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- Thread starter racinnut15xm
- Start date

they advertise it as 425 because that is the advertised size of the plate class that is allowed to race. That doesn't mean the plate is 425. It just means that if you want to race in that 425 class they'll sell you a 425 plate.they aren't selling you a 425 size plate they're selling you a plate for the 425 class

Wow, All the horstman plates were sized correctly.

You do realize you're talking .005" of an inch? and because it is a circle, its actually only .0025" of the diameter. If you plan to get a perfect plate, I'd be getting a certified No-Go tech tool to check with and not your harbor freight calipers...

Huh?You do realize you're talking .005" of an inch? and because it is a circle, its actually only .0025" of the diameter. If you plan to get a perfect plate, I'd be getting a certified No-Go tech tool to check with and not your harbor freight calipers...

His plate is .005" smaller than MAX. Since it is a round hole, the actual difference is .0025" overall circumference is it not? Maybe I'm not typing it out correctly, but I know what I'm trying to say! LOLHuh?

We could talk about a % change in cross sectional area of the hole and that would be meaningful.

Where area = Pi x R^2

So the ideal area, right at limit (.425") would be 3.14159 x ((.425/2)x(.425/2)) = .14186 square inches.

The area of the plate you have (.420") would be 3.14159 x ((.420/2)x(.420/2)) = .13854 square inches.

The difference in area is .14186 - .13854 =

The difference expressed as a percentage is .00318/.14186 x 100%=

We could talk about a % change in cross sectional area of the hole and that would be meaningful.

Where area = Pi x R^2

So the ideal area, right at limit (.425") would be 3.14159 x ((.425/2)x(.425/2)) = .14186 square inches.

The area of the plate you have (.420") would be 3.14159 x ((.420/2)x(.420/2)) = .13854 square inches.

The difference in area is .14186 - .13854 =.00318 square inches

The difference expressed as a percentage is .00318/.14186 x 100%=2.3%

Yep, and thats why I flunked out of getting my ME many years ago! lol

I like the % number that's how I like to think of things when we get on the dyno. .2 hp doesn't seem like much on the dyno but you start stacking a couple of those together and you've got something. These things produce such low numbers that a possible 2% increase in plate size would give a potential for mo powa. But alas Dover doesn't do this with the plates anymore. Back to the drawing board

We could talk about a % change in cross sectional area of the hole and that would be meaningful.

Where area = Pi x R^2

So the ideal area, right at limit (.425") would be 3.14159 x ((.425/2)x(.425/2)) = .14186 square inches.

The area of the plate you have (.420") would be 3.14159 x ((.420/2)x(.420/2)) = .13854 square inches.

The difference in area is .14186 - .13854 =.00318 square inches

The difference expressed as a percentage is .00318/.14186 x 100%=2.3%