I need to re-wire the mast of my pearson 30' coaster. Lights on the mast are a strobe and tri-color on mast head, white running light just above the spreaders on the mast, and two spreader lights. Has anyone done this, and can you give me any help/direction. Thanks
Mast re-wire
- Thread starter dreamcatcher13400
- Start date
- Status
What do you need to know?
I have rewired many masts, Not a big thing. For the life of me I can't really give you an answer unless I write a book starting from un-stepping the mast. You have done that haven't you? What you are calling a running light just above the spreaders is called a steaming light. You cannot run the steaming light alont with the tri-color masthead light. When the steaming light is on the Port, Starbord, and stern light must be lit and not the tri-color. YOu cannot display both sets of lights at the same time. Not my rule but COLREGS. How is this for a start. Fair Winds The Cap'n
I have rewired many masts, Not a big thing. For the life of me I can't really give you an answer unless I write a book starting from un-stepping the mast. You have done that haven't you? What you are calling a running light just above the spreaders is called a steaming light. You cannot run the steaming light alont with the tri-color masthead light. When the steaming light is on the Port, Starbord, and stern light must be lit and not the tri-color. YOu cannot display both sets of lights at the same time. Not my rule but COLREGS. How is this for a start. Fair Winds The Cap'n
Mast wiring
You’ll want a separate switch for each of: 1. strobe 2.tri-color on mast head, 3.white running light just above the spreaders on the mast, 4/5. and two spreader lights (could be switched separately or together) Hence you’ll need at least 5 wires (4 switched positives, and minimum 1 oversized negative). Assuming a 20 Watt load at about 100 ft. return circuit length, the positive wires should be sized at #12AWG minimum, and the negative #10 AWG (or you could run a separate #12) negative for each circuit). HTH, Gord
You’ll want a separate switch for each of: 1. strobe 2.tri-color on mast head, 3.white running light just above the spreaders on the mast, 4/5. and two spreader lights (could be switched separately or together) Hence you’ll need at least 5 wires (4 switched positives, and minimum 1 oversized negative). Assuming a 20 Watt load at about 100 ft. return circuit length, the positive wires should be sized at #12AWG minimum, and the negative #10 AWG (or you could run a separate #12) negative for each circuit). HTH, Gord
Gord, 20 watts isn't much, even at 100ft.
Are you thinking 'amps'? If the total mast wiring load were 20 amps, then yes, 10 or 12 ga. wire would be needed.
Are you thinking 'amps'? If the total mast wiring load were 20 amps, then yes, 10 or 12 ga. wire would be needed.
Yes 20 W
Yes, I did mean 20 WATTS in my example. A 20 AMP load would require #2 AWG wire at 100 Ft cct. length. These figures are based upon a maximum desired (permitted) voltage drop of 3%, and a return circuit length being the sum of the Positive Supply & Negative Return (to & from) wire lengths. Goto: http://cruisersforum.com/showthread.php?s=&threadid=372 Where you’ll find a tutorial on voltage drop & wire-sizing (“Ohms Law & Boats” excerpted from 'Good Old Boat' magazine - Issue 29 - Mar/Apr '03)), including: The industry standard formula for calculating minimum Wire Size for a given Voltage Drop is: CM = (K x A x L) ÷ VD Where: CM = The Circular Mil cross sectional area of the wire (See Table 1, and ABYC Section E-8, Table III) K = 10.75 Representing the Mil-Foot Resistance of Copper Wire @ 78 deg. F. L = The Total Length of the wire in Feet (This is the ‘Round Trip’ length of both Positive + Negative wires) VD = Permitted Drop in Volts (Ie: 3% @ 12.5V = 0.375 ED or 3% @ 25V = 0.750 ED) Isn’t it scary how undersized much of our boat wiring is? Regards, Gord
Yes, I did mean 20 WATTS in my example. A 20 AMP load would require #2 AWG wire at 100 Ft cct. length. These figures are based upon a maximum desired (permitted) voltage drop of 3%, and a return circuit length being the sum of the Positive Supply & Negative Return (to & from) wire lengths. Goto: http://cruisersforum.com/showthread.php?s=&threadid=372 Where you’ll find a tutorial on voltage drop & wire-sizing (“Ohms Law & Boats” excerpted from 'Good Old Boat' magazine - Issue 29 - Mar/Apr '03)), including: The industry standard formula for calculating minimum Wire Size for a given Voltage Drop is: CM = (K x A x L) ÷ VD Where: CM = The Circular Mil cross sectional area of the wire (See Table 1, and ABYC Section E-8, Table III) K = 10.75 Representing the Mil-Foot Resistance of Copper Wire @ 78 deg. F. L = The Total Length of the wire in Feet (This is the ‘Round Trip’ length of both Positive + Negative wires) VD = Permitted Drop in Volts (Ie: 3% @ 12.5V = 0.375 ED or 3% @ 25V = 0.750 ED) Isn’t it scary how undersized much of our boat wiring is? Regards, Gord
Amp/Feet
My comments on using a “common” negative wire were incomplete. I recommend using a discrete Negative wire for each Positive - the wires being the same size. Sometimes, especially when 2 (or more) circuits will NOT likely be “on” at the same time, you can combine two circuit Negatives on a single “common” Negative. It’s usually prudent to “upsize” a common negative, to accommodate a potential emergency situation where both circuits are “on” simultaneously. In this case, the Negative wire size should be calculated for the combined load of both circuits. I have prepared a little chart which includes pre-calculated figures for required wire sizes for Amp/Feet of circuit (at 3% Voltage Drop), and other useful information required for for circuit design. Send me a FAX number, and I’ll fax a copy to anyone interested. Gord@BoatPro.zzn dot com Ie: For maximum 3% Voltage Drop @ 12 Volt #12 AWG Copper wire ~ 150 A/Ft #10 AWG ~ 300 A/Ft #8 AWG ~ 450 A/Ft #6 AWG ~ 900 A/Ft #4 AWG ~ 1,250 A/Ft #2 AWG ~ 2,200 A/Ft #1/0 AWG ~ 3,500 A/Ft etc. An Amp/Foot is 1 Amp over a distance of 1 Foot (return circuit length = Pos + Neg lengths) Hence: 20 Watts @ 12.75 V = 1.57 Amps (actually 1.5686 A) 1.5686 A x 100 Ft = 156.86 A/Ft From the chart, #12 AWG is only suitable for 150 A/Ft - so I very slightly undersized my previous example - could have selected the next higher wire size - #10 @ 300 A/Ft, and perhaps #8 for common negative (requires 314 A/Ft). More info’ on the CruiserForum.com FWIW, Gord
My comments on using a “common” negative wire were incomplete. I recommend using a discrete Negative wire for each Positive - the wires being the same size. Sometimes, especially when 2 (or more) circuits will NOT likely be “on” at the same time, you can combine two circuit Negatives on a single “common” Negative. It’s usually prudent to “upsize” a common negative, to accommodate a potential emergency situation where both circuits are “on” simultaneously. In this case, the Negative wire size should be calculated for the combined load of both circuits. I have prepared a little chart which includes pre-calculated figures for required wire sizes for Amp/Feet of circuit (at 3% Voltage Drop), and other useful information required for for circuit design. Send me a FAX number, and I’ll fax a copy to anyone interested. Gord@BoatPro.zzn dot com Ie: For maximum 3% Voltage Drop @ 12 Volt #12 AWG Copper wire ~ 150 A/Ft #10 AWG ~ 300 A/Ft #8 AWG ~ 450 A/Ft #6 AWG ~ 900 A/Ft #4 AWG ~ 1,250 A/Ft #2 AWG ~ 2,200 A/Ft #1/0 AWG ~ 3,500 A/Ft etc. An Amp/Foot is 1 Amp over a distance of 1 Foot (return circuit length = Pos + Neg lengths) Hence: 20 Watts @ 12.75 V = 1.57 Amps (actually 1.5686 A) 1.5686 A x 100 Ft = 156.86 A/Ft From the chart, #12 AWG is only suitable for 150 A/Ft - so I very slightly undersized my previous example - could have selected the next higher wire size - #10 @ 300 A/Ft, and perhaps #8 for common negative (requires 314 A/Ft). More info’ on the CruiserForum.com FWIW, Gord
Gord, thanks.
I guess that the requirement of less/equal to a 3% voltage drop is a standard. But is that really necessary with a light bulb? How is visibility reduced with, say, 16AWG wire in our example?
I guess that the requirement of less/equal to a 3% voltage drop is a standard. But is that really necessary with a light bulb? How is visibility reduced with, say, 16AWG wire in our example?
1% VD = 4% Light Output Drop
The maximum Voltage Drop permitted by ABYC is: 3% for Navigation Lighting, and other “essential” services 10% Convenience Lighting etc. Incansescent Lamps are very sensitive to Voltage Variations, greatly reducing luminous efficacy (but also increasing lamp life) as applied voltage decreases. As a rough approximation, light output is reduced approximately 4 percent for each 1 percent drop in rated Voltage. This means that 5% reduction in operating voltage will double the life of the bulb, at the expense of reducing its light output by 20%. A 20W Nav. Light, fed with 100 Feet (50 Pos + 50 Neg) of #16AWG Cu. wire (0.0041 Ohms/Ft, or 4.1 Ohm/1000'), will draw about 1.57 Amps. V Drop = I x R = 1.57 x (0.0041 x 100) 0.644 Volts 0.644 ÷ 12.75 = 5% Voltage Drop 12.75 - 0.644 = 12.106 Volts at Lamp (12.106 ÷ 12.75)to the 3.4Th Power (more than cubed) = about 80% output The light output, if wired with #16, will be about 80% of that if wired /w #10. That’s over 25% wasted energy, and ‘lost’ light output (1.00 ÷ 0.80). Formula: Actual Luminous Intensity = Rated Luminous Intensity x (Applied Voltage ÷ Rated Voltage) raised to the power of 3.5 HTH, and I hope this is enough (I hate math’) ... Gord
The maximum Voltage Drop permitted by ABYC is: 3% for Navigation Lighting, and other “essential” services 10% Convenience Lighting etc. Incansescent Lamps are very sensitive to Voltage Variations, greatly reducing luminous efficacy (but also increasing lamp life) as applied voltage decreases. As a rough approximation, light output is reduced approximately 4 percent for each 1 percent drop in rated Voltage. This means that 5% reduction in operating voltage will double the life of the bulb, at the expense of reducing its light output by 20%. A 20W Nav. Light, fed with 100 Feet (50 Pos + 50 Neg) of #16AWG Cu. wire (0.0041 Ohms/Ft, or 4.1 Ohm/1000'), will draw about 1.57 Amps. V Drop = I x R = 1.57 x (0.0041 x 100) 0.644 Volts 0.644 ÷ 12.75 = 5% Voltage Drop 12.75 - 0.644 = 12.106 Volts at Lamp (12.106 ÷ 12.75)to the 3.4Th Power (more than cubed) = about 80% output The light output, if wired with #16, will be about 80% of that if wired /w #10. That’s over 25% wasted energy, and ‘lost’ light output (1.00 ÷ 0.80). Formula: Actual Luminous Intensity = Rated Luminous Intensity x (Applied Voltage ÷ Rated Voltage) raised to the power of 3.5 HTH, and I hope this is enough (I hate math’) ... Gord
- Status