":{"self":[{"href":"https:\/\/helicalcutgears.top\/tr\/wp-json\/wp\/v2\/posts\/2284","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/helicalcutgears.top\/tr\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/helicalcutgears.top\/tr\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/helicalcutgears.top\/tr\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/helicalcutgears.top\/tr\/wp-json\/wp\/v2\/comments?post=2284"}],"version-history":[{"count":2,"href":"https:\/\/helicalcutgears.top\/tr\/wp-json\/wp\/v2\/posts\/2284\/revisions"}],"predecessor-version":[{"id":2288,"href":"https:\/\/helicalcutgears.top\/tr\/wp-json\/wp\/v2\/posts\/2284\/revisions\/2288"}],"wp:attachment":[{"href":"https:\/\/helicalcutgears.top\/tr\/wp-json\/wp\/v2\/media?parent=2284"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/helicalcutgears.top\/tr\/wp-json\/wp\/v2\/categories?post=2284"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/helicalcutgears.top\/tr\/wp-json\/wp\/v2\/tags?post=2284"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}