Column C

First off you'll notice the text "Column C to be used in all cases except as otherwise permitted in Note 3.)", in the title of table 220.55. This makes it simple. You have 2 appliances, so follow that over in the table, and you'll see 11 in Column C. So there you go, you can just use 11 kW. Done.

11,000 W / 240 V = 45.8333 A

So you'll need a 50 ampere breaker, and wire sized appropriately for the load.

Note 3

Note 3 says:

3. Over 1 3⁄4 kW through 8 3⁄4 kW. In lieu of the method provided in Column C, it shall be permissible to add the nameplate ratings of all household cooking appliances rated more than 1 3⁄4 kW but not more than 8 3⁄4 kW and multiply the sum by the demand factors specified in Column A or Column B for the given number of appliances. Where the rating of cooking appliances falls under both Column A and Column B, the demand factors for each column shall be applied to the appliances for that column, and the results added together.

Perfect, so instead of just using the value from column C you can do math. Let's step through it.

...it shall be permissible to add the nameplate ratings of all household cooking appliances rated more than 1 3⁄4 kW but not more than 8 3⁄4 kW...

8.4 kW + 5.0 kW = 13.4 kW

...and multiply the sum by the demand factors specified in Column A or Column B for the given number of appliances...

Let's check the table again... You have 2 appliances, both between 3 1/2 and 8 3/4 kW. So You'll look at column B, and find 65%.

13.4 kW * 65% = 8.71 kW

8710 W / 240 V = 36.2916 A

So using this method you can use a 40 ampere breaker, and appropriately sized wire. However, keep in mind that if you change the equipment, you'll have to do the calculation again. So while you can use this value, you may have to upgrade the circuit later if you change equipment.

Note 4

I'm not exactly sure how note 4 comes into play, but I think it can be used if this is the only equipment on the branch circuit. Just for fun, let's run through that one too.

4. Branch-Circuit Load. It shall be permissible to calculate the branch-circuit load for one range in accordance with Table 220.55. The branch-circuit load for one wall-mounted oven or one counter-mounted cooking unit shall be the nameplate rating of the appliance. The branch-circuit load for a counter-mounted cooking unit and not more than two wall-mounted ovens, all supplied from a single branch circuit and located in the same room, shall be calculated by adding the nameplate rating of the individual appliances and treating this total as equivalent to one range.

You're only concerned with the second half of this note, since you have one counter-mounted cooking unit, and one wall-mounted oven, all supplied by a single branch-circuit, and located in the same room. So you can add the nameplate values, and treat it as a single range.

8.4 kW + 5.0 kW = 13.4 kW

So you can treat the units as a single 13.4 kW range. Check the column C again, this time for a single range. You'll find a value of 8 kW. But wait... The column header says "(Not over 12 kW Rating)". Your range is 13.4 kW. That's bigger than 12 kW. Now you'll have to check note 1

1. Over 12 kW through 27 kW ranges all of same rating. For ranges individually rated more than 12 kW but not more than 27 kW, the maximum demand in Column C shall be increased 5 percent for each additional kilowatt of rating or major fraction thereof by which the rating of individual ranges exceeds 12 kW.

That's easy enough.

13.4 kW - 8 kW = 5.4 kW

Since .4 is not a "major fraction", you can just use 5 kW. So you'll have to add 5% 5 times.

5% * 5 = 0.25
8000 W * 0.25 = 2000 W
8000 W + 2000 W = 10,000 W

That means you'll have to use 10 kW as your demand.

10,000 W / 240 V = 41.666 A

Which means you can use a 50 ampere breaker, and appropriately size wire.

Does the sub-panel seem over loaded? If so, I could keep the water-heater in the main panel and free up space in the panel another way.

Seems reasonable to me. Most of the equipment won't draw anywhere near the overcurrent rating, at least not during normal operation. Motor loads will draw a higher current on start, but you shouldn't have a problem.

I know I need four-strand wire to run to the sub-panel (2 hot, neutral, ground) but copper or aluminum and what gauge?

You can use either copper or aluminum, though I recommend copper for DIYers. Copper is quite a bit more expensive, but it's easier to work with (in my opinion). If you feel confident working with aluminum conductors, you can save some money using it.

I've covered the topic of feeder sizing here, so I won't go into detail. If you're using copper, you'll want to use 3 AWG conductors. If you choose to use aluminum, you'll need 1 AWG conductors.

If you want to run a single cable, instead of individual conductors in conduit. You can buy what's called 3-3-3-5 SER cable (1-1-1-3 for aluminum), which will contain three 3 AWG conductors (hot,hot,neutral) and a 5 AWG grounding conductor.

When I run the wire along the floor joist, does it need to be secured to the joist or can it just hang there and rest on the drop ceiling? Seems like it should be secured to the joist with wire hanger or something.

You'll have to attach the cable to the joists, using 1 - 1 1/4" staples or other approved means. Check the packaging, to make sure they are rated for the size cable you're using.

What are the things about this project that I don't know that I don't know. :) These are the scary things IMO...i.e. the questions I don't know enough to ask.

The cable you'll be working with is thick and heavy, and it's not going to be fun pulling it. You'll probably want a couple helpers, to help you wrangle it.

Make sure all your connections are tightened to the manufacturer's specified torque.

If you choose aluminum conductors, make doubly sure you tighten the connections. And don't forget the anti-oxidant.

Come back a day or two after the panel has been put into service, and tighten any connections that need it.

Don't forget to remove the bonding jumper between the grounded and grounding bus bars.

You'll need clamps big enough for the cable, to secure it to the panels.

should I put a 100 amp breaker in the sub-panel to act as the "main" for the sub-panel? Or is the 100 amp breaker in the main panel sufficient?

You can usually pick up a main breaker panel, for about the same price as a main lug only (MLO) panel. In my opinion, unless the secondary panel is next to; or within sight of, the main panel. You're better served to install a main breaker panel. It simply offers better protection during maintenance, or other work within the panel.

For example. If you turn off the feeder breaker in the main panel, and start working in the secondary panel. Somebody could easily come along, and flip on the feeder breaker. Since you can't keep an eye on the breaker, you can never be sure the panel will be dead. (unless of course you're using a lockout like you should).

If the secondary panel is in a separate building or structure, then you either need a main breaker, a main disconnect, or the ability to disconnect all ungrounded conductors within 6 or less hand moves.