How to calculate speed loss from lbs./sec.

[Camillo]

Hi. I'm planning to install two RAMI 529 antennas.
I studied a post from gmcpilot where he explains how to calculate speed loss from antenna surface.

I now managed to deduce such antenna produces approx. 2,4 lbs. @ 187 mph., which is approx. 1,22 engine horse power loss.

I can't make the final step to calculate mph loss.
Formula was 1/3^[(HP-HP loss)/HP]*speed.

So, if in my RV9A with 160 HP I can reach 187 mph., I should say:
1/3^(158,78/160)*187

It is not clear to me what "1/3^" means

Don't think I'm crazy. My calculations give me more than 0,25 mph. drop per antenna! That's why I'd like to solve this issue.
Thanks.
Camillo

[az_gila]

It is not clear to me what "1/3^" means

This indicates the cube root of the part in parentheses...

gil A

[RBurn]

I got 5.7 mph of loss although I find this hard to believe from two antennas.

(158.78/160)=0.99
0.99*187=185.6
185.6^(1/3)=5.7

The way you have your equation written is slightly confusing because it doesn't follow the Order of Operations from Algebra. A better way to write it would be...

{[(HP-HP loss)/HP]*speed}^(1/3)

Sorry to be nit picky, but we've all seen little errors like this turn up badly. Kinda like when one company uses Metric and the other uses English, and spacecraft end up crashing into planets.

[hevansrv7a]

The change in speed, if you were trying to go faster, would be the cube root of the change in HP. You'd need 1.01^3 HP to go 1% faster. While this is the reverse, the same relationship should apply. For speed change, I'd take the cube root of the power change. What do you think?

Originally Posted by AlexPeterson

Speed is generally proportional to the power cubed in our speeds of interest.

So, double the speed takes 2^3 = 8 times the power.

So, if power loss to drag is 1.2 hp, the power to pull the plane is now 158.8 (assuming sea level and a bunch of other stuff).

The ratio of power is 158.8/160 = .9925

The cube of this power ratio is .9925^3 = .9777

This is the speed ratio, so new speed is .9777 x 187 = 182.8

Speed loss (if the drag is correct) is 4.2 mph.

I've left in a lot of significant digits just so others can follow the calculations. There are a lot of assumptions in this...

I hope this helps (and, it's Friday so I hope I'm right on this!) Beer time

[Norman CYYJ]

Unless you are into racing you won't notice the difference. You would need an extremely accurate ASI to notice it and also be able to fly very accurately.

[Kyle Boatright]

I getg 186.5 mph after the antenna installation.

I believe the new speed = Original speed/(original hp/recalculated hp)^1/3


187/(160/158.75)^1/3 = 186.5


To back-check, pretend you had a 1.25 hp increase in a 158.75 hp airplane.

New speed = 186.5 * (160/158.75)^1/3 = 187

[AlexPeterson]

Originally Posted by Kyle Boatright

I getg 186.5 mph after the antenna installation.

I believe the new speed = Original speed/(original hp/recalculated hp)^1/3


187/(160/158.75)^1/3 = 186.5


To back-check, pretend you had a 1.25 hp increase in a 158.75 hp airplane.

New speed = 186.5 * (160/158.75)^1/3 = 187

Kyle, you are correct. I retracted a post I made, obviously too late on a Friday!

[Camillo]

Thank-you very much. I will study again the formula.

Original formula was taken here http://www.vansairforce.com/communit...ighlight=balun

"Fluid Dynamic Drag, by Hoerner, assume a drag coefficient (Cd)
of approximately 0.5 for a whip antenna
(test data in figure 13 on page 3-9).

Drag is equal to the dynamic pressure times frontal area times the Cd.

Dynamic pressure (lb per square foot) at sea level equals the speed
in mph squared, divided by 391 (from Hoerner, eqn. 16 on page 1-10).

Frontal area of the antenna in square inches, we get:
20 inches long, diameter of 3/16 inch, frontal area of 20 * 0.1875 = 3.75
square inches. (actual area really 2.88 in sq, bent whip including base)

drag = speed squared times frontal area times Cd divided by 56,304
(drag in pounds, speed in mph, frontal area in square inches)

At 200 mph the drag would be: 200 x 200 x 3.75 x 0.5 / 56,304 = 1.3 lb.


Power required = speed x force.
200 mph x 5280 ft/mile x 3600 sec/hour = 293 ft/sec.


So, the power in ft-lb/sec = 293 x 1.3 = 381.


One hp = 550 ft-lb/sec, so it takes 381/550 = 0.69 hp.

Prop efficiency assumed as 0.8,
that means we need 0.69/0.8 = 0.87 engine horse power
to drag that antenna.


Assume it takes 160 hp to go 201 mph (RV-6)
Speed loss, 1/3^(159.31/160)*201=200.71
So it takes .29 MPH.

The above is conservative. It is closer to 0.50hp /.25 mph."

[Kevin Horton]

Originally Posted by Camillo

Hi. I'm planning to install two RAMI 529 antennas.
I studied a post from gmcpilot where he explains how to calculate speed loss from antenna surface.

I now managed to deduce such antenna produces approx. 2,4 lbs. @ 187 mph., which is approx. 1,22 engine horse power loss.

I can't make the final step to calculate mph loss.
Formula was 1/3^[(HP-HP loss)/HP]*speed.

So, if in my RV9A with 160 HP I can reach 187 mph., I should say:
1/3^(158,78/160)*187

It is not clear to me what "1/3^" means

Don't think I'm crazy. My calculations give me more than 0,25 mph. drop per antenna! That's why I'd like to solve this issue.

I agree that 2.4 lb drag at 187 mph TAS would require 1.2 hp to push it through the air. But, the propeller has about 80% efficiency, so the engine will need to develop 1.5 hp to push this antenna.

It isn't completely clear what flight condition you are using to do the calculations. Looking at Van's published performance data, you are probably using the 187 mph at 75% power at 8000 ft point. Taking Van's data as gospel, the engine is producing 120 hp at that condition.

So, our new speed would be approximately 187 * the cube root of ((120 - 1.5)/120), or 186.2 mph TAS.

Note: If the 2.4 lb drag number had been calculated at a sea level condition, then the drag at the 8000 ft condition would be lower, due to the lower air density. At the same TAS, the drag varies with the air density. The density at 8000 ft is roughly 79% of sea level density (assuming standard temperature at both conditions). Thus the drag would be about 1.9 lb, and loss in speed would be slightly less (0.6 mph vs 0.8 mph).

[AlexPeterson]

Originally Posted by hevansrv7a

The change in speed, if you were trying to go faster, would be the cube root of the change in HP. You'd need 1.01^3 HP to go 1% faster. While this is the reverse, the same relationship should apply. For speed change, I'd take the cube root of the power change. What do you think?

H, you are absolutely correct. Now my erroneous post is forever captured in your quote of it! Keeps one humble...