question. simple starter motor conversion to 12v

[woofa.express]

I have busted 2 shafts on my starter motor, being 6v and running 12v system.


I have researched and there seems many ways to convert. The rest of the car has been converted. what is the most simple way. emphasis on simple. that's how I like things and that's why I like Model A's


thankyou in anticipation. gary

[Mark DeRoseau]

Replace the starter drive to a modern Bendix https://www.snydersantiqueauto.com/modern-starter-drive

[MikeK]

There is a reason things are breaking! A 6V starter on 12V will deliver four times the torque, not just twice as much. Power =amperage squared times resistance. The amperage will double for a 6V starter on 12V. When you square that, you get 4X engagement slam force.

You can buy new 12V configuration starter field coils to make the conversion from the vendors. With the starter motor fields rewired from parallel to series, (making it a 12V starter) the average converted starter will still show almost 2X the power, but not so much as to severely abuse the drive.

As far as 'barrel' drives go, they are not without their own problems. The can and do fail and are extremely difficult to change on-the-road unless you carry a special tool to compress and expose the shrouded set screw.

When a barrel drive fails and locks in the extended position because the much finer helix shatters from excessive motor torque and poor off-shore manufacture and heat treating, the rear of the drive often cannot be compressed, even with the special removal tool, to access the set screw. It becomes hack saw and abrasive cutoff wheel time.

The motor field conversion is the only way to go. If you continue to slam the flywheel ring gear with a 6V starter on 12V you risk walking the ring gear off the flywheel.

[steve s]

Originally Posted by MikeK

There is a reason things are breaking! A 6V starter on 12V will deliver four times the torque, not just twice as much. Power =amperage squared times resistance. The amperage will double for a 6V starter on 12V. When you square that, you get 4X engagement slam force.

I don't know what the correct formula is but I'm pretty sure this is not correct. Power is not torque (they have different units). Torque is the relevant quantity here, corresponding to "slam" force.

[MikeK]

Originally Posted by steve s

I don't know what the correct formula is but I'm pretty sure this is not correct. Power is not torque (they have different units). Torque is the relevant quantity here, corresponding to "slam" force.

Steve, torque would only apply to a gear tooth engagement load if the rotational inertia / moment of inertia of the starter armature and Bendix assembly was zero. Since this assembly has mass, and in a specific three-dimensional physical arrangement, it gains rotational inertia as it accelerates. That becomes the "relevant quantity" as the gear teeth crash (moving object meets stationary with a huge inertia component). I should have differentiated as to whether, in this starter motor problem, it was average power, instantaneous power, or mechanical power that I made reference to. All three have a time component, but differ in how that is applied.

In the case of this starter motor problem you are looking at near-instantaneous power with the time factor approaching zero, something in the millisecond range as the rotational inertia of the starter and Bendix is overcome and the angular velocity of the huge, massive flywheel and crank assembly increases. The difference between how many milliseconds to gear tooth engagement (slam!) for 6V vs12V is a function of force, in this case the amperage creating the magnetic field. 4X more amperage will result in faster armature/Bendix acceleration. You will have 4X more rotational inertia (function of RPM) at the point of engagement if voltage doubles.

OK, lets look at this slightly differently, as mechanical power. That is a product of force and movement. Movement will always contain a time element (exception- some particle physics). For a running motor, that's the product of torque on the shaft and angular velocity (RPM). Now, for this "slam" problem, movement instantaneously becomes near zero. The kinetic energy in the rotational mass as the gear teeth crash (moving object meets stationary with a huge inertia component) becomes the primary factor in the mechanical force applied to the gear teeth.

That said, I believe I am correct. Power will be a product of the amperage squared times the motor's resistance, you will have increased kinetic energy in the rotational inertia of the starter, and thus 4X the instantaneous 'crash'.

If you want to look at this a bit deeper, reverse EMF in a motor from a moving armature will decrease the amperage as the motor speed increases from starting (zero RPM). This reduces the power output. This is clearly evident in the MTFCA graph, amperage goes up/down as the starter motor slows (compression stroke) and increases (compressed mix passes TDC). At near zero elapsed time on that graph (X axis) the amperage is purely a product of Ohm's law. At about 25ms you see a strong amperage decline as the armature accelerates and creates an increasing reverse EMF. Then you see the up/down/up/down amperage as the starter runs fast/slow/fast/slow across the compression strokes.

Also on that graph you see an interesting huge 'down' spike in the amperage between T=0 and the few milliseconds to the initial start of motor rotation. That is the reverse EMF created by the buildup of the magnetic field in the iron cored motor armature and field. When the field stops building (max, but motor still not moving) the amperage spikes back up. That graph would be extremely interesting and revealing if a third component, angular velocity of the armature, were shown along the same time line.

[steve s]

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Originally Posted by MikeK

Steve, torque would only apply to a gear tooth engagement load if the rotational inertia / moment of inertia of the starter armature and Bendix assembly was zero. Since this assembly has mass, and in a specific three-dimensional physical arrangement, it gains rotational inertia as it accelerates. That becomes the "relevant quantity" as the gear teeth crash (moving object meets stationary with a huge inertia component). I should have differentiated as to whether, in this starter motor problem, it was average power, instantaneous power, or mechanical power that I made reference to. All three have a time component, but differ in how that is applied.

In the case of this starter motor problem you are looking at near-instantaneous power with the time factor approaching zero, something in the millisecond range as the rotational inertia of the starter and Bendix is overcome and the angular velocity of the huge, massive flywheel and crank assembly increases. The difference between how many milliseconds to gear tooth engagement (slam!) for 6V vs12V is a function of force, in this case the amperage creating the magnetic field. 4X more amperage will result in faster armature/Bendix acceleration. You will have 4X more rotational inertia (function of RPM) at the point of engagement if voltage doubles.

OK, lets look at this slightly differently, as mechanical power. That is a product of force and movement. Movement will always contain a time element (exception- some particle physics). For a running motor, that's the product of torque on the shaft and angular velocity (RPM). Now, for this "slam" problem, movement instantaneously becomes near zero. The kinetic energy in the rotational mass as the gear teeth crash (moving object meets stationary with a huge inertia component) becomes the primary factor in the mechanical force applied to the gear teeth.

That said, I believe I am correct. Power will be a product of the amperage squared times the motor's resistance, you will have increased kinetic energy in the rotational inertia of the starter, and thus 4X the instantaneous 'crash'.

If you want to look at this a bit deeper, reverse EMF in a motor from a moving armature will decrease the amperage as the motor speed increases from starting (zero RPM). This reduces the power output. This is clearly evident in the MTFCA graph, amperage goes up/down as the starter motor slows (compression stroke) and increases (compressed mix passes TDC). At near zero elapsed time on that graph (X axis) the amperage is purely a product of Ohm's law. At about 25ms you see a strong amperage decline as the armature accelerates and creates an increasing reverse EMF. Then you see the up/down/up/down amperage as the starter runs fast/slow/fast/slow across the compression strokes.

Also on that graph you see an interesting huge 'down' spike in the amperage between T=0 and the few milliseconds to the initial start of motor rotation. That is the reverse EMF created by the buildup of the magnetic field in the iron cored motor armature and field. When the field stops building (max, but motor still not moving) the amperage spikes back up. That graph would be extremely interesting and revealing if a third component, angular velocity of the armature, were shown along the same time line.

Mike ,
I feel like we're speaking two different but related languages: you, engineering and me, physics (although perhaps somewhat out of date). Perhaps you could answer a question to see if we have some common ground.
Do we agree that Ohm's Law, V=IR, cannot reliably used for circuits with inductance or capacitance elements?

[MikeK]

Originally Posted by steve s

Mike ,
I feel like we're speaking two different but related languages: you, engineering and me, physics (although perhaps somewhat out of date). Perhaps you could answer a question to see if we have some common ground.
Do we agree that Ohm's Law, V=IR, cannot reliably used for circuits with inductance or capacitance elements?

Steve, Good grief, kind of a loaded question. Even what appears to be a simple purely resistive circuit, like a battery, two wires, and a light bulb will have some, albeit near immeasurable elements of reactance/inductance and capacitance. That said, Ohm’s law is never an absolute.

Anything that functions on the relationship between electricity and magnetism, like any motor, will be highly inductive and, while running will not simply follow Ohm’s law. Case in point: The starter motor in this discussion. If it draws 125A @ 6V while running at cranking speed it WILL NOT draw twice that @12 while running. For most parallel dual field DC series motors that number will be about 1.4X the amperage if you double the voltage. Now, does this mean you will not get 4X the power? Yes and no. While running, no. Stalled, yes. At that point it will draw near 2X the amperage. The motor will transition from 4X “power” (actually ~3.8X if you want to set up a dyno on a test bench for the starter in question) to about 2X as the inductive reactance changes in a non-linear relationship with armature rotational velocity. The near-stall speed 4X (OK, ~3.8X) is what imparts the increased momentum energy in the armature/Bendix that creates the nasty ‘crash’ engagement.

On anything I’ve posted on Model A forums I’ve always tried to keep things in perspective of the audience. Not too many engineers here, so I have rarely delved into application of the work of Henry, Farad, Steinmetz, etc. . . for electrical problems. Even if I did, this forum lacks a math editor/font, and if I started posting anything with calc or functions derived from the work of anyone beyond Volta, Ampere, and Ohm I’d be talking to an audience of one (You!). I spent too many years scribbling that kind of stuff on blackboards in front of wannabe EE’s and ME’s with blank looks.

[Jim/GA]

Originally Posted by woofa.express

I have busted 2 shafts on my starter motor, being 6v and running 12v system.


I have researched and there seems many ways to convert. The rest of the car has been converted. what is the most simple way. emphasis on simple. that's how I like things and that's why I like Model A's


thankyou in anticipation. gary

The "6 volt field coils" get replaced with "12 volt field coils". That's it. It's actually pretty easy to do. If you have a good automotive starter and generator/alternator rebuilding shop near you (we have a couple in Houston) they can do it for you. You get the field coils from a Model A parts supplier, like Bratton's. If you have a good soldering iron to solder the field coils in, you can do it yourself.

Check your initial spark timing. That's will break starters.

[Synchro909]

I once bought a set of those 12v coils. Never again. I convert using the original ones. It is as simple as rearranging them so that they are all in series rather than two sets of two in series each connected in parallel. That’s not easy to put in words but I hope you get it. This gives a field coil resistance 4 times what it was and reduces the power and torque to about the equivalent of running the in modified stater on 8 or 9 volts, I estimate.
Another way is to reduce the voltage getting to the starter. I’ve seen this done a number of ways including using a Bendigo spring in the +ve cable (I assume you converted to negative earth). Another way is to go to a wrecker and find a set of leads from a really small car - the ones that are not much thicker than a pencil. The longer the better. The small cables have a higher resistance and will cause a voltage drop, making life easier on the starter. There are several ways to go - these are just a couple for you to consider.
PS. Have you collected that gearbox yet? All Good?

[Jim Brierley]

I've been running 12v on a stock starter since 1960, never broke anything except 1 Bendix spring.

[Purdy Swoft]

I agree with Jim Brierley. I've also been running a completely stock original 6 volt starter on 12 volts for many years without a problem , I did the same with Y block engines with good results . Back in the day 1954-55 six volt starters were to kill for .

[chap52]

Ran the 6 volt starter on 12 volts for many years without any problems.
Chap

[wensum]

I'm running a 6 volt starter on 12 volts with a modern Bendix drive and so far no problems. It's spinning a modified motor with 7:1 head

[Purdy Swoft]

I also run the modern starter drive with good results for the past 12 years .

[lake_harley]

I'm totally new to Model As, in fact I finally drove the '31 Coupe I bought last Summer for the first time today. I promised myself to finish a number of other started projects before I would do anything on the A. I'm now at 2.1 miles and counting.

It might be considered heresy that my plans are to do some hot rod type things like might have been done in the 40's or 50's but I'll give it a go anyway. I plan to retain the stock engine with mild mods and drive the car as much as possible so I want bright lights and reliability.

Since my question is starter related I thought I'd ask it here rather than starting another thread. I'm considering converting the car to 12V (I know - not 40's stuff) and since it is an older "survivor" car I'd plan to rewire everything on the car. While changing the starter to 12V, is there any big deal to convert to negative ground while I have it at my local starter/generator shop? I was afraid I messed up something today when I was getting the car started. I absent-mindedly hooked a 6V battery charger up as if it was negative ground. I finally realized my error and hooked it up correctly. With that little deal it made me think it would be nice to have it 12V negative ground just like everything else I own.

Thanks, in advance. Next I'll be asking about leaks that appear to be coming from both the front and back crankshaft seals.

Lynn

[Synchro909]

With respect, the fact that some of you have run 12v on a 6 v starter for years doesn’t alter the fact that Woofa has broken 2 in a short time. I run 12 v on my cars but I’ve played with the starter too to prevent what has happened to Woofa. I’ve made my suggestions and answered a query from him by email. Let’s see how he goes - hope he reports back.

[woofa.express]

Originally Posted by Synchro909

With respect, the fact that some of you have run 12v on a 6 v starter for years doesn’t alter the fact that Woofa has broken 2 in a short time. I run 12 v on my cars but I’ve played with the starter too to prevent what has happened to Woofa. I’ve made my suggestions and answered a query from him by email. Let’s see how he goes - hope he reports back.

yeh, that's all correct mr Synchro and thankyou. however I must correct you. you see Woofa is my hound and can be viewed on my page. Woofa. express is my log on site. I say Woofa is my dog. Well that's not quite correct. Woofa was my dog until he was sadly retired by a Nissan Patrol. Never liked Nissan's since.

[Purdy Swoft]

If you want to run negative ground, first reverse the battery cables to negative ground. reverse the wires at the coil and at the ammeter and you will be wired negative ground . The starter doesn't know the difference between negative or positive ground . if you feel that it is necessary go ahead and have 12 volt field coils installed in the starter or install them yourself if you can . you will either need to install a ballast resister at the coil or change to a 3.0 OHM coil . For the brightest head lights you will need to install at least a 60 amp alternator . change out the headlamp bulb to the 60-55 watt halogen bulbs The original type wiring harnesses will work just fine with the wires switched as mentioned above . the larger gauge wire will carry the higher voltage and amps with no problem . All of the parts are available in the common model A parts catalogs . If you choose to use a three OHM coil instead of a ballast resistor , you should be able to get one at NAPA . For better performance , I use the Pertronix flame thrower coil . I hope this helps .

[lake_harley]

That's the info and advice I was hoping to get. Thanks, Purdy Swoft!


Lynn

[Purdy Swoft]

Originally Posted by lake_harley

That's the info and advice I was hoping to get. Thanks, Purdy Swoft!


Lynn

Lynn , you are very welcome .