Electrical Question

[frenchek]

to those who know: I am planning on mounting two fans on my 2013 rt. each fan draws 3.5 amps, total of 7 amps. will the system take this safely, other than the stock draws and a gps, that's all that is on the system now.

 

[Chupaca]

Should work..!!

like Roger said have them fused, have them operating only when needed. Let us know how they work out..!! :thumbup:

 

[Kmort44]

Its a go

The electrical system on the Spyder which while running is powered by the magneto is rated for 500watts @12 volts. The fans your wanting to install would only draw 84 watts of that so keeping in mind other equipment ie lights,accessories,12 volt power outlets etc... you should be well under it's rating. I would recommend fusing the fans separately if they are on long runs from your power source. IMHO

 

[Kmort44]

correction

actually the rt is rated ar 650 watts

The electrical system on the Spyder which while running is powered by the magneto is rated for 650watts @12 volts. The fans your wanting to install would only draw 84 watts of that so keeping in mind other equipment ie lights,accessories,12 volt power outlets etc... you should be well under it's rating. I would recommend fusing the fans separately if they are on long runs from your power source. IMHO




Better Roger?

 

[Phil]

I agree with all of these answers. We assume that you don't have lots of other unusual current drawing items, like lots of heated clothing, heavy lighting, etc. If you do, think about things like not running the fans while using heated clothing (duh...).

You should be fine.

 

[frenchek]

electric

The electrical system on the Spyder which while running is powered by the magneto is rated for 650watts @12 volts. The fans your wanting to install would only draw 84 watts of that so keeping in mind other equipment ie lights,accessories,12 volt power outlets etc... you should be well under it's rating. I would recommend fusing the fans separately if they are on long runs from your power source. IMHO




Better Roger?

thanks to all

 

[Kmort44]

almost.

Wattage is never specified at a voltage.

then you would have no problem if I shocked you with 650 watts @120 VAC, that would only be 5.4 amps you should be fine

 

[Phil]

then you would have no problem if I shocked you with 650 watts @120 VAC, that would only be 5.4 amps you should be fine

About 6 amps. Problem.

650 watts @ 100,000 volts.. .0065 amps. Not so bad!


Sent from my iPhone7 using Tapatalk

 

[Kmort44]

where did you come up with 5.4 amps?
it actually works out to 7.7 amps.
I'll leave up to you to figure out why.

It's called "OHM's Law" power equals voltage times current

P=VxI P=power(wattage) V=Voltage(pressure) I=current(flow)

(648 actual) 650 watts= 120x5.4

using your values P=IXR

924 = 7.7 x 120
Shocking isn't it:lecturef_smilie:

 

[Phil]

Unless Rodger is looking and some forward looking PF


Sent from my iPhone7 using Tapatalk

 

[scarecrow]

Or it's 650watts /14.4 vdc is 45 amps but your deal with 12vdc at 1200 rpm.
The trick is to use a thremo sensor on the fans and have them come on at a temperature.
Also have to remember are the fans free wheeling or are they still until turned on.
Having them make a venture and ram air through the bike when drive and come on when stopped.

 

[Kmort44]

Yes, and below is the correct way to calculate it in an AC circuit, since you specified 120vac

his average value we use for the voltage from a wall socket is known as the root mean square, or rms, average. Because the voltage varies sinusoidally, with as much positive as negative, doing a straight average would get you zero for the average voltage. The rms value, however, is obtained in this way:

• first, square everything (this makes everything positive)
• second, average
• third, take the square root of the average

Here's an example, using the four numbers -1, 1, 3, and 5. The average of these numbers is 8 / 4 = 2. To find the rms average, you square everything to get 1, 1, 9, and 25. Now you average those values, obtaining 36 / 4 = 9. Finally, take the square root to get 3. The average is 2, but the rms average is 3.
Doing this for a sine wave gets you an rms average that is the peak value of the sine wave divided by the square root of two. This is the same as multiplying by 0.707, so the relationship between rms values and peak values for voltage and current is:
V[SUB]rms[/SUB] = 0.707 V[SUB]o[/SUB] and I[SUB]rms[/SUB] = 0.707 I[SUB]o[/SUB]
In North America, the rms voltage is about 120 volts. If you need to know about the average power used, it is the rms values that go into the calculation.



so, when you specify 120 vac the actual voltage used in the calculation is .707 X 120.

And that ain't Ohm's Law

Ohm's law says nothing about power.
It is E is IR

Roger, there's always actual and calculated values. Those who calculate engineer and those who measure get the job done and don't get hung up on technicalities . Your not being helpful here just argumentative.......your right.......I'm wrong............go crush others with your wisdom I'm done with you.
Take care and take a chill pill.
.