Yes, it's true. I was able to obtain a very rare unbored, "solid" FE block. I then instructed my builder to bore and stroke it to exactly 122 cubic inches. You can't tell that from the outside though.:cool:

I have a Painless Wire harness/fuse block in my 65 fastback....the wiring goes something like this: Battery + cable to the starter selonoid, then a large wire with an in-line BAF 50amp fuse to the fuse block to feed the entire system from the same terminal.....

I'm sure circuit breakers work fine, I just don't care for them, I'd rather have a fuse, but that's just me.... I'm sure Painless or any other wiring suppler could supply a large enough wire with an in-line BAF to go to your main fuse block....

Mine is mounted on the inner fender well, not far from the starter selonoid, where it is easy to access should that need arise, so far, 15 years later, I have never touched it, knock on wood......

David

In rush current is typically 6 times steady operating amp on a motor. Much higher on large inertia loads. I would put the rotating mass on an FE as higher than typical. So I would say the battery likely limited the inrush.

Since I would have expected closer to 350 amp (about 5hp) to turn an FE, I could be wrong. 200 amp would only be 3hp. It still sounds like that FE is on the wimpy side to me! :LOL:

Uhhh, what do you expect for 122 cubic inches?:p

If I had a GM style starter/solenoid combo I'd consider a fusible link over a circuit breaker.

Pat, inductive amp meter? Wait, GOTTA be, no way you could pass 900 amps through that thing! :) Mine is limited to 10 amp direct input.

Yep, inductive DC ammeter. Nothing but the best to help out my fellow Club Cobra members.:cool: Plus, the clamps on the end of the meter are perfect for pulling Cheetos out of a bag without breaking them (when your fingers have anti-seize all over them). The ammeter still works properly even with orange cheese powder on the ends.

Just thinking out loud,,, I would suspect the amperage to go up with a longer rather than a shorter battery cable. Not to mention a good or great ground strap path.

Battery in the trunk? More amps than under the hood. The starter will attempt to draw every amp it needs, no matter what, even if that means overheating the battery cable.

Well quit, as resistance goes up, amperage goes down.;)

You say you have a bad history with batteries & cables ?:LOL:

Here you go Ernie.....


To make a current flow through a resistance there must be a voltage across that resistance. Ohm's Law shows the relationship between the voltage (V), current (I) and resistance (R). It can be written in three ways:
V = I × R or
I = V/R
or
R = V/I

where:
V = voltage in volts (V)
I = current in amps (A)
R = resistance in ohms (R)
or:
V = voltage in volts (V)
I = current in milliamps (mA)
R = resistance in kilohms (k)

Ah yes, the math, but the question remains...

If Patrick had a shorter cable, of the same size, would the amps go down from 210?

If Patrick had a smaller cable, same length as now, would the starter still draw 210 amps and over heat the cable in the process? As resistance goes up (cable gets hot) I'm wondering if the starter will simply draw the amps it needs (210), the cable be damned.

Of course "time" is also a factor, cranking for to long and over heating the cable to much would be a serious problem. But were talking about a short time frame of cranking, as recommended by manufacturers.

No. In theory, the resistance would go down, so the amperage would go up.


A smaller cable, with the same length, would result in higher resistance, and thus, less amperage.

Ernie, here is a handy-dandy little resistance and voltage drop calculator http://www.stealth316.com/2-wire-resistance.htm I think most of our battery cables are 4 gauge, right? If you add in the additional few feet of wire that it takes to move a cable to the back (my current battery-to-solenoid is 32 inches long) it will calculate the additional resistance and voltage drop. It even has a "battery move to trunk" example.

The theory, and the math, are solid of course. OK, so here's your next weekend project, Pat, move your battery to the fender and rig up a short cable of same size and take some readings. :) :) :)

Bonus points: Then use a smaller cable!

Oh, I thought your battery WAS in the trunk! Mine is and my cable is certainly longer than 32". Don't know the size, but like me, it's really big.

I'll get right on it.:D

Ernie, this is a really nice little article on "voltage drop." It's what you want to know and it is explained very well. http://www.engine-light-help.com/voltage-drop.html

Voltage, and drop, is easy to measure. Amps are much more difficult without a good inductive amp meter. Especially in the 1,000 amp range! :)

I tried the fancy geared starter that cost a ton, it cranked to slowly for my taste. Back to a stock Ford OEM NAPA starter (50 bucks) and it works just fine. I sure wish I had taken amp readings before and after with those two starters (just for the heck of it).

Dang, I'm tempted to go out and buy a clamp-on ammeter that will work with my Fluke-87. :) In fact, I may not only do that, but also rig up a short cable to the starter (solenoid actually) to compare amps to my trunk mounted cable setup!

Although Patrickt is quoting the laws correctly he may be incorrectly picking which variables are constant. An electric motor is turning a load and the torque required to turn that load is not changing. If the wire is longer and thus has more resistance, there will be more voltage drop across the wire (IR drop). So less voltage gets to the motor. If the battery can supply more current, which it can, the motor will pull more current to make the power needed to turn the load.

On AC motors what I have said is absolutely 100% correct because motor speed is a factor of the Hz. When voltage dips in a plant the ac motors pull more current because the work or wattage is constant. W = E * I so if W is constant and E goes down then I must go up. Now with that said DC is a little different because Motor speed is a function of the voltage. So when voltage goes down the motor rpm will be less and the motor will be doing less work. Now with both the work and the voltage changing it is more difficult to predict what the current will do, as it depend on how much less power is needed to turn the motor at the lower speed. Since the voltage drop in the wire is going to be slight, I suspect the work to turn the motor will drop an insignificant amount relative to the voltage drop and the motor will pull more amps.

PS
On second thought. On a dc motor current is proportional to the torque the electric motor is putting out. In the rpm range change of the voltage drop in the wire, the torque required to turn the gasoline engine is going to be negligible, although theoretically it would be less. Therefore, the current draw will remain unchanged. Other variables such as oil temp and compression will be bigger than the voltage drop in the wire. However if the voltage drop in the wire were to be large enough to substantially reduce the rpm of the motor and cranking engine, the torque require to rotate the engine would go down and the dc motor would pull less amps, because the torque load was reduced.

Now THATS what I'm talking about! That starter gonna do what it has to do, cable be damed, melt that sucker if it has too. :)

No, your DC starter motor doesn't "take what it wants," it uses what it's given. You can actually vary the speed of a DC motor by just using a large enough potentiometer. In that case, the motor doesn't say "Oh, I see you've turned up the potentiometer but I'm still going to rip more current through it and spin at top speed anyway." (And the torque of a DC motor is also proportional to the current that it's drawing.)