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  • 匿名
关注:1 2013-05-23 12:21

求翻译:当给定2%(换算成电流是0.34MA)的开启角度时,定位器不能执行动作,只有开启角度值给定高于4%(换算成电流是0.68MA)时定位器才执行动作是什么意思?

待解决 悬赏分:1 - 离问题结束还有
当给定2%(换算成电流是0.34MA)的开启角度时,定位器不能执行动作,只有开启角度值给定高于4%(换算成电流是0.68MA)时定位器才执行动作
问题补充:

  • 匿名
2013-05-23 12:21:38
When given 2% (converted into current 0.34ma) in the opening angle, the locator can not perform the action, only to open the angle values ​​given above 4% (converted into current 0.68ma locator) to perform an action
  • 匿名
2013-05-23 12:23:18
When a given 2% (0.34 is converted to a current MA) when the opening angle, positioning is not action, it is only open for a given angle value higher than 4 per cent (0.68 is converted to a current position when MA) server is executing the move.
  • 匿名
2013-05-23 12:24:58
正在翻译,请等待...
  • 匿名
2013-05-23 12:26:38
When given 2% (conversion into an electric current is 0.34MA) opening angle, locators could not perform the action, only the opening angle values given above 4% (conversion into an electric current is 0.68MA) Locator only when performing an action
  • 匿名
2013-05-23 12:28:18
When when assigns 2% (to convert electric current is 0.34MA) the opening angle, the locator cannot carry out the movement, when the opening angle value assigns is higher than 4% (to convert electric current is 0.68MA) the locator only then to carry out the movement
 
 
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