When I stated that "there is no advantage to having the dual [Sata] configuration versus the single [Sata] configuration", I was assuming -- without bothering to do any calculations -- that the transmission losses and heat dissipation of either configuration would not differ significantly.

Ohm's Law can be used to calculate:

- the voltage drops on the three wire sectors from the PSU to the Vidcard's 8-pin connector

- the associated power transmission losses, and resulting heat generation

Closed-DC-circuit diagram for the +12VB rail with the Dell PD145 adapter installed on the P3 connector, and the single-Sata [sS] adapter plugged into the Vidcard's 8-pin input:

|------R1 [hot]------<>------R2 [hot]------<>------R3 [hot]------|

| . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .|

+12VB Rail . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Vidcard_8pin

| . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .|

|------R1 [gnd]------<>------R2 [gnd]-----<>------R3 [gnd]-----|

|______________| . |_____________| . |_____________|

. PSU to P3 . . . . . . . . . P3 to sS . . . . . . . . . . . sS to 8-pin

. 1 x 16(?)awg . . . . . . 1 x 16(?)awg . . . . . . . 3 x 18awg

. A = 1.31 mm2 . . . . . . A = 1.31 mm2 . . . . . . A = 3 x 0.82 mm2

. 1.2 ft (two-way) . . . . 1.6 ft (two-way) . . . . 1.0 ft (two-way)

R1 [hot] - represents a single hot wire

R1 [gnd] - represents the complementary single ground wire

R2 [hot] - represents a single hot wire

R2 [gnd] - represents the complementary single ground wire

R3 [hot] - represents the 3 parallel hot wires ["3-lane highway"]

R3 [gnd] - represents the 3+ complementary parallel ground wires

The lengths are rough estimates, and can be corrected as needed.

Note that the cross-sectional area [ A ] shown for the single-Sata [ sS ] sector equals 3 times the cross-sectional area of a single-wire, and computes to 2.46 mm2.

To calculate the voltage drop associated with the resistance in the wiring, assume the card is drawing 12.5A at approximately 12V, thus drawing approximately 150 Watts of power at its 8-pin connector.

[I write "approximately" 150 Watts, because -- if the voltage from the rail holds steady at 12V, and the current drawn by the card is steady at 12.5A -- then there is a voltage drop over the wiring of 0.16V (see the results further below), and the card's 8-pin input only sees a voltage difference of 11.84V. . Thus the wattage actually delivered turns out to be only about 148 watts, with about 2 watts being dissipated in the wiring. . The card and the wiring together would have to draw around 12.67A (through the 8-pin) at to get a true 150W (11.84V * 12.67A) for the card.]

V = I * R

I = constant at 12.5A, i.e. the current flow is the same through every element of the circuit

V1 = 12.5A * R1

V2 = 12.5A * R2

V3 = 12.5A * R3

The voltage drops can be calculated using this online-calculator:

http://yeroc.us/calculators/wire-resistance?material_resistivity=1.68E-08&web_resistivity=&gauge=18&web_area=&area_factor=1E-06&web_length=1&length_factor=0.304878¤t=12.5&submit=Calculate
This calculator uses the two-way or "round-trip" format. . It gives the following results for resistance, voltage drop, and power dissipation:

2*R1 = 0.005 Ohms . 2* R2 = 0.006 Ohms . 2*R3 = 0.002 Ohms

2*V1 = 0.059 Volts . . 2* V2 = 0.078 Volts . . 2*V3 = 0.026 Volts

2*D1 = 0.734 Watts . 2* D2 = 0.978 Watts . 2*D3 = 0.325 Watts

In the single-Sata adapter configuration, the total watts dissipated between the +12VB rail and the Vidcard is about 2 Watts. . At 6 Amps, it would be about 1 Watt, whereas at 18 Amps it would be about 3 watts.

Doubling or tripling the number of "wire-lanes" all the way to the rail (and/or rewiring with 14awg or 12awg) would be able to cut the losses by a half or a third, but.... how serious is a loss of 2 watts?

Small terrarium heaters start at around 4 watts. . .

http://www.amazon.com/Zilla-Terrarium-Heater-4-Watt-4-Inch/dp/B005SRVNU6
Suppose a 12V PSU powers a 12.5 Amp load through a pair of 6"-long hook-up wires -- of unspecified AWG. . How thin would the wires have to be, to become "too hot to touch"?. . Maybe if they were thin enough to dissipate about as much heat as an incandescent light bulb? . . With that in mind, I played with the online calculator, and found that if the AWG is dialed down to 36, the pair of 6" long 36awg wires (0.005" diameter) will dissipate about 80 Watts between them, or 40 Watts each.

An 18awg wire is 1.024mm in diameter and has an Area of 0.82 mm2.

A 36awg wire is 0.027mm in diameter, and has an Area of 0.0127 mm2 -- about 40 times thinner in diameter, with about 65 times less Area, compared to an 18awg wire.