Timeline for How to calculate forces on bearings in a chain drive
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 16, 2023 at 23:15 | vote | accept | php_nub_qq | ||
| Jun 10, 2022 at 6:18 | comment | added | drC1Ron | The force from the chain must be supported somewhere | |
| Jun 9, 2022 at 12:58 | comment | added | php_nub_qq | It is exactly as you have drawn it, but those bearings bear only the weight of the shaft and sprocket, not the entire 200kg. The sprocket drive does however move the 200kg which is the total weight of the vehicle, and it is driven by a motor | |
| Jun 9, 2022 at 10:17 | comment | added | drC1Ron | You should include a drawing of the system | |
| Jun 9, 2022 at 10:06 | comment | added | php_nub_qq | So basically the entire 3200N force will also be applied radially on the bearings? I don't think this is correct. I think the confusion is perhaps caused from my explanation, trying to simplify things I actually complicated them more. The bearings in question are not carrying the weight of the system (which is a vehicle, 200kg), they are only carrying the weight of the drive unit, which is negligible | |
| Jun 9, 2022 at 10:05 | history | answered | drC1Ron | CC BY-SA 4.0 |