- Joined
- Feb 11, 2002
- Location
- Seattle area
Here's the situation. I have a Swiftech Storm G4, and an ASRock 939Dual-SATAII motherboard. The two don't quite mix, cuz the mobo doesn't come with backplate spec'd by AMD. That backplate is required for stock mounting of the G4 on a 939 socket.
To remedy the situation, I'm going to remove the stock 939 bracket from the mobo, exposing those 4 mounting holes. I'll then use the Xeon mounting method and hardware supplied by Swiftech (or mostly so). The important part is the 4 black springs provided with the Storm. I believe AMD specifies 75 pounds of mounting force, which seems absurdly high, but I can't argue. I've calculated that I need to compress each of the 4 springs a total of 9.5 turns on a 6-32 bolt. That's 9.5 turns passed just touching.
Someone please double check my work. Here's how I came up with that:
I created a jig that would allow me to measure the spring compression with a dial caliper while applying a known weight. I started with just the weight of the jig, and recorded spring heights of 0.509 and 0.502 inches. I needed to use 2 springs, side by side, to make it balance. After applying 8 pounds (an unopened gallon of milk), I measured spring height of 0.444 and 0.440 inches. The sum of the two compressions = 0.127 inches; the amount that 1 spring should have compressed if I could have done it with just 1 (this simultaneously averages the values). This gives a spring constant of 63 pounds per (linear) inch, per spring. When multiplied by 4 parallel springs, that gives 252 pounds (rounded) per inch. 75 / 252 = 0.2976. Multiplied by 32 threads per inch (a 6-32 bolt), gives 9.5 turns.
To remedy the situation, I'm going to remove the stock 939 bracket from the mobo, exposing those 4 mounting holes. I'll then use the Xeon mounting method and hardware supplied by Swiftech (or mostly so). The important part is the 4 black springs provided with the Storm. I believe AMD specifies 75 pounds of mounting force, which seems absurdly high, but I can't argue. I've calculated that I need to compress each of the 4 springs a total of 9.5 turns on a 6-32 bolt. That's 9.5 turns passed just touching.
Someone please double check my work. Here's how I came up with that:
I created a jig that would allow me to measure the spring compression with a dial caliper while applying a known weight. I started with just the weight of the jig, and recorded spring heights of 0.509 and 0.502 inches. I needed to use 2 springs, side by side, to make it balance. After applying 8 pounds (an unopened gallon of milk), I measured spring height of 0.444 and 0.440 inches. The sum of the two compressions = 0.127 inches; the amount that 1 spring should have compressed if I could have done it with just 1 (this simultaneously averages the values). This gives a spring constant of 63 pounds per (linear) inch, per spring. When multiplied by 4 parallel springs, that gives 252 pounds (rounded) per inch. 75 / 252 = 0.2976. Multiplied by 32 threads per inch (a 6-32 bolt), gives 9.5 turns.