Load at the range?

Gatehouse

Gatehouse

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I'v been taking loading gear to the range for years.

When we built our NEW range rifle shack, iinsited that we had 2 layers of 3/4" plywood counters, to bolt your gear to.

The biggest issue, normally, is the scales. You have to set them up ,check weight them etc...

I had this brainiac moment, and took my truck inverter, attached an extension cord, and VOILA! MY RCBS Chargemaster was up and DOING it!!!
 
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Gatehouse said:
I'v been taking loading gear to the range for years.

When we built our NEW range rifle shack, iinsited that we had 2 layers of 3/4" plywood counters, to bolt your gea to.

The biggest issue, normally, is the scales. You have to set them up ,check weight them etc...

I ahd this brainiac moment, and took my truck inverter, attached a extensioon cord, smnd zvoils11 MY RCBS Chargemaster was up and DOING it!!!
Click to expand...

I have been using my truck inverter for years now for range reloading. It is a very good idea to mention for some that have never thought of it in the past.

Good to hear that you are getting with the times Gatehouse! Still thinking of using that spear ;) :D
 
Gatehouse said:
I'v been taking loading gear to the range for years.

When we built our NEW range rifle shack, iinsited that we had 2 layers of 3/4" plywood counters, to bolt your gea to.

The biggest issue, normally, is the scales. You have to set them up ,check weight them etc...

I ahd this brainiac moment, and took my truck inverter, attached a extensioon cord, smnd zvoils11 MY RCBS Chargemaster was up and DOING it!!!
Click to expand...
if you have a princes auto near by you can get one of those "booster packs" that have either the 12 volt cig lighter thingy in it or the more expensive version has the actual inverter built in wit ha 110volt receptacle that way theres no need for extension chords as they drain juice faster, and you can use it any where. i paid i think $35 +tax for my booster pack charge ti over night and i have the same amount of juice as a car battery and will run a 12volt cooler for about 4-5 hours off the cig lighter thingy and with my inverter attached i can run a 110 volt coffee pot for about 2 hours before needing to recharge
 
The amount of voltage drop on a given wire(conductor) is the function of the current drawn, that's why you may see you lights dim when you turn on your table saw or another large appliance, but as the motor gets up to speed the current drops and the lights go back to normal.
The pitance of current that a digital scale uses wont cause enough voltage drop in 100 fteet of #16 wire to even worry about.
BTW it's called I(squared) R losses, where I is the current going through the wire, and R is the nominal resistance of the wire. The end product is Watts or power, and from the power equation P=E*I(P power in Watts, E voltage in volts, I current in Amps) you can determine the voltage drop.
If anyone is wondering I learnt this back in Grade 10, when I took electronics in High School.
 
I²R losses are watts.

Let's take a 50' #16 guage cord. 100 feet of #16 wire has a resistance of about .408 ohms. RCBS Chargemaster has a maximum power consumption of 220mA.

Total voltage drop on the 50' cord would be about .09 volts.

Total power lost through heat by the extension cord is .0366 watts or about 1/10 of the heat given off by a good quality LED flashlight bulb spread over 50' of cord.

Sorry for the hijack Gatehouse.:redface:
 
BTW it's called I(squared) R losses, where I is the current going through the wire, and R is the nominal resistance of the wire. The end product is Watts or power, and from the power equation P=E*I(P power in Watts, E voltage in volts, I current in Amps) you can determine the voltage drop.

If you have no way of detemining the resistance of a length of wire you look at the current drawn, and the power rating of the device, divide the power rating by the square of the current and voila you have the real resistance of the system , conductor, plugs grit grime corrosion etc.
By using basic math skills you can manipulate the I (squared) formula to determine the real resistance of your system, then plug the resistance back in to the Ohms law formula and voila you have the voltage drop you will see.

Of course if you know the resistance of your conductor, and everything being about equal, you could add a bit to the resistance for the connections and crunch it right from the voltage divider principal.

I know, a simple question and all the apparent geniuses chime in, but really quite the bit of electrical theory at play here.

Hi Jack over.
 
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Lefty #### said:
BTW it's called I(squared) R losses, where I is the current going through the wire, and R is the nominal resistance of the wire. The end product is Watts or power, and from the power equation P=E*I(P power in Watts, E voltage in volts, I current in Amps) you can determine the voltage drop.


Is there an echo in here.
Click to expand...

There sure is.:)

2ef99fe754f522f2479c7f935e07257b.png


Do you always take the long way?:D
 
Lefty #### said:
The amount of voltage drop on a given wire(conductor) is the function of the current drawn, that's why you may see you lights dim when you turn on your table saw or another large appliance, but as the motor gets up to speed the current drops and the lights go back to normal.
The pitance of current that a digital scale uses wont cause enough voltage drop in 100 fteet of #16 wire to even worry about.
BTW it's called I(squared) R losses, where I is the current going through the wire, and R is the nominal resistance of the wire. The end product is Watts or power, and from the power equation P=E*I(P power in Watts, E voltage in volts, I current in Amps) you can determine the voltage drop.
If anyone is wondering I learnt this back in Grade 10, when I took electronics in High School.
Click to expand...
Amazing that you learnt this in grade 10, I learned it in grade 11. Same sheet.
 
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