Wave drag of a Sears-Haack Body

Kryptid

Kryptid

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I've been doing some work on an original aircraft design. As part of the feasibility process, I decided to try calculating the wave drag of a Sears-Haack body with a length of 49.2 feet (15 meters) and a fineness ratio of 14 (a recommendation given by Dr. Raymer's aircraft design book). This results in a maximum radius of 1.76 feet (0.54 meters). I used an equation I found on Wikipedia to calculate wave drag at Mach 1 at 36,000 feet. The result I got seemed unreasonably low, however: about 186.5 pounds of force (829.6 newtons). Granted, this is a very skinny object compared to just about any existing airplane, but that drag force still seems awfully small. There were two equivalent forms of the equation and both of them give this same answer. Unfortunately, Wikipedia doesn't cite a source for where it got its equation.

Does this seem wrong to anyone else?

Here is the math in case anyone wants to check over it:

Here are the calculations to find the volume of the Sears-Haack body:

Volume = ((3 x pi^2)/16) x maximum radius^2 x length
Volume = ((3 x pi^2)/16) x 1.76 feet^2 x 49.2 feet
Volume = 282 cubic feet

These are the calculations to find the drag force at Mach 1 at 36,000 feet (660 miles per hour/968 feet per second)

Wave drag = ((64 x Volume^2)/(pi x Length^4)) x air density x velocity^2
Wave drag = ((64 x 282 cubic feet^2)/(pi x 49.2 feet^4)) x 0.00072 slugs/cubic feet x 968 feet per second^2
Wave drag = 186.5 pounds

I saw in Raymer's design book that slugs per cubic foot are the go-to measurement for air density when using imperial units. Should I have used pounds per cubic foot instead?
 
After taking another look at Dr. Raymer's book, it says that the wave drag of a Sears-Haack body can be calculated using the equation (D/q)wave = ((9 x pi)/2) x (Amax/length)^2. This is on page 339 of Aircraft Design: A Conceptual Approach Fourth Edition.

For the selected body length of 49.2 feet (15 meters) and maximum cross-sectional area of 9.7 square feet (0.9 square meters), I get...

(D/q)wave = ((9 x pi)/2) x ((9.7 square feet)/49.2 feet)^2
(D/q)wave = 0.55 pounds per square foot.

Now, this isn't the drag itself, but rather the "drag area". I wasn't sure how to turn that into a drag number. Do I multiply it by the surface area of the Sears-Haack body (which from another reference, is about 72% that of a cylinder of the same length and width)? Or do I multiply it by the maximum cross-sectional area? Doing the former results in a total drag force of 218.74 pounds, which is surprisingly close to my previous calculation of 186.5 pounds. Multiplying it by the maximum cross-sectional area yields only 5.335 pounds of drag force, and there's no way that's correct. So would I assume that I'm right on the money with dividing it by the surface area of the Sears-Haack body because it's in the same ballpark as the first calculation? I know the total drag will be greater than just the wave drag, but I still need to calculate wave drag (or at least wave drag coefficient) to do a drag build-up method to find total drag.

I see that Raymer's book states that dividing the drag area by the wing reference area yields the drag coefficient, but I can't use that because a Sears-Haack body doesn't have wings.
 
Kryptid said:
After taking another look at Dr. Raymer's book, it says that the wave drag of a Sears-Haack body can be calculated using the equation (D/q)wave = ((9 x pi)/2) x (Amax/length)^2. This is on page 339 of Aircraft Design: A Conceptual Approach Fourth Edition.

For the selected body length of 49.2 feet (15 meters) and maximum cross-sectional area of 9.7 square feet (0.9 square meters), I get...

(D/q)wave = ((9 x pi)/2) x ((9.7 square feet)/49.2 feet)^2
(D/q)wave = 0.55 pounds per square foot.

Now, this isn't the drag itself, but rather the "drag area". I wasn't sure how to turn that into a drag number. Do I multiply it by the surface area of the Sears-Haack body (which from another reference, is about 72% that of a cylinder of the same length and width)? Or do I multiply it by the maximum cross-sectional area? Doing the former results in a total drag force of 218.74 pounds, which is surprisingly close to my previous calculation of 186.5 pounds. Multiplying it by the maximum cross-sectional area yields only 5.335 pounds of drag force, and there's no way that's correct. So would I assume that I'm right on the money with dividing it by the surface area of the Sears-Haack body because it's in the same ballpark as the first calculation? I know the total drag will be greater than just the wave drag, but I still need to calculate wave drag (or at least wave drag coefficient) to do a drag build-up method to find total drag.

I see that Raymer's book states that dividing the drag area by the wing reference area yields the drag coefficient, but I can't use that because a Sears-Haack body doesn't have wings.
Click to expand...

Kryptid, Raymers formula gives an area, not pounds per area. You just have to multiply your result (0.55 square feet) by dynamic pressure, q, in pounds-force per area, at the desired altitude, to obtain drag D in pounds-force.

Best regards!
 
FrankM said:
Kryptid, Raymers formula gives an area, not pounds per area. You just have to multiply your result (0.55 square feet) by dynamic pressure, q, in pounds-force per area, at the desired altitude, to obtain drag D in pounds-force.

Best regards!
Click to expand...

Okay, that makes sense. When I do it the way you say, I get the same answer as the first equation. Sounds like that's the right answer. Thank you!
 
Scott Kenny said:
You're finding the drag for a 1.08m diameter, 15m long projectile. Something you'd launch out of a Gustav railroad gun. Or maybe drop out of orbit to see how deep a hole it makes.
Click to expand...

I know it's not realistic for an actual aircraft design, but I was using it as a baseline for an aircraft I'm designing.
 

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