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quinto173
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樓主  發(fā)表于: 2022-04-28 10:21
電機出軸是8mm,1.8步距角，要求是1秒4圈，齒輪是5M18齒和5M36齒的，細分6400
怎么換算成PLC的脈沖和速度啊
就是DDRVI的兩個(gè)K
隱姓埋名
電氣工程從業(yè)者，愛(ài)好者。廣交天下同行VX：em5480011
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1樓  發(fā)表于: 2022-04-29 10:44
1，步進(jìn)的一圈脈沖數 通過(guò)撥碼設定  。 假設 10000 脈沖一圈。
2.  主動(dòng)輪的周長(cháng)是知道的，查齒輪的參數。   不懂你這個(gè)5M18，就假設 周長(cháng)為100MM 吧。
3. 假設PLC按照10000個(gè)脈沖每秒速度，
4.那速度就是100MM/S     假設沒(méi)有減速機的情況下。
求真務(wù)實(shí)，厚積薄發(fā)。VX:em5480011

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