Can't scale up BlinkM network.

Hi Will,

Short answer: try smaller pull-up resistors (1.2k) first, then try using an "I2C bus driver" chip.

Longer answer:
At 30 feet, you're approaching the theoretical limit for I2C bus length using standard ribbon cable. Making reliable long I2C busses is a bit tricky. This is because I2C is more sensitive than most protocols to increased bus capacitance. And any time you get a long strands of multiple wires next to each other, they start looking more like a capacitor (what's a capacitor than two pieces of metal separated by a gap)

There's some pretty good discussion of these issues on this page:
http://www.i2cchip.com/i2c_connector.html#Bus%20Pullups%20and%20Drive

Where the main take-away might be: try using smaller pull-up resistors (1.2k) on longer bus lengths.

To really solve the problem, people use "I2C bus driver" chips. These are constant-current sources that are more accurate than a simple pull-up resistor. There are many different vendors of these chips, and they're all about the same. A real-world example of using these chips is Dave Vondle's big BlinkM LED wall:
http://labs.ideo.com/2009/08/19/glass...

For *really* long I2C runs, there are various types of "I2C bus extender" chips. These typically need to go in front of each I2C device, so probably aren't applicable for use with BlinkM. But some of these kinds of devices even do a kind of wire protocol modification to allow I2C to work up to several hundred meters.

The Arduino board can only provide about 500mA of current, so you can't use the power supply on the Arduino to power your 50 BlinkMs. But if you have a power supply that can provide up to 3A. For instance, here's a $15 wall-wart that provides up to 2.5A at 5VDC.
http://www.jameco.com/webapp/wcs/stores/servlet/ProductDisplay?productId=252736
You could get two of those to give you about 5A, which should be enough head room, and you can use it to power the Arduino.

(If you're not used to doing power calculations, typically you want a power supply that can provide about twice the current needed at a given voltage. If you had a system that consisted of 50 BlinkMs (60mA each), an Arduino (100mA at most), and some support circuitry (maybe another 100mA), you could estimate that at most the current draw you'd see would be (50 * 60mA + 100mA + 100mA) = 3200 mA = 3.2A. Double that to get 6.4A. Now in practice you might know that you won't be powering all your BlinkMs at full brightness at once, so you won't need the full 3.2A, which is why I said 5A should be sufficient.)