| sonof19 | -- 11-19-2010 @ 8:58 AM |
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I have a 1948 ford super deluxe v8 12v system.before I knew that the heater reducer was suppose to get hot I change the ceramic one to a 10 amp metal one. (see pic.and this one was so hot it put a blister on my finger is this normal? and is there any way to reduce this heat? I am afraid the car will catch fire.also I put the old one back on and now the heater will not work it act like it want to but don't.what have I done? thanks son of 19
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| supereal | -- 11-19-2010 @ 9:25 AM |
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Your voltage dropper appears to be a simple ceramic resistor. It is usual for these to get very hot when there is a heavy load, such as a heater motor. Voltage drop is determined by current multiplied by resistance (Ohm's Law). But motors provide a reactive load, quite different than just a resistive load, particularly when they are operating at a higher speed, which easily exceeds the capacity of just a resistor. At one time, a 12 amp dropper was available, but most suppliers don't carry them now. As heaters are rarely used in old cars, you will probably have to live with the resistor, but be sure it is not near anything combustible.
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| sonof19 | -- 11-19-2010 @ 10:08 AM |
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Thanks supereal" do thank I could put two of thes 10amps togather run them concurent would that help or no?
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| supereal | -- 11-19-2010 @ 12:36 PM |
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Have you checked to see what kind of voltage is presented to the heater motor when it is operating? You can check by placing your voltmeter between ground and the wire feeding the heater motor. If is not in excess of 7 volts, you can place another resistor in parallel to increase the current carrying power. If it is greater than 7 volts, you can double the resistance by placing the resistors in series. Then, recheck the voltage with the motor on high speed. If resistance remains constant, voltage will fluctuate with the current draw, but can't exceed battery voltage, in any case. Voltage reducers are designed to operate items with a constant current requirement, such as gauges, lights, and radios. Motors are a different class because the load varies widely.
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| sonof19 | -- 11-19-2010 @ 1:47 PM |
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Thanks again" I will do some checking.you have given me some good advice. son of 19
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| Early46 | -- 11-19-2010 @ 2:39 PM |
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The resistor in the photo looks like the item that Speedway sells as a voltage reducer for fans, etc. It is a 1 Ohm, 50 Watt resistor. They also sell a 1.5 Ohm, 25 Watt ceramic device in a different package. The actual device required does depend upon the current drawn at steady state by the fan motor running at high speed. You must then calculate the required resistance using Ohm's law ( Resistance = Voltage / Current). I would not connect two resistors in parallel as this will cut the resistance in half and thus raise the voltage the fan motor "see's" and therefore increase the power the fan motor must dissipate. In changing back to the ceramic resistor, I expect he may have a loose connection, they must be tight otherwise they will also act like a resistor. Make sure any terminal lugs are properly crimped! Remember, if the heater motor draw 5 Amps, then the resistor must dissipate 30 Watts. This is like a 30 Watt light bulb, it WILL get HOT!
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| proxie35 | -- 11-19-2010 @ 2:48 PM |
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I,ve had very good luck with the ceramic reducer, CM used on the 55-57 Chev. It was mounted on the firewall & reduced the voltage to the coil. I'm planning to try one for the Overdrive on the 49 that I have almost compled. I don't know what amp. this reducer is. I'll post the results on the overdrive in a few weeks.
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| supereal | -- 11-19-2010 @ 3:14 PM |
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It is true that placing the resistors in parallel will halve the resistance, and placing them in series will double it, but without knowing the rating of the resistors, it will double the current carrying ability. That is why the resultant voltage during the operation of the motor needs to be measured. The problem is addressing the impedance of the motor, a different quantity than pure resistance, and the type of motor (permanent magnet, etc) in calculating the effect of any voltage dropping device.
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| TomO | -- 11-20-2010 @ 8:13 AM |
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If you are committed to leaving your car 12 Volts, spend the money and have your heater motor converted or look for a 12 volt motor the will fit in your heater. Conversion cost should be around $100 a rebuilt motor will be about the same. Tom
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| sonof19 | -- 11-20-2010 @ 9:36 AM |
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I want to thank each and ever one of you for your help.no one should ever have any problem operating these old fords as long we have folks like the v-8 club.I thank I will buy me a 12v heater motor any one know where I locate one?
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| supereal | -- 11-20-2010 @ 3:07 PM |
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Any good full service electric motor shop should be able to find one that will fit the mounting. We have one of those next door to our shop, and they have found all kinds of motor for us. Pull your motor and take it along.
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| 51f1 | -- 11-21-2010 @ 7:43 AM |
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Twelve volt heater motors are available from the usual parts suppliers. LMC has one for $109.95 on page 66 of their latest truck catalog. Phone 800-562-8782. Richard
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| Early46 | -- 11-21-2010 @ 8:17 AM |
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I would agree, installing a 12V motor would be much better solution, just much more expensive. If you want multiple speeds, you will also have to find a switch/resistor combination designed for the motor. Super, I would agree that trying to measure the DC resistance of a motor to determine full load current is meaningless, but measuring the voltage drop across the motor or the voltage reducing resistor will only tell you if the voltage resistor is properly sized (i.e. correct resistance value). If the voltage reading is not to your satisfaction ( i.e. between 6 and 7 volts across the motor) then you should change the reducer. To determine the resistance value of the correct reducer, you will either need to determine the current draw or the value of the currently installed voltage reducing resistor. Hopefully the resistance value of the reducing resistor is marked on the device, but if it is not then you must otherwise determine it's value. As the nominal value is typically around 1 Ohm, making an accurate resistance measurement will be problematic, as most consumer grade VOM's or DVM's will be very inaccurate at this range. The measurement must be made with care, and you need to account for lead resistance and minimize probe contact resistance. Even using the Fluke 88, a high end automotive DVM requires the use of the Relative mode and then the accuracy will be around 20%. Once you know the resistance though, you essentially know the current and can compute an approximate value for the new resistor. This will only be an approximate value as the current drawn by the motor will change with applied voltage. That is why measuring the actual current at the nominal operating conditions is the best way! An aside to everyone, the "voltage reducing" resistor sold by most after-market automotive vendors for over $12.00 is nothing more than a aluminum housed, wire-wound, chassis mount power resistor, available from most electronic suppliers for around $5.00!! FINALLY, to the question of heat dissipation, the reducer will become warm, as it must dissipate around 30 Watts. To reduce the case temperature, I would recommend mounting the device on the firewall or frame so that the heat can be conducted away, essentially using the car as a heat sink. Make sure that there is very little between the resistor and the metal, as you need a good heat conducting path.
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| supereal | -- 11-21-2010 @ 10:48 AM |
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Can't quarrel with that, 46. Most non-professional meters can't measure current draw above milliamps, so we sometimes use a 1 ohm 20 watt resistor in series with the part to be tested, and put a voltmeter across the resistor. The voltage reading will be about 1 volt per amp. We have compared this with our Fluke clamp-on, and they are close. Once you know the current, you can approximate the amount of resistance to produce the nominal voltage. As even these small motors require more current than just a wire wound resistor can handle, a better voltage regulator/dropper would use a Zener diode and power resistor. A more cost effective solution is just to get a 12 volt motor, of course.
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| 37RAGTOPMAN | -- 11-21-2010 @ 4:08 PM |
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I am with TOM O, put a 12 volt fan motor in it,with a 3 way switch, also I also HOPE you have ALL NEW WIRING in the car,PLUS disconnect switch,for safty sake my 3 cents worth,37 RAGTOPMAN, and keep on FORDIN,,,!!!
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| sonof19 | -- 11-24-2010 @ 8:18 AM |
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Thanks for all the good advice. I think a 12v motor is the way to go. yes I do have new wireing.son of 19
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| trjford8 | -- 11-24-2010 @ 11:43 AM |
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You might also try Napa Auto Parts. They have a good selection of heater motors. You need to know the rotation of your current motor(clockwise or counter).Somtimes you may have to shorten the shaft if you find one that will fit your mounting. Putting the 12 volt motor in your heater is the best solution.
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| sonof19 | -- 11-27-2010 @ 7:54 AM |
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I think facing the heater switch the first notch to the left defrost high. the next notch is low.and the motor is running backwards the first notch to the right is hi heat. and the next is low heat and the motor is running forward.any thoughts on this? son of 19
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| TomO | -- 11-27-2010 @ 8:36 AM |
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son of 19. Turn the switch to the right for defrost and to the left for direct heat. This agrees with your description of the switch. If you are going to use the original switch with your 12 volt motor, be sure to replace the bulb with a 12 volt bulb. Tom
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