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Looking for help Computing Drag Impact

lr172

Well Known Member
I will be adding an ads-b antenna, similar to my transponder antenna. I am wondering if it is worth the cost/hassle of going to blade style antennas.

According to Rami, the rod and ball antennas has .41 lbs of drag at 250mph and the blade has .09 lbs at 250. I would be reducing .8 lbs of drag at 250 mph.

Not being an aeronautical engineer, I have no idea how that number translates to improvement in airspeed on an RV cruising at 160 knots.

Can anyone help me ballpark the speed improvement?

Thanks,

Larry
 
drag x velocity = power

250 mph = 367 ft/s

The equivalent power for the small drag you describe is 0.8 lb x 367 ft/s = 293 ft-lb/s.

1 Hp = 550 ft-lb/s. So your drag change is equivalent to a power change of 1/2 Hp.

Thats at 250 mph.

160 kts = 184 mph = 270 ft/s.

At that speed, the equivalent power is 0.8 lb x 270 ft/s /550 = 0.4 Hp.

For constant drag coefficient, power varies in proportion to velocity^3, so assuming you are using 180 hp to go 160 kt at sea level, the speed increase is approximately, (0.4/180)/3 x 160 = 0.12 kts.
 
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250 mph = 367 ft/s

The equivalent power for the small drag you describe is 0.8 lb x 367 ft/s = 293 ft-lb/s.

1 Hp = 550 ft-lb/s. So your drag change is equivalent to a power change of 1/2 Hp.

Thats at 250 mph.

160 kts = 184 mph = 270 ft/s.

At that speed, the equivalent power is 0.4 Hp.

Thanks for the help and detail. Can I assume that the speed obtained with that added .4 hp is a small fraction of .4 in MPH?

Larry
 
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Drag

Looks like a great project for our VAF bros. with the 3D printers to make a blade cover for transponder antenna.I want one. Hope Steve Melton sees this.
Bob
 
Looks like a great project for our VAF bros. with the 3D printers to make a blade cover for transponder antenna.I want one. Hope Steve Melton sees this.
Bob

Why not just buy a real blade antenna from Delta Pop? They're super cool, reasonably priced, and work as expected.

website
 
Curious about 2 things

1. Steve said, "speed increase is approximately, (0.4/180)/3 x 160 = 0.12 kts." Not sure where the "/3" term comes in. Feel like there should be a ^-3 function in here somewhere. But I'm NOT a math guy.

2. Have noted for some while the auto industry's use of spiral-fluted automobile antennas, presumably for reduction in vibration seen in smooth rod broadcast receive antennas on car fenders. Is there a drag reduction to be gained from tripping the boundary layer with a spiral flute like this? Might be a compromise design between blade and rod-and-ball antennas that we could implement?
 
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2. Have noted for some while the auto industry's use of spiral-fluted automobile antennas, presumably for reduction in vibration seen in smooth rod broadcast receive antennas on car fenders. Is there a drag reduction to be gained from tripping the boundary layer with a spiral flute like this? Might be a compromise design between blade and rod-and-ball antennas that we could implement?

The spiral flute on the auto antenna is there to prevent vortex shedding which will cause the antenna to whip back and forth perpendicular to the airstream.
 
1. Scott said, "speed increase is approximately, (0.4/180)/3 x 160 = 0.12 kts." Not sure where the "/3" term comes in. Feel like there should be a ^-3 function in here somewhere. But I'm NOT a math guy.

2. Have noted for some while the auto industry's use of spiral-fluted automobile antennas, presumably for reduction in vibration seen in smooth rod broadcast receive antennas on car fenders. Is there a drag reduction to be gained from tripping the boundary layer with a spiral flute like this? Might be a compromise design between blade and rod-and-ball antennas that we could implement?

I bet he just got lazy with the keyboard when he typed the formula. I did a cube root of that ratio, & got around .13.
 
I bet he just got lazy with the keyboard when he typed the formula. I did a cube root of that ratio, & got around .13.

Its just a quick gouge. (i.e. easy approximation)

If power requirement increases as the cube of the velocity, then a 10% increase in velocity requires approx 30% increase in power. Multiply by 3.

In reality, its 33%, because (1.1)^3 = 1.331.

Similarly, if something varies in proportion to the square of something else, then multiply a small percentage change by 2. So a 10% change x 2 = 20%. Again, more precisely, (1.1)^2 = 1.21

Multiplying (or dividing) is a quick approximation for ratios near 1.0.

Back to the antenna drag change, if the drag change was the equivalent of 0.4 hp out of 180, then (180.4/180)^(1/3) = 1.00074 so 160 x 1.00074 = 160.118.

There was a question of whether spiral wrap might reduce drag of a rod antenna. Spiral wrap breaks up the coherency of the vortex shedding so that there is no periodic force causing motion, vibration or fatigue. But it is such a large-scale feature that I suspect the drag is probably slightly higher, not lower. If you want to reduce the drag on the rod, you have to trip the boundary layer without adding drag from the trip itself. Coating it with grit, or knurling it might help.
 
In my sailing days we would often get mast pump while sitting at anchor.
wrapping the halyard around the mast would often solve the issue. Just like the new car antennas. Way before the cars starting doing it.
 
novel marketing idea

Grit-coated belly antennas - great for low-drag ribbon-cutting. Out buzzing the neighbor's cattle? Our carbide models cut fencing wire, too :D
 
250 mph = 367 ft/s

The equivalent power for the small drag you describe is 0.8 lb x 367 ft/s = 293 ft-lb/s.

1 Hp = 550 ft-lb/s. So your drag change is equivalent to a power change of 1/2 Hp.

Thats at 250 mph.

160 kts = 184 mph = 270 ft/s.

At that speed, the equivalent power is 0.8 lb x 270 ft/s /550 = 0.4 Hp.

Hi Steve,

Thanks for these great posts. I enjoy the opportunity to think about these topics. I have a question regarding your post, above:

I interpreted from the original post, a potential drag reduction of 0.8 lbs at 250 mph. Assuming the drag varies with the square of velocity, would the drag delta at 160 kts be much less than that given for the higher speed?

If this is so, I would calculate a drag of about .43 lbs at 160kts, compared to .8 lbs at 250 mph. Would this figure into the equivalent power calculation?

Thanks for the help.
 
Hi Steve,

Thanks for these great posts. I enjoy the opportunity to think about these topics. I have a question regarding your post, above:

I interpreted from the original post, a potential drag reduction of 0.8 lbs at 250 mph. Assuming the drag varies with the square of velocity, would the drag delta at 160 kts be much less than that given for the higher speed?

If this is so, I would calculate a drag of about .43 lbs at 160kts, compared to .8 lbs at 250 mph. Would this figure into the equivalent power calculation?

Thanks for the help.

Dean, generally that's right. If the drag "coefficient" stays constant, then the drag force would decrease with the square of the velocity. A good assumption for the blade antenna, but not necessarily the rod. drag coefficient for cylinders varies a lot with speed change because of changes in the basic flow topology, which we characterize by "Reynolds no." Depending on where you are compared to the "critical Reynolds number" the drag coefficient can change quite a bit, increasing as you get slower. A quick check for an 1/8" rod at 160 kts, the Re=200,000. That is just below the critical Reynolds no. which means the drag coefficient is about 1 and relatively constant for slower speeds, but is starting to drop dramatically at speeds a little above. Google 'drag coefficient of cylinder' and you will see what I mean.
 
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