Timeline for Calculating power measurement error from V and I error margins
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 14, 2016 at 14:00 | comment | added | carloc | Yes your quite right, any "power rectangle" is scaled just the same, so you do not loose their areas ratio. BTW one might also think that current and voltage scales are dimensional quantities, say they are A/mm and V/mm, so again they can't be compared. | |
| Aug 12, 2016 at 7:42 | history | edited | Transistor | CC BY-SA 3.0 |
Removed "scaling" sentence from Figure 1.
|
| Aug 12, 2016 at 7:41 | comment | added | Transistor | I think @carloc is right. I was thinking that different scaling might make \$ V \delta i \$ look more significant than \$ I \delta V \$ but it doesn't - they'll all scale the same amount. All the areas will still be proportional so if we divide the volts scale by two all the areas become VA/2. I'll remove the sentence. | |
| Aug 12, 2016 at 7:25 | comment | added | WhatRoughBeast | @carloc - If Transistor had used a square rather than a proportional rectangle, it would be impossible to see that the V term is much larger than the i term (for the specific V and i which were used here). | |
| Aug 12, 2016 at 6:18 | comment | added | carloc | @Transistor Your picture caption says "cut... V and I scales are identical as is required to make the areas meaningful".I really cannot follow you, could you clarify? | |
| Aug 11, 2016 at 18:46 | vote | accept | skrrgwasme | ||
| Aug 11, 2016 at 7:47 | comment | added | Transistor | You've made me think about this a little. I've taken the liberty of adding Figure 1 and its caption. Please feel free to delete / edit as you see fit or if I've totally missed your point. | |
| Aug 11, 2016 at 7:45 | history | edited | Transistor | CC BY-SA 3.0 |
Added Figure 1 and caption.
|
| Aug 11, 2016 at 0:43 | history | answered | WhatRoughBeast | CC BY-SA 3.0 |