Okay. Im confused on this whole injector thing.
10-09-02, 12:31 PM
#26
What R U thinking self?
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Originally posted by mac_dad6
how do you know if your primaries and secondaries are high or low impedance
how do you know if your primaries and secondaries are high or low impedance
10-18-02, 10:33 AM
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What R U thinking self?
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Thanks J-Rat I got the resistor pack in the mail today.
Okay this thing has 5 wires, my guess is; one for each injector and the middle\fifth wire is a ground?? does this sound correct? Can anyone please share some insight on where to mount this? and where & how to hook it up? Its the stock injector pack for a low imp injector RX7.
Thanks alot
Okay this thing has 5 wires, my guess is; one for each injector and the middle\fifth wire is a ground?? does this sound correct? Can anyone please share some insight on where to mount this? and where & how to hook it up? Its the stock injector pack for a low imp injector RX7.
Thanks alot
10-18-02, 11:53 AM
#29
Banned. I got OWNED!!!
some injector info: www.rotaryresurrection.com tech injector info
10-18-02, 04:31 PM
#30
I'm a boost creep...
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Re: resistance
Originally posted by J-Rat
Actually RETed, resistance in parallel is the sum of total resistance divided by the number of resistors in parallel.
Actually RETed, resistance in parallel is the sum of total resistance divided by the number of resistors in parallel.
1/Rt = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn
If you have two 6 ohm resistors and a 3 ohm in parallel, the total resistance is 5 ohms.
1/Rt = 1/6 + 1/6 + 1/3 = 0.667, so Rt = 1.5
10-18-02, 04:46 PM
#31
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Re: Re: resistance
Originally posted by NZConvertible
Nope. Using the above equation:
1/Rt = 1/6 + 1/6 + 1/3 = 0.667, so Rt = 1.5
Nope. Using the above equation:
1/Rt = 1/6 + 1/6 + 1/3 = 0.667, so Rt = 1.5
10-18-02, 06:41 PM
#32
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Originally posted by Rpeck
Damn! thats all I needed to know...
Damn! thats all I needed to know...
Last edited by NZConvertible; 10-18-02 at 06:43 PM.
10-18-02, 08:19 PM
#33
Re: Re: resistance
Originally posted by NZConvertible
Nope. Using the above equation:
1/Rt = 1/6 + 1/6 + 1/3 = 0.667, so Rt = 1.5
Nope. Using the above equation:
1/Rt = 1/6 + 1/6 + 1/3 = 0.667, so Rt = 1.5
Crap...you are right...Thats why when you put 2 eight ohm loads in parallel, you see 4 ohms of resistance at the source. Three 8 ohms yields 2 (.6??) ohms of resistance. What was I thinking?...
Rat
Last edited by J-Rat; 10-18-02 at 08:23 PM.
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