#26
IP: 68.190.216.151
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Regarding resistance in the coil: Single digit ohms is almost no resistance. Most resistors in electronics are measured in Kohms - 1000s of ohms. Megohms (millions) is not unusual.
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"Halcyon" 36' custom sloop. 8 tons. Glass over strip plank mahoghany. Spruce mast and booms, launched 1969. Original A4. |
#27
IP: 137.200.0.112
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Ballast resistors are usually in the 0.5 to 1.5 ohm range. IIRC, the resistors I got for my coolant pumps are 1 ohm 25 watt units.
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#28
IP: 137.200.0.112
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True.
Also the lower resistance coils have presumably thicker wire and less turns to give higher voltage output. The less turns in the primary, the higher the ratio to the secondary and thus higher voltage out. What I don't know for a fact is that the difference between a 4 ohm, a 3 ohm, a 1.6 ohm, and a 0.6 ohm coil, etc. etc. are all ENTIRELY down to the amount and size of primary wire or if some of them have an internal ballast as well. A little poking around shows this: A 0.7 ohm coil has a turns ratio of 100:1. http://www.summitracing.com/parts/msd-8203/overview/ A 0.6 ohm coil has a turns ratio of 108:1. http://www.summitracing.com/parts/pnx-45011/overview/ A 3 ohm coil has a turns ratio of 50:1 http://www.summitracing.com/parts/pnx-28010/overview/ Clearly there is more wire in the 3 ohm coil (lower turns ratio = more wire on the primary side), but without sawing the thing open I cannot say 100% of the resistance increase is due to the wire. I may call a coil company and ask, now I just have to know. I think it is, but not 100% sure. A visualization aid for those lacking EE backgrounds - A dimmer switch acts like a ballast resistor. The more resistance, the less total current, and the less heat and light coming out of the light bulb. Switching the light on and off rapidly is like the points. The more ON time vs. OFF time is the dwell. The higher the dwell, the hotter the bulb. Electronic ignition leaves the light ON more than points, so the coil gets hotter. Last edited by joe_db; 04-07-2016 at 08:20 AM. |
#29
IP: 24.152.132.65
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Decent analogy Joe. About how different internal resistance coils are constructed, the Bosch Blue Coil #00-012US says right in its description "Stock Bosch blue coils have the 3 ohm ballast resistor INSIDE the coil" Further, here's a drawing of a vintage Autolite canister coil that should help clear up the speculation.
Is there a reason for this exercise? It seems it's headed down the path of how to build a watch when you only asked what time it was.
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Neil 1977 Catalina 30 San Pedro, California prior boats 1987 Westsail 32, 1970 Catalina 22 Had my hands in a few others |
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Administrator (04-07-2016) |
#30
IP: 107.0.6.242
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Quote:
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#31
IP: 137.200.0.112
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I am building virtual machines this week, which is kind of like watching paint dry.
If anyone just wants a coil, as I have posted a ton in these threads, the Moyer coil works fine and you can skip all theory and practice of ignition threads Quote:
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#32
IP: 68.190.216.151
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Quote:
In any case, these ignition systems (coil,points), like carbs, are arcane, and there are lots of people, even people who fix engines, who don't know how they work.
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"Halcyon" 36' custom sloop. 8 tons. Glass over strip plank mahoghany. Spruce mast and booms, launched 1969. Original A4. |
#33
IP: 73.255.216.151
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hanleyclifford (04-07-2016) |
#34
IP: 70.192.4.68
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Theory
1. EVERY metal has resistance. So do many nonmetals.
2. Every metal's resistance increases with temperature (I know of no exceptions to this). 3. The heat generated by passing current through a wire is due to the resistance of the wire. Power is dissipated by the resistance, and this power is proportional to the square of the current: From Ohm's Law, V = I x R Power: P = V x I (watts) substituting for V, P = I x I x R or P = I^2 x R or, substituting for I, P = V x V / R or P = V^2/R 4. Energy (of which one form is heat) is E = P x t watt-hours (Wh) or if you prefer to measure heat in BTUs (and I do), then: 1 Wh= 3.4 BTU Therefore the energy (heat) dissipated by the resistor is: E = I^2Rt 5. Whether you string the wire out straight, wind a coil with the wire, scrunch it in a ball, stomp on it, using copper, or aluminum, or gold, or spaghetti (without sauce), when a current passes through it, its resistance will create heat. This is called resistive heating and is characteristic of ALL resistance. 6. For #28 copper wire at 77°F, the resistance is 66.2Ω per 1000 ft, or .0662 per foot. (See http://www.engineeringtoolbox.com/co...re-d_1429.html). So if we have a wire which measures a total of 4Ω resistance, then it's length is, L = 4 / .0662 = 60.4 feet. If this wire is connected to a battery at 13.2V, a current of: I = 13.2V/4Ω = 3.3A flows The power generated by this circuit is one of these (they're equivalent): P = 13.2V x 3.3A = 43.6W, or P = 3.3A x 3.3A x 4Ω = 43.6W, or P = 13.2Vx 13.2V / 4Ω = 43.6W Then after one hour, the wire produces 43.6W x 3.4 BTU/Wh, E = 148BTU of heat 6. What happens to this heat? Some is transferred to its surroundings (air, oil, epoxy, spaghetti sauce, vinyl insulation, etc), and much of it goes to heating up the wire. The wire heats up from its own resistive heating. The above referenced table also shows how the resistance changes with temperature. A #28 wire at 77°F is .0662Ω per foot, and .0764Ω/ft at 149°F. Assume this wire connected to the battery heated up to 149°F. Redoing the above calculations for 60.4 ft of this wire at 149°F gives a resistance of: R = .0764Ω/ft x 60.4ft = 4.6Ω Recalculating for the wire heated to 149°F, I = 13.2V/4.6Ω = 2.9A The power generated is: P = 2.9A x 13.2V = 38.3W, or P = 2.9A x 2.9A x 4.6Ω = 38.3W, or P = 13.2V x 13.2V / 4.6Ω = 38.3W Then after one hour, the wire produces 38.3W x 3.4 BTU/Wh E = 130BTU of heat 7. The above deals ONLY with resistance of a wire, and does not address inductance. Note NOTHING was said about a coil, although forming this wire into a coil and stuffing it into a can, and pouring oil over it has NO EFFECT ON THE ABOVE CALCULATIONS. They only deal with resistance and the heating due to that resistance. 8. Some ways to change a wire's resistance: Change the diameter (gauge) Change the material (copper to aluminum) Change its length Cool or heat it Last edited by tac; 04-07-2016 at 09:50 PM. Reason: Fixed boo-boo |
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