If your looking for ANY type of mathematical formula to calculate any type of heat dissipation, or quantum singularity, or space time continuum, PM lehpron

__NaZ
searching though the forum i found this  interressting formula:
(OC TDP/ stock TDP) = (OC multi/ stock multi) x (OC Bclk/ stock bclk) x (OC Vcore/ stock Vcore)²

Yeah that's my work there, specifically its an interpretation of Ohm's law applicable to computer processor power and heat variances.  The only truly unknown quantity is stock TDP. 
 
With that said, how did you estimate 270W, did you use the TDP rating for your CPU (95W) or estimate actual TDP?  Intel's ratings aren't the real heat values of each CPU, they are just ranges a bunch of CPUs will fall under to make it easier to cooling companies to tailor heatisinks towards. 
 
Frequency ratios affect each CPU core and coresponding Vcore, but you should add the tiny portion used by the integrated memory controller at least in the new Intel "Core i" CPU's and AMD CPUs.  Note however, Vdimm affects the RAM's heat, not the CPU's heat.  For LGA1156 CPUs, the PCIe controller is also in the CPU, so consider that extra heat.  "Affect" is a certainty, it will happen; whether you can measure by thermal probe or sensor it is a different story.
 

__NaZ
knowing that I would want to know how much heat 1 120mm rad can get rid off. i know one rad isnt enough.

is there some kind of "rule of thumb" how much heat a rad can roughly handle ?

 
Any single 120mm radiator is actually capable of removing more than 200W of heat by itself, but this is a design limit and the heat isn't removed at any particular stable temperature that suits us, which varies person.  More than likely heat removed at a design limit is at a temperature we as overclockers and enthusiasts aren't comfortable with, hence needing more radiators to compensate.
 
How many more depends on you, rule of thumb is generally perogative but IMO there is no such thing as overkill provided you did your homework, you already know what you want and whether what you get will work for you, for the most part.  Choose a minimum of 2-3 ratio gap, take your requirements, multiply by 2 or 3 and get radiators for that new value.  It is safer to have a larger gap that way you aren't at the mercy of ambient/case temperatures, fan CFM or pump flow rate varations which will force wildly varying load temperatures.  Since load temps are everything to a stable overclock, it is best to plan ahead with more.
 
All the components in your liquid loop absorb some heat and dump the rest, this is how all coolers work actually, when an object absorbs heat, its own temperature rises.  All components have a thermal resistence factor, its basically a percent of what is absorbed.  Lower the better, but can't go below zero.  For example, a typical high-end air cooler can have a factor of 0.15^oC/W (degrees centigrade per Watt of heat) up to about a 210W load (which is about what six 6mm heatpipes can handle by themselves), this means for a 200W load, 15% will raise temps above ambient and you get load temps, or about 30^oC above.  A typical high-end liquid can get as low as a factor of 0.08^oC/W up to as much heat removal as radiators available.
 
I'll use Swiftech's radiators to estimate, as they are the only vendor I know that gives detailed specs of their stuff away.  The thermal resistence factor of their MCR120, MCR220 and MCR320 radiators at a flow rate of 2gal/min and 80CFM fans are 0.055^oC/W, 0.033^oC/W and 0.025^oC/W respectively (to get the total factor for the loop, add in the factor for the water block).  Their total amounts of heat removal under the same conditions are 175W, 325W and 450W, respectively.  Doesn't seem like much, I think those by HWlabs are able to handle twice the load.  But then we are talking about a flow rate of 2gal/min, doubling that will give you results by deminishing returns.  In otherwords, even if you get the rads for your target load, if you don't have the pump for the job, forget it.
 
When plotting these on a graph, you'll see with evey additional radiator, the thermal resistence drops exponentially while the heat removal curve looks almost straight, as expected though after a few dozen it could level out:
 

 
Depending on which brand of rads you go with, the amplitude of the plots will change, but the curves will have the same form.
 
I threw in those formulas so you can make your own estimatations, but I'm sure some may regard that I had too much fun with this post.

Regular Guy my a$$, that man is as close to Albert Einstein as we get here in these forums.

lehpron 

...rule of thumb is generally perogative but IMO there is no such thing as overkill provided you did your homework...

That, I can understand!  Hooray!
 
Most of the time lehpron you go way over my head.  Not that it's hard to do...  I'm just glad that you are in these forums.

I can't tell if you guys are making fun of me or putting yourself down, just ask questions, don't hesitate, anything.  But you have to do it, I can't initiate as I can't read minds.

__NAZ, when you get back in here I'd like to know your stock Vcore, I'm still have a hard time with how you figured your 860 to be dissipating 270 Watts, seems almost double what it could be. 

Going by the formula:

(OC TDP÷stock TDP) = (19÷21) x (200÷133) x (1.3÷stock Vcore)²

if assuming 95W TDP as your stock (which it isn't btw), and 270 for the OC TDP, then your stock Vcore was way down at 0.89v... 
 
This was done by taking 270÷95 = 2.84, then divide the multi ratio and divide the bclk ratio, then take the square root.  Lastly, divide 1.3 from this number and you get the value for stock Vcore.  When you do the TDP equation, do the division first, then raise the voltage exponent, them multiply everything together.
  
Fact is, the typical i7 CPU will have a stock value between 1.1 and 1.3, OC TDP will not be higher than 160W at your settings for 3.8GHz -- you could use an H50 with two high-flow fans and call it a day.

The impact of hyperthreading according to many reviews have shown that enabling it increases temperature by 5%; all things equal, this accounts for a 5% increase in heat wattage.

As for estimating 860's TDP, I set the fastest of a series/type as the hottest part and scale down by clockspeed (or process tech or cache).  There is an i7 880 rumored to appear, so I'll tentatively assume 880 is 95W TDP to qualify for the rating.  860 has the 21x stock multi, 880 is supposed to get the 23x stock multi, so 860 is 9% slower.  Assuming the same, or very similar stock voltage, then that's 9% less heat, or about 86W -- this is closer to true i7 860 stock heat load.  Redoing the OC TDP equation and using this review's reference stock Vcore (since I don't know yours):

(19÷21) x (200÷133) x (1.3÷ 1.247)² = 1.478

86W x 1.478 = 127W --- this is closer to your heat load, not 270.

Above I stated best to have a gap between what you need and what you have, don't get minimum.  So multiplied by 2 or 3 and we get a radiator set from 260 - 390W, any dual 120 radiator will suffice, more than that will bring temperatures lower but with diminishing returns.

If you pruchased those dual rads from Swiftech, MCR220, that's a thermal resistence factor of 0.033^oC/W supporting 325W at 2gal/min and a pair of 80CFM fans.  A 2gal/min pump is a their 350/355 pump and if you choose Swiftech's Apogee GT CPU block which is rated for about 0.05^oC/W at 2 gal/min.  Total thermal resistence factor is the sum, 0.05+0.033 = 0.083^oC/W.

What does this mean and why should you care?  At a load of 127W, a resistence of 0.083^oC/W means the cooler as a whole will raise temperatures by 127W x 0.083^oC/W = 10.5^oC above ambient.  Which ambient, air?  Not necessarily, the liquid in the loop doesn't act as just a medium, not all the heat is transfered from the block to the radiator, its the whole reason the liquid heats up.  For Swiftech's radiator tests, they maintained a 10^oC difference between air temeprature and the liquid temperature throughout.  So the 10.5^oC actually adds to the liquid temperature.

Your theoretical minimum full load temperature with dual rads would then be 10 + 10.5 = 20.5^oC above the reference air temperature, which is the air entering the radiator (not air measured anywhere inside or outside your case or general room temperature).  If the air temperature entering your dual 120's is 35^oC, then the CPU IHS full load temperatures could be 55.5^oC.  If you went for Swiftech's triple 120 rad instead your full load temps dip into the upper 40's.
 
Here's another example:
 
When K|ngp|n ran his QX6700 to a WR 4.9GHz back in 2006, I memorized his numbers: 15 multi from stock 10, 327Mhz FSB from stock 266, and Vcore of 1.792 from stock 1.25.  His full load temps were -191^oC centigrade, used LN₂.  His TDP ratio was = (15÷10) x (327÷266) x (1.792÷1.25)² = 3.79 times higher than stock, or almost 500W of heat from a 65nm quad!  LN₂ boils at -196^oC, so the thermal resistence of his copper pot was the change in temperature divide by heat wattage or (-196 - (-191))÷ ~500 = 0.01^oC/W.

First, the power and heat may be measure in Watts, they aren't the same number.  The difference is energy efficiency, for electrical components, 80-85% of power input is heat dissipated.  The remaining 10-15% of power in is used to run your programs, for instance.

Power is voltage times current (P=IV), but since voltage is current times resistence (V=IR), we can say voltage squared divided by resistence is also power (P=V²÷R) -- this is the definition of Ohm's Law.  When we change the frequency of a processor, we mess with the inverse of the internal resistence of the circuit. 

A process called electromigration is already occuring in many processors, that electrons leak out of the silicon at a predesigned steady rate that for a certain stock voltage, if the processor is kept at a certain temperature, it maintains a certain lifespan.  But when we overclock, more electrons leak out at a higher rate, this i why we add voltage to maintain circuit stability (we are trying to compensate by putting electrons back in), because stock voltage was only tailored for the stock electron leakage.

The above TDP formula has no temperature factor, it is a plug'n chug manipulation based on numbers that may have been gathered at a certain temperature.  Only if a series of points were plotted with variations in temperature would that trend point towards a variation in heat/power since circuit resistence is temperature dependent and thus the circuit would not need as much voltage the lower those temperatures became.  This pretty much explains the advantages of better cooling from stock, whether overclocked or not. 

In effect, you're right, if K|ngp|n's QX6700 was stable 4.9GHz at any temperature, then the room temp Vcore would be much higher than 1.792v because this was the requirement at -191^oC, and thus the ambient heat dissipation would rise significantly.  If 2v was required at ambient, that's a 25% increase from LN₂.  But such speeds cannot be maintained at higher temperatures, so the only real way to detect variations in heat/power is if a colder cryogenic was used (like LHe₂) to see how much less Vcore was needed for the same clockspeed, since with LN₂ 4.9Ghz was stable with 1.792v. 
 
This can also be done with a stock processor too, with each succesive drop in temperature tried with better cooling, see how far the CPU can be undervolted stably. 

YerBuddy

lehpron

I can't tell if you guys are making fun of me...

Quite the contrary. 

+1

lehpron

just ask questions, don't hesitate, anything.  But you have to do it, I can't initiate as I can't read minds.

Teach me your ways

Watts X 3.41 = Btu

lephron you have my permission to date my mom.