7 ms·
None of this affects the cell current though. If your motors draw W Watts at peak, and you have N cells at V volts then the peak per-cell current will be (appr
by aidenn0 8d ago
None of this affects the cell current though.
If your motors draw W Watts at peak, and you have N cells at V volts then the peak per-cell current will be (approximately -- there are internal losses) W/(NV) regardless of the geometry of your stacks.
If N is small, your designs may be restricted, but I'll hazard a guess that N will be large for a 30 passenger plane capable of 2 hour flights (e.g. a Tesla model 3 is almost 3000 cells; 31s96p).
- ActorNightly 8d agoA plane needs to be able to takeoff in a given runway length, with a given payload, and climb at that power to a safe altitude for an option of an aborted landing. And just Tesla set to a high power mode, if when starting to drive it you do max acceleration for 1-2 minute you drain a significant portion of your battery very quickly. The overall point that Im trying to make is that slight battery specific energy density improvements don't matter when compared against the power losses which are proportional to square root of the current.
- aidenn0 5d ago> And just Tesla set to a high power mode, if when starting to drive it you do max acceleration for 1-2 minute you drain a significant portion of your battery very quickly. An airplane under typical operation has a much narrower ranges of power than a Tesla in high power mode. Takeoff power is a low single-digit multiple of cruise power. For my math earlier I used a 3:1 ratio. A Model S Plaid is more like 50:1, or 30:1 for the "regular" Model S. > The overall point that Im trying to make is that slight battery specific energy density improvements don't matter when compared against the power losses which are proportional to square root of the current. You've completely missed making that point, since I still haven't seen an argument so far that the current must be very high.