> ----- Original Message -----
> Sent: Monday, March 01, 2004 11:11 PM
> Subject: RE: [M]: Help With Deep Cycle Battery - Some real numbers
>

> Hey Roger,
>
> One question. On getting the adapter for the Dell with the higher
> current (since that is what my laptop is) I was wondering if it
> would be better to not get it and get the one with the lower current.
> The other day I was in a meeting and I plugged into one of the
> adapters we had in the conference room. It was one for the lower
> current. The laptop worked great. The only thing I got was a message
> when I plugged it in saying that it knew it was the lower current and
> that it would continue to run on the battery profile.
>
> From the point of view of expending less power wouldn't it be the
> better way to go? I'm sure there is some EE answer as to why it
> wouldn't, but just thought I'd ask.
Interesting. It sounds at though Dell have improved the 'intelligence' in
this regard. I have seen a couple of the older adapters destroyed by
connecting the later laptops to them, but these were the 'first generation'
Inspirons that needed the high power adapter (the 8200 I think). However you
can turn down the power used in 'mains' mode, to match the settings in
'battery mode' yourself, so there is no gain, and often the higher power
adapters will actually have slightly lower switching losses, so there is no
advantage to risking it.

Best Wishes

-----Original Message-----
Roger Hamlett
Sent: Monday, March 01, 2004 2:34 AM
Subject: Re: [M]: Help With Deep Cycle Battery - Some real numbers


> Hey Guys,
>
> Thanks so much for helping me understand how things work.
> I can't believe I've forgotten this much from my EE classes
> when I got my CS degree. Haven't done pretty much any EE
> since, except for installing car stereos in my cars and stuff
> like that.
>
> Anyways, can you guys help me with some real numbers then.
> I am clear on the fact that batteries should not be run down
> to the ground, so I want to distribute my loads well. I am
> going to get another battery. Hopefully a 115 AH battery.
> Here is what I am thinking of doing. Can someone help me
> figure out how long I can run these things before I get
> either of the batteries below 30%:
>
> What I have to run is:
>
> - Laptop: Dell Inspiron. The power supply says it consumes 4.5A.
It says that the supply can _deliver_ a maximum of 4.5A (probably at about
18v). This is different from the consumption. The supply only has to deliver
this when the laptop batteries are flat, and it is having to charge them at
the same time as the laptop is being used. You can keep the actual
consumption of the laptop lower, using the options to switch the processor
speed down, keep the display dimmer, etc. etc..
Ideally, get a 12v supply for the laptop. These are available from Dell
themselves, and (cheaper) from a lot of third party companies (given that
your unit has the higher current supply - there are two versions of Dell
supply, according to the Inspiron model involved, make sure that the supply
you get is for the higher current unit). This gets rid of the double
inefficiency of converting to 120v, and then back down to 18v, and means
that the whole system in this area is running at the lower (safer if
conditions are damp) voltage.

- SBIG ST-7: From the SBIG website for the ST-7 it says: 5 VDC at
> 1.5 amps, ±12 VDC at 0.5 amp desktop power supply included. So
> which one do I use?
Again ideally, get the 12v supply. The same comment applies as for the
laptop. The figures given are the _outputs_ from the supply, which is having
to deliver a maximum of 19.5W (5*1.5 + 12*.5 + 12*.5).

> - Thousand oaks digital dew controller: Unfortunately, their site
> has very little info on current consumption and the unit has
> nothing printed on its back. I'd be running a 10" strap, a telrad
> heater, and a 2" strap for EPs.
The unit itself, draws practically nothing. This is why it does not have a
figure on it. It is the _straps_ that draw the power, and the amount varies
according to how high the heaters are set. Typically, a 10" strap, will draw
perhaps 3A on full, while the eyepiece strap (and the Telrad heater), will
only draw about 0.2A at the same setting.

> Here's the site: http://www.thousandoaksoptical.com/newproducts.html
>
> - LX200: As advised, I'll turn slewing speed down to maybe 4?
Perhaps 2... :-)
I suspect the reason for the laboured sound, is that the _peak_ current
delivery of the 1812 module, is less than the mains supply. High speed is of
no advantage really (if you are going to another part of the sky, the scope
taking a few seconds longer is not a great loss).

> That's it. I have a COleman 800 W inverter that does up to 7
> amps, and I'll probably get another COleman 400 W that does up to 3.5
amps.
You do not need more inverter power here. If you are buying extra stuff, go
for low voltage supplies instead, and get away from running high voltages in
the conditions round the scope, and the inefficiencies involved.

> I was thinking of running the laptop and sbig on one battery (to keep
> them separate from the scope) and since they are the ones that consume
> the most, run them on the big battery (115 ah) and the scope and
> digital dew system on the other battery (the 75 ah).

> Do you guys agree with this? SHould I get the smaller coleman
> inverter or another big one? They are both on sale right now. $29.99
> for the small one and $50 for the big one.
>
> Will this be the best use of these batteries? How do I calculate how
> long I should run them before getting to 30% charge?
>
> Thanks so much for the help!
Work like this:
Take the quoted battery power, and _halve_ it. (this allows for the battery
ageing, and the fact that you will be using it at low temperatures), while
also avoiding fully discharging the battery.
So for the 115Ah battery, treat it as if it can deliver 57.5Ah.
Now work out the total power needed.
Laptop say perhaps (working at 3/4 the 'peak') 3A, at 18v = 54W.
ST7 = 20W
Heaters = (take 3/4 power) 32W
Scope (look at an average between the 'peak', and 'continuous' power, at
perhaps 0.7A at 18v) = 13W
Assuming you have switched to using all '12v' inverters, instead of 'double'
converting, then add these together, and add an 'inefficiency' for the
inverters at 20%.
Total = (54+20+32+13)*1.2 = 143W
Current (at 12v) = 143/12 = 11.9A.

The battery can then give about 5 hours.
In fact the figure for the laptop, heaters, and scope, are all higher than
'real'. My own Inspiron, only draws about 2A, provided the internal battery
is charged, while the heaters typically operate nearer to perhaps 30% power.
I see more like 10hours operation, with a similar setup, on this sort of
battery. The figure represents a 'worst case', for a really cold night.
However if you work through the 'double inverters' (using the Coleman to go
up to 120v, then down again to the voltages required by the individual
units), you would need to think in terms of:
Total = 143*1.2 (to allow for the inefficiency of the second inverter) =
170W
Notice that this is still well short of what your 800W inverter can handle.

Battery current= 170/12 = 14.2A
The 'worst case life', then falls to only 4 hours!...
The same comment then applies that the figures are worse than real, but does
begin to bring home how much power is actually involved...

Best Wishes

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