5 min once every two weeks should be sufficient assuming the
battery and charging system is good. Turn everything off except
engine. It will need to be run at least 1,000 rpm for that time
OTH, Harbor Freight has cheap little float chargers but 120v power
will be needed. A friend uses one of these when out of the country:
On Mon, 4 Feb 2019 16:45:34 -0500, Tom Del Rosso wrote:
Having said that, here's how I arrived at 72 seconds, bearing in mind
there's a complexity to your question which, outside of the engineering
specs of both the battery & engine (and parasitics), we can only help you
guess at it mathematically, where empirical results would seem to be more
accurate than our guestimates.
Starting with the basics, a quick search for a Buick Regal Alternator nets
which says the alternator outputs 100 amps at idle (if needed) and 150 amps
output at max rpm (again, if needed as alternators adjust output based on
Running a direct search for the power needed to start an 87 Buick Regal,
it's easy to find the vehicle, but hard to find the power needed to start
We're kind of stuck with the "generic" stuff, such as this:
o How Many Amps Does It Take to Start a Car?
Which says an average car needs 400 to 500 amps but doesn't say how long.
Let's assume it takes five to ten seconds to start it, at 500 amps, where
the maximum power would be 10 seconds times 500 amps, which means you
sucked out 5,000 Coulombs (i.e., 5000 amp seconds) if the math is right.
If I did the math right, that's less than 1.5 amp hours, and since we
guessed high, I'd say the amount used is roughly about 1 amp hour to 1.5
amp hours, but since we want to "be safe" and have "easy math", I'd use 2
amp hours as the amount to add back.
If you put back two amp hours (to cover for inherent losses, mostly in
heat), you're back to where you started, where we have to "assume" that the
battery sense circuit allows the alternator to output enough current to
charge the battery after just one start.
At idle, if we assume the battery sense allows you to get those 100 amps we
saw in the spec, to generate 2 amp hours would take only about 0.02 hours,
or about 72 seconds (if I did the quick math right) - which -
coincidentally - is about how long it took to run the quick math. :)
If that 72 second answer is wrong, I welcome someone who can tell us how to
arrive at the better answer.
Having authored peer-reviewed papers myself (in a different field), I took
a quick peek at the abstract of that paper, titled:
o Computer Simulation of an I.C. Engine During Cranking by a Starter Motor
"A mathematical model is developed to study the transient behavior of a two
stroke or four stroke, single cylinder I.C. engine during cranking and
starting by a starter motor. The engine model includes forces due to
inertia of reciprocating and rotating parts of engine, gas pressure,
frictional loss while starter motor dynamics is determined by the motor's
torque versus speed behavior. The numerical results of the analysis when
compared with the experimental results showed close correlation.
Engine starting by three models of starter motor is presented for a given
battery. Effect of different parameters like engine inertia and reduction
ratio between engine and a starter motor is described. It is shown here how
this analysis can be effectively used as a first step by an engine designer
for determining a suitable starter motor characteristic and its related
Hmmm... they _might_ cover the charge payback component, but I suspect
likely it will only be an ancillary input to the mathematical model, and
certainly it won't apply _directly_ to an 87 Buick Regal.
We should note that the given "battery" is seemingly incidental in this
paper, which seems to be aimed more toward designing starter motors, and,
specifically between choosing among three different types of fundamental
starter motor designs.
Still, it may be an interesting read, where, I'd be curious how the three
types of starter motors affected the model - but - I hazard a guess that we
won't find a direct answer for our charge component in that paper.
It's just an equation. (I suppose a 12V car battery).
3s of starter, means 3*(900/12) C = 225C = 62.5mAh=0.0625Ah (900W is the
Just add this to the loss of the battery.
Knowing that the alternator charges at 13.8V about 500W (500/(13.8-12)
i.e. 200As=0.55Ah , it's easy.
Tom Del Rosso a écrit le 04/02/2019 à 22:45 :
Here's a modern review of a 1989 Buick Century:
"Boomers were in their early 40s and at the height of their power - they
understood the world, and the world turned according to their whims.
They understood computers because floppy disks were goddamn floppy, and
they knew the HIV virus was out there doing the good Lord's work."
My '94 Saturn SL2 failed to start once, at 0 degrees and only having taken 4 mile trips for some weeks. After some nice girl gave me the jump start I soon took it on a long trip & it's back to normal.
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