Green plate

Kart43

Member
Are your measuring tools calibrated? What are you using to measure? The difference between the .421" and .425 is .0026 sq inch that is 2.6 thousandths of a square inch of area. Do you feel there will be a performance difference? What is the spec listed in your rules that it must measure?
 

1fasttiller

Member
They are purposely undersized to keep you from getting tossed in tech. If you have one that is EXACT or one that is .0001 under then you stand a huge chance of getting tossed in tech because the tech mans tools are calibrated perfectly. You can usually order "Max" size plates from guys like Dover and such but you risk a DQ if the wrong tech man gets ahold of it.
 
I mean its advertised at .425 and it measures .420. Trying to eeek every last bit out of these things and I'm sure it will make a little difference. I didn't know if only one company was allowed to make them. I will contact dover on the next one.
 

DynoDon

Moderator
I mean its advertised at .425 and it measures .420. Trying to eeek every last bit out of these things and I'm sure it will make a little difference. I didn't know if only one company was allowed to make them. I will contact dover on the next one.
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
 

1fasttiller

Member
I mean its advertised at .425 and it measures .420. Trying to eeek every last bit out of these things and I'm sure it will make a little difference. I didn't know if only one company was allowed to make them. I will contact dover on the next one.

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...
 
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?
 

flattop1

Dawg 89
Diameter or radius?
Diameter is .420
Radius is .210
In the end the solution is buy a max size or blueprinted plate . Install that and move on . Also be aware any mark made with the calipers while measuring is cause for disqualification .
 

Big Chris

Member
No you have that wrong, the diameter is .005" smaller than max. That's it, no other math is required, unless you are talking about the area of the hole. The circumference has nothing to do with the conversation of the hole being small.

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%
 

1fasttiller

Member
No you have that wrong, the diameter is .005" smaller than max. That's it, no other math is required, unless you are talking about the area of the hole. The circumference has nothing to do with the conversation of the hole being small.

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
 
No you have that wrong, the diameter is .005" smaller than max. That's it, no other math is required, unless you are talking about the area of the hole. The circumference has nothing to do with the conversation of the hole being small.

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%
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
 
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