RE: [MV] Windshields etc...

From: Rikk Rogers (rkltd@fullnet.net)
Date: Sun Oct 22 2000 - 11:50:07 PDT


Richard
This is why I like forums like this, to gain a deeper understanding of the
practical applications of what we take for granted every day.

I thank you all, for doing what we are doing, both MV and this list.
When I bought my M35, I was just buting a work on the farm truck (wife was
planning to paint it pink)
After following this list, I am quite happy to restore this beast and join a
new community in doing so.

Well done all!

Rikk Rogers
RK Lion LTD.
(580)762-3157
rkltd@fullnet.net
MVPA -22345-

-----Original Message-----
From: Military Vehicles Mailing List [mailto:mil-veh@mil-veh.org]On
Behalf Of Richard Notton
Sent: Sunday, October 22, 2000 3:40 AM
To: Military Vehicles Mailing List
Subject: Re: [MV] Windshields etc...

----- Original Message -----
From: <kbernste@us.ibm.com>
To: "Military Vehicles Mailing List" <mil-veh@mil-veh.org>
Sent: Sunday, October 22, 2000 3:23 AM
Subject: [MV] Windshields etc...

> Hey, an electrical engineer can't let a question like this go -
>
Indeed I can't !

> If all you have available is a 12V bulb, then take a volt/ohm meter and
> measure the
> equivalent resistance of the bulb you have. Then, in series with the bulb
> put a resistor of value
> equal to that of the bulb. It will voltage-divide the 24V so that half the
> drop
> is across the resistor, and half across the bulb.
>
Errrrr, no it wont, not even nearly.

Doing this you will have some stunningly bright lamps, for the few moments
before they go pop.

There is a substantial difference between the _cold_ and _hot_ resistance.
The cold resistance of a filament lamp is very low compared to its hot,
operating resistance and this becomes more and more different as bulb size
goes up.

I have noticed that the US tends to use CP - candle power for most/many
bulbs smaller than headlight size, whereas we always use a watt rating, this
use of CP denies you the opportunity to do a simple sum where the wattage is
marked on the bulb eg.,

(for the non-electricals, remember we use I for amps)

Watts (W) = V x I . . . . . . . . . . . .(1)
V = I x R . . . . . . . . . . . . . . . . . . .(2)

Now you can do this in two steps, re-arrange (1) into I = W / V get the
current and use this value in (2) as R = V / I.

Alternatively in one step, as you are after R (resistance) we could
substitute I in (2) for W / V as we know I = W / V, so then V = W / V x R
and this simply re-arranges to
R = V squared / W.

(I'm sorry it looks messy but if you use the appropriate symbols from
character map the list software just strips them out)

Just to illustrate how wrong you can be by measuring the cold resistance of
a filament lamp, I have some 24V spares for the FV623 here, 21W for the
indicators (turn signals) and 5W for the side (running) lights.

So, a 24V 21W lamp is from R = V squared / W, R = 576 divided by 21, which
is 27.43 ohms. The digital meter says its 1.8 ohms cold !!!!!!!!!!!!!!!!!
Similarly the 24V 5W lamps calculate as 115.2 ohms but measure cold just 9.3
ohms.

Richard
Southampton - England

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