From the data available here and at the manufacturer of the heatsink in question, you can make a very reasonable estimate of the size required. From the spreadsheets that are flying all over this board for each of the various chips pull the total watts used and the %efficiency for the design you are estimating. Figure out the watts of heat that your solution will be emitting. Then take a look at the specs page of the sink in question and find the thermal resistance. It will be shown as c/w or degrees c rise over ambient per watt. Then its just simple math to figure the expected rise above room temp you can expect for the solution.
http://www.led-heatsink.com/seepricesandbuy.php?cateid=1276
Something like this will work very well for a single chip 3070 or V29 @ 1.4a or less.
3000k 3070 AB @1.4a = 52.2 w total dissipation @ 42% efficiency @ 50c.
So @ Tj of 50c it will be putting out 0.58 x 52.2 = 30.3 watts as heat
The sink above is rated at 0.88 c/w
30.3 x 0.88 = 26.7 c expected rise above ambient for this solution.
For your project you can still work the math to get a reasonable number.
http://www.heatsinkusa.com/12-000-wide-extruded-aluminum-heatsink/
These have the thermal resistance listed as c/w/3" which means the number applies to a 3" length of sink.
3000k CXB-3590 CB 72v @ 700ma = 48.8w total dissipation @ 51.7% efficiency @ 50c Tj
48.8 x .483 = 23.5 watts of heat x 3 chips = 70.7 watts of heat output
The heatsink above has a c/w/3" of 0.9
A 12" length of this model can expect to see 70.7 / 4 x 0.9 = 15.9 c rise above ambient.
Now this is for the worst case scenario as it assumes no air movement over the unit. A light breeze from canopy fans will help significantly.
It is always a good idea to err on the high side when it comes to sizing of heatsinks. But you do not have to make wild guesses about where to begin.