I’ve touched on this before, but I thought it might be nice to have this information all in one spot, and since I addressed this in another forum, I thought it would be worthwhile to cut and paste it here.

The question was asked, more or less, how many 4″ speakers does it take to equal an 8″ speaker-

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Find the Piston area of the speakers using the area of a circle. If you do this, you will get a chart that looks something like this -

4.5″ – 1

5.0 – 1.3

6.5 – 2.6

8.0 – 4.3

10 – 6.8

12 – 9.9

15 – 16.0

This was actually given to me when I asked a similar question in another forum.

Area of a Circle equals PI times the Radius Squared.

A = (PI) R²

So, let’s work this out.

A = (pi) R² = (3.14159) x 2² = 3.14159 x 4 = 12.566 square inches *(4″ driver) *

A = (pi) R² = (3.14159) x 4² = 3.14159 x 16 = 50.265 square inches* (8″ driver)*

So, if we divide the two 50.265 / 12.566 = 4.00

So, one 8″ speakers equal FOUR 4″ speakers.

Here is another chart that I calculated myself -

4.00″ = 12.57 sq.in. = 1.00

5.00″ = 19.64 sq.in. = 1.56

5.25″ = 21.65 sq.in. = 1.72

6.00″ = 28.27 sq.in. = 2.49

6.50″ = 33.18 sq.in. = 2.64

7.00″ = 38.49 sq.in. = 3.06

8.00″ = 50.27 sq.in. = 3.999 = 4.0

10.0″ = 78.54 sq.in. = 6.25

12.0″ = 113.1 sq.in. = 8.997 = 9.0

15.0″ = 176.7 sq.in. = 14.06

Using this chart, we divide 4 (for an 8″ speaker) by 1 (for a 4″ speaker) and we get the same result; FOUR 4″ speakers equals ONE 8″ speaker.

Also notice that a 6.5″ speaker using this last chart, is 70% larger than a 5″ speaker. (2.64/1.56 = 1.70)

A 12″ speaker is 125% larger than a 8″ speaker. (9/4 =2.25)

And so on…

If you want the difference between two specific speakers rather than a generalization, look up the full specs on the speakers, and look at the *Sd *specification, which is the surface area of the actual cone.

Here is a chart I makes using the *Sd* of Dayton speakers. Keeping in mind that the cones could be very different. One could be shallow and flat, another very deep. Some could be curved, other very straight.

Using Dayton woofers as an example, and using their *Sd *spec, we determine the relative effective size of various speakers.

4.00″ = 37.4 cm² = 1.00

5.00″ = 52.8 cm² = 1.41

5.25″ = 93.3 cm² = 2.49

6.00″ = 84.9 cm² = 2.27

6.50″ = 134.8cm² = 3.60

7.00″ = 132.7cm² = 3.55 (more typical 125cm²)

8.00″ = 211.2cm² = 5.65

10.0″ = 343.1cm² = 9.17

12.0″ = 498.8cm² = 13.3

15.0″ = 843.7cm² = 22.6

Further because the surround is not part of the Pistonic action of the driver, we have to subtract the surrounds. Now a typical surround is about 0.75″. But part, about half of the surround, does have some pistonic action; about half.

To allow for this, we subtract 0.75″ from the rated dimension of the speakers.

So, an 8″ speaker minus 0.75″ is now 7.25″. To find the area we still use the same formula (7.25/2 = 3.626) -*A = (pi) R² = 3.14159 x 3.625² = 41.28 square inches.*

Now a 4″ driver, which would now be 3.25″ (3.25/2 = 1.625) -*A = (pi) R² = 3.14159 x 1.625² =8.296 square inches
So, we divide the value for the 8″ by the value for the 4″ and we have –*

*41.28 / 8.296 = 4.96*

Using this method, it takes 4.96 FOUR inch drivers to equal ONE 8″ driver.

This is probably the most accurate as a general estimation for all drivers, as it uses the actual cone diameter plus half the Surround, and assumes Pistonic action.

Sorry haven’t worked this chart out yet.

The corrected formula, though more confusing, would be – where “D” is diameter –

*A = (pi) ((D – 0.75) / 2)²*

A = (pi) ((8″ – 0.75) / 2)²

A = (pi) ((7.25) / 2)²

*A = (pi) (3.625)² = 3.14159 (3.625)² = 41.28*

*Which is exactly what we had before.*

This last method, is the most accurate, but simply comparing the area of a circle of the rated size of the driver, is probably sufficient to make an estimation.

from：http://www.avforums.com/threads/relative-size-of-drivers-how-many-of-these-to-equal-one-of-those.1899701