Measuring Total Plane Drag
04-03-2009 | 05:09 PM
#1
Measuring Total Plane Drag
I know this is a tough nut but I'd like to hear your ideas. I'm building a modified Duellist twin that I'd like to do some speed tests with. I've tried to reduce frontal area and improve the nose and nacelle transition. I'd like to be able to measure drag so I know how any future changes enhance or hurt the performance.
04-03-2009 | 06:03 PM
#2
RE: Measuring Total Plane Drag
Yes, it is a tough one!
Drag is a force that depends on the projected area, the air density, the air velocity, and also a coefficient of drag.
You have improved the area and that coefficient.
In order to measure the improvement, you need to know the value of a drag force that has been measured before, and under the very same conditions of air velocity and lift (AOA), which is a major contributor to the coefficient of drag.
Drag is a force that depends on the projected area, the air density, the air velocity, and also a coefficient of drag.
You have improved the area and that coefficient.
In order to measure the improvement, you need to know the value of a drag force that has been measured before, and under the very same conditions of air velocity and lift (AOA), which is a major contributor to the coefficient of drag.
04-03-2009 | 11:44 PM
#4
RE: Measuring Total Plane Drag
Not actually, the drag due to the lift will, in 'S&L' flight always be about the same, no matter what the speed, cause lift will always equal the weight. This only changes at high AOA's and low speeds. The ratio of this lift drag to the rest of the airframe drag will change with speed, as you suggest, but the differences between the two, at model sizes, is probably not great. It is likely that lift drag is greater than airframe drag at low speeds, and the ratio gradually shifts to parasite drag as the speed increases. That's why, to go faster, you need to go lighter, in preference to cleaning up the airframe and increasing power, both of which tend to add weight. All things together, of course, will give the best results.
Evan, WB #12.
Evan, WB #12.
04-04-2009 | 01:41 AM
#5
RE: Measuring Total Plane Drag
Ah, but the tailplane carries a variable "load" depending on the CG location and the pitching moment of the airfoil.
Cruncher, if you can locate the CG and select a wing airfoil such that the pitching moment and the "stability lift" that the tail needs to generate is at or darn close to zero for a lift coefficient then the tail will be only producing the profile drag. Selecting a good, thin, known performing high speed airfoil will go a huge way towards minimizing the tailplane's drag.
There was an old DOS based program called "Sailplane Design" many years ago written by David Fraser. That program not only provided the wing's lift coefficients for a range of speeds but also the tails lift coefficient. It was very interesting to see this stuff in action. It didin't take me long to realise that for my optimum speed for moving around the sky I wanted a CG location that generated the least tailplane drag at the best L/D speed.
Cruncher, if you can locate the CG and select a wing airfoil such that the pitching moment and the "stability lift" that the tail needs to generate is at or darn close to zero for a lift coefficient then the tail will be only producing the profile drag. Selecting a good, thin, known performing high speed airfoil will go a huge way towards minimizing the tailplane's drag.
There was an old DOS based program called "Sailplane Design" many years ago written by David Fraser. That program not only provided the wing's lift coefficients for a range of speeds but also the tails lift coefficient. It was very interesting to see this stuff in action. It didin't take me long to realise that for my optimum speed for moving around the sky I wanted a CG location that generated the least tailplane drag at the best L/D speed.
04-04-2009 | 09:11 AM
#6
Senior Member
RE: Measuring Total Plane Drag
CrateCruncher, you are correct. Above stall, induced drag is inversely proportional to the square of airspeed, and is relatively unimportant at several times stall airspeed. For example, at five times stall airspeed, induced drag is only about 1/25th as great as at near-stall airspeed.
Here is a NASA blurb on induced drag that may be useful:
http://www.grc.nasa.gov/WWW/K-12/airplane/induced.html
Measuring drag is quite a tough nut to crack, unless you can get ahold of a wind tunnel, and test a reduced scale model of your airplane. Another way to measure drag would be to use onboard airspeed telemetry, kill the engine, dive your airplane straight down from high altitude until it reaches its maximum airspeed. This may produce a dangerously high airspeed and cause the airplane to break up from flutter, turning it into an unguided missile. The drag of the stationary propeller can be calculated reasonably accurately; a windmilling propeller produces tremendous drag that is also difficult to estimate. At maximum speed, vertically down, drag is equal to weight.
Here is a NASA blurb on induced drag that may be useful:
http://www.grc.nasa.gov/WWW/K-12/airplane/induced.html
Measuring drag is quite a tough nut to crack, unless you can get ahold of a wind tunnel, and test a reduced scale model of your airplane. Another way to measure drag would be to use onboard airspeed telemetry, kill the engine, dive your airplane straight down from high altitude until it reaches its maximum airspeed. This may produce a dangerously high airspeed and cause the airplane to break up from flutter, turning it into an unguided missile. The drag of the stationary propeller can be calculated reasonably accurately; a windmilling propeller produces tremendous drag that is also difficult to estimate. At maximum speed, vertically down, drag is equal to weight.
04-05-2009 | 12:46 PM
#8
RE: Measuring Total Plane Drag
ORIGINAL: Rotaryphile
A way to measure drag would be to use onboard airspeed telemetry, kill the engine, dive your airplane straight down from high altitude until it reaches its maximum airspeed. This may produce a dangerously high airspeed and cause the airplane to break up from flutter, turning it into an unguided missile. The drag of the stationary propeller can be calculated reasonably accurately; a windmilling propeller produces tremendous drag that is also difficult to estimate. At maximum speed, vertically down, drag is equal to weight.
A way to measure drag would be to use onboard airspeed telemetry, kill the engine, dive your airplane straight down from high altitude until it reaches its maximum airspeed. This may produce a dangerously high airspeed and cause the airplane to break up from flutter, turning it into an unguided missile. The drag of the stationary propeller can be calculated reasonably accurately; a windmilling propeller produces tremendous drag that is also difficult to estimate. At maximum speed, vertically down, drag is equal to weight.
TERMINAL DIVE: mg = F(drag)
vs.
LEVEL FLIGHT Vmax: P(engine) x Eff(prop) = F(drag)
Assuming I could keep the dive perfectly straight and reach terminal velocity (TV) before hitting the ground this is far simpler than a level flight approach. It's conceivable I could have the plane towed to altitude allowing me to remove the prop beforehand. The plane should come out to about 10 pounds dry. If I put it on a starvation diet for the test that should allow a shorter dive duration, lower TV, and reduce airframe stress during pullout. I'll crunch some numbers based on a ballpark Cd range and see what kind of start height and TV I could expect.
04-05-2009 | 01:20 PM
#9
RE: Measuring Total Plane Drag
ORIGINAL: topspeed
I used the AR-5 site for the Bruce Carmichael example following drag calculations for my pusher ac.
I used the AR-5 site for the Bruce Carmichael example following drag calculations for my pusher ac.
His analysis seems to work backward from the maximum level flight velocity recorded by the NAA using the level fight relation I described above. I think I need that before I can do any component analysis. Or am I missing something?
04-08-2009 | 06:01 PM
#10
RE: Measuring Total Plane Drag
ORIGINAL: Rotaryphile
Another way to measure drag would be to use onboard airspeed telemetry, kill the engine, dive your airplane straight down from high altitude until it reaches its maximum airspeed. This may produce a dangerously high airspeed and cause the airplane to break up from flutter, turning it into an unguided missile. At maximum speed, vertically down, drag is equal to weight.
Another way to measure drag would be to use onboard airspeed telemetry, kill the engine, dive your airplane straight down from high altitude until it reaches its maximum airspeed. This may produce a dangerously high airspeed and cause the airplane to break up from flutter, turning it into an unguided missile. At maximum speed, vertically down, drag is equal to weight.
Turns out its pretty easy because theres no vortex drag in an unpowered vertical dive with a symmetrical airfoil. So at equilibrium of gravity and parasitic/form drag we get:
mg = F(drag) OR: mg = V^2 [rAC/2]
m = 10# airplane
g = gravity (32.2 ft/s^2)
r=air density (0.0705 #/ft^3 @ 90F and 500 ft asl)
A = frontal area of airplane (1.076 ft^2)
C = drag coefficient (I guessed 0.1 like a smooth sphere) Cessna 172 is 0.024 so I think this is conservative. Anybody know?
Solving for V and plugging the numbers I got 291 fps or 198 mph! Thats with a conservative drag coefficient! A Cd of .05 results in a TV of 412 fps or 281 mph!
Next I wanted to determine how high I would need to start the descent from zero initial velocity to arrive at TV of 291 fps. A simple assumption often done to solve this problem is to assume constant acceleration. Using the derivative form for acceleration a = dv/dt and velocity v = ds/dt results in the kinematic relation:
a ds = v dv , where ds is the change in distance we want and dv is the change in velocity that we now know.
Since a is just gravity (g) if things are assumed constant we get:
g ds = v dv
Solving the integral and evaluating up to a maximum of TV (291 fps) I got 1,315 feet! Yikes! Thats with no air resistance!
OK, in the real world we have air resistance so the equation for a really is:
a = g - F(drag) OR a = g - .0038V^2
I won't burden you with the kinematic solution but it winds up being a natural logarithmic function which is bad, really bad. Total height needed to achieve TV using air resistance is 5270 feet!
If you've ever tried to see how fast your car will go you'll appreciate whats going on here. The last 10 percent take 3 times as much distance as building the first 80% of the total. Thats why log functions are bad for us here.
One other interesting thing I discovered was while calculating the frontal area of the plane. It turns out the fuselage is only 9% of the total while the Duellist wing comprises 81% of the total!!! So if you want to go fast use a thin wing with a long chord and short wingspan - in other words a JET!
I did a run through with the engines removed and found I could lighten the plane by a couple of pounds. I couldn't save all the weight because I have to add ballast to the nose for balance. The TV changed to 175 mph but the logarithm is relentless with very little change to vertical height. I'm going to continue to work on this to see if I can start the dive at a high horizontal speed but that rules out my no-prop hopes.
04-09-2009 | 09:57 AM
#11
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From: Bloomington, MN,
RE: Measuring Total Plane Drag
Hi CrateCruncher,
A Cd of 0.1 is very conservative. The value of .025 seem like a typical value for full-sized aircraft. Models might be a little higher due to the lower Reynolds numbers at which they operate. One important thing to note is that the Area used in Cd for aircraft is not the frontal area, but rather the wing projected area. You need to modify your calculations to reflect this. This is different from the practice in characterizing the drag of other bodies, so drag results are not generally directly comparable between aircraft & other bodies.
banktoturn
ORIGINAL: CrateCruncher
I spent last night looking into this compelling suggestion. I wanted to find out if it could be done using a data logger at a typical r/c field so the first step was to figure out the value for Terminal Velocity (TV).
Turns out its pretty easy because theres no vortex drag in an unpowered vertical dive with a symmetrical airfoil. So at equilibrium of gravity and parasitic/form drag we get:
mg = F(drag) OR: mg = V^2 [rAC/2]
m = 10# airplane
g = gravity (32.2 ft/s^2)
r=air density (0.0705 #/ft^3 @ 90F and 500 ft asl)
A = frontal area of airplane (1.076 ft^2)
C = drag coefficient (I guessed 0.1 like a smooth sphere) Cessna 172 is 0.024 so I think this is conservative. Anybody know?
Solving for V and plugging the numbers I got 291 fps or 198 mph! Thats with a conservative drag coefficient! A Cd of .05 results in a TV of 412 fps or 281 mph!
Next I wanted to determine how high I would need to start the descent from zero initial velocity to arrive at TV of 291 fps. A simple assumption often done to solve this problem is to assume constant acceleration. Using the derivative form for acceleration a = dv/dt and velocity v = ds/dt results in the kinematic relation:
a ds = v dv , where ds is the change in distance we want and dv is the change in velocity that we now know.
Since a is just gravity (g) if things are assumed constant we get:
g ds = v dv
Solving the integral and evaluating up to a maximum of TV (291 fps) I got 1,315 feet! Yikes! Thats with no air resistance!
OK, in the real world we have air resistance so the equation for a really is:
a = g - F(drag) OR a = g - .0038V^2
I won't burden you with the kinematic solution but it winds up being a natural logarithmic function which is bad, really bad. Total height needed to achieve TV using air resistance is 5270 feet!
If you've ever tried to see how fast your car will go you'll appreciate whats going on here. The last 10 percent take 3 times as much distance as building the first 80% of the total. Thats why log functions are bad for us here.
One other interesting thing I discovered was while calculating the frontal area of the plane. It turns out the fuselage is only 9% of the total while the Duellist wing comprises 81% of the total!!! So if you want to go fast use a thin wing with a long chord and short wingspan - in other words a JET!
I did a run through with the engines removed and found I could lighten the plane by a couple of pounds. I couldn't save all the weight because I have to add ballast to the nose for balance. The TV changed to 175 mph but the logarithm is relentless with very little change to vertical height. I'm going to continue to work on this to see if I can start the dive at a high horizontal speed but that rules out my no-prop hopes.
ORIGINAL: Rotaryphile
Another way to measure drag would be to use onboard airspeed telemetry, kill the engine, dive your airplane straight down from high altitude until it reaches its maximum airspeed. This may produce a dangerously high airspeed and cause the airplane to break up from flutter, turning it into an unguided missile. At maximum speed, vertically down, drag is equal to weight.
Another way to measure drag would be to use onboard airspeed telemetry, kill the engine, dive your airplane straight down from high altitude until it reaches its maximum airspeed. This may produce a dangerously high airspeed and cause the airplane to break up from flutter, turning it into an unguided missile. At maximum speed, vertically down, drag is equal to weight.
Turns out its pretty easy because theres no vortex drag in an unpowered vertical dive with a symmetrical airfoil. So at equilibrium of gravity and parasitic/form drag we get:
mg = F(drag) OR: mg = V^2 [rAC/2]
m = 10# airplane
g = gravity (32.2 ft/s^2)
r=air density (0.0705 #/ft^3 @ 90F and 500 ft asl)
A = frontal area of airplane (1.076 ft^2)
C = drag coefficient (I guessed 0.1 like a smooth sphere) Cessna 172 is 0.024 so I think this is conservative. Anybody know?
Solving for V and plugging the numbers I got 291 fps or 198 mph! Thats with a conservative drag coefficient! A Cd of .05 results in a TV of 412 fps or 281 mph!
Next I wanted to determine how high I would need to start the descent from zero initial velocity to arrive at TV of 291 fps. A simple assumption often done to solve this problem is to assume constant acceleration. Using the derivative form for acceleration a = dv/dt and velocity v = ds/dt results in the kinematic relation:
a ds = v dv , where ds is the change in distance we want and dv is the change in velocity that we now know.
Since a is just gravity (g) if things are assumed constant we get:
g ds = v dv
Solving the integral and evaluating up to a maximum of TV (291 fps) I got 1,315 feet! Yikes! Thats with no air resistance!
OK, in the real world we have air resistance so the equation for a really is:
a = g - F(drag) OR a = g - .0038V^2
I won't burden you with the kinematic solution but it winds up being a natural logarithmic function which is bad, really bad. Total height needed to achieve TV using air resistance is 5270 feet!
If you've ever tried to see how fast your car will go you'll appreciate whats going on here. The last 10 percent take 3 times as much distance as building the first 80% of the total. Thats why log functions are bad for us here.
One other interesting thing I discovered was while calculating the frontal area of the plane. It turns out the fuselage is only 9% of the total while the Duellist wing comprises 81% of the total!!! So if you want to go fast use a thin wing with a long chord and short wingspan - in other words a JET!
I did a run through with the engines removed and found I could lighten the plane by a couple of pounds. I couldn't save all the weight because I have to add ballast to the nose for balance. The TV changed to 175 mph but the logarithm is relentless with very little change to vertical height. I'm going to continue to work on this to see if I can start the dive at a high horizontal speed but that rules out my no-prop hopes.
04-09-2009 | 01:45 PM
#12
RE: Measuring Total Plane Drag
Hey Banktoturn!
Thanks for the info. I've never even seen a Cd for a model airplane so I have no clue what to expect but you've confirmed my hunch in which case the Terminal Dive idea is just not going to work for the Duellist.
I have to admit my only drag experience is with full size cars and that was a long time ago. I assumed frontal area and projected area were the same thing for parasitic drag purposes.
The way I found area for the wing was by finding the largest cross-sectional area of the wing panel perpendicular to the airflow. I found the thickest point of the root rib, added skin thickness and multiplied that number by the length of the panel. I then subtracted a triangular area to compensate for change in thickness from root to tip. Is this incorrect? How should I do it?
Thanks for the info. I've never even seen a Cd for a model airplane so I have no clue what to expect but you've confirmed my hunch in which case the Terminal Dive idea is just not going to work for the Duellist.
I have to admit my only drag experience is with full size cars and that was a long time ago. I assumed frontal area and projected area were the same thing for parasitic drag purposes.
The way I found area for the wing was by finding the largest cross-sectional area of the wing panel perpendicular to the airflow. I found the thickest point of the root rib, added skin thickness and multiplied that number by the length of the panel. I then subtracted a triangular area to compensate for change in thickness from root to tip. Is this incorrect? How should I do it?
04-09-2009 | 02:05 PM
#13
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Joined: Aug 2002
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From: Bloomington, MN,
RE: Measuring Total Plane Drag
I'm sorry, I wasn't very clear. The area used for Cd is the planform area of the wing, rather than the frontal area. That is, the area as seen from the top of the plane, not the front.
banktoturn
ORIGINAL: CrateCruncher
Hey Banktoturn!
Thanks for the info. I've never even seen a Cd for a model airplane so I have no clue what to expect but you've confirmed my hunch in which case the Terminal Dive idea is just not going to work for the Duellist.
I have to admit my only drag experience is with full size cars and that was a long time ago. I assumed frontal area and projected area were the same thing for parasitic drag purposes.
The way I found area for the wing was by finding the largest cross-sectional area of the wing panel perpendicular to the airflow. I found the thickest point of the root rib, added skin thickness and multiplied that number by the length of the panel. I then subtracted a triangular area to compensate for change in thickness from root to tip. Is this incorrect? How should I do it?
Hey Banktoturn!
Thanks for the info. I've never even seen a Cd for a model airplane so I have no clue what to expect but you've confirmed my hunch in which case the Terminal Dive idea is just not going to work for the Duellist.
I have to admit my only drag experience is with full size cars and that was a long time ago. I assumed frontal area and projected area were the same thing for parasitic drag purposes.
The way I found area for the wing was by finding the largest cross-sectional area of the wing panel perpendicular to the airflow. I found the thickest point of the root rib, added skin thickness and multiplied that number by the length of the panel. I then subtracted a triangular area to compensate for change in thickness from root to tip. Is this incorrect? How should I do it?
04-10-2009 | 04:00 PM
#14
Senior Member
RE: Measuring Total Plane Drag
Is it not the case that an airplane flying along level at a constant speed has thrust equal to drag? If so, it might be easier to determine thrust than to determine drag directly.
04-10-2009 | 04:57 PM
#15
RE: Measuring Total Plane Drag
ORIGINAL: Jim Thomerson
Is it not the case that an airplane flying along level at a constant speed has thrust equal to drag? If so, it might be easier to determine thrust than to determine drag directly.
Is it not the case that an airplane flying along level at a constant speed has thrust equal to drag? If so, it might be easier to determine thrust than to determine drag directly.
04-10-2009 | 06:27 PM
#16
RE: Measuring Total Plane Drag
By measuring the drag
Seriously, the only practical way of doing this is to find a university handy that has an aeronautical thread and a tunnel big enough to put the whole model in. Knowing the tunnel corrections they can the measure all the parameters at various airspeeds and come up with Cd and Cl for the model. They can do lots of other things, stall speed and stall aoa etc. Could be a useful trade off here, it gives the students a practical set of problems, and you all you need to know about your model, and the cost of doing this, which could otherwise be horrendous, may be within some sort of bargaining reach. It may mean you lose the model for a month or two, but the results could be well worth the time. Based on what you find, the answers could be extrapolated across a wide range of models, and answer quite a few questions.
Evan, WB #12.
Seriously, the only practical way of doing this is to find a university handy that has an aeronautical thread and a tunnel big enough to put the whole model in. Knowing the tunnel corrections they can the measure all the parameters at various airspeeds and come up with Cd and Cl for the model. They can do lots of other things, stall speed and stall aoa etc. Could be a useful trade off here, it gives the students a practical set of problems, and you all you need to know about your model, and the cost of doing this, which could otherwise be horrendous, may be within some sort of bargaining reach. It may mean you lose the model for a month or two, but the results could be well worth the time. Based on what you find, the answers could be extrapolated across a wide range of models, and answer quite a few questions.Evan, WB #12.
04-11-2009 | 08:23 AM
#18
Senior Member
RE: Measuring Total Plane Drag
Measuring static thrust is easy - I just use a fish scale tied to the tail. Thrust in flight is normally lower than static thrust, and needs a wind tunnel for accurate measurement.
Consider a fairly clean airplane that can achieve 100 mph (146.7 feet/sec.) straight and level, with a 2 hp engine. Note that one hp = 550 foot-pounds per second.
At 70% propeller efficiency, thrust = (2x550x0.7)/146.7 = 5.25 pounds.
Getting back to my proposal for measuring drag by finding terminal velocity in a vertical dive, with a dead engine, when drag equals weight: If the airplane weighs 7 pounds, it would clearly go faster in a vertical dive than at full power, straight and level.
Its actual speed in vertical dive, neglecting the drag of the stationary prop, would be the square root of (7/5.25), multiplied by the maximum powered speed, straight and level of 100 mph, which equals 115.5 mph.
If the airplane weight is increased from 7 to 10 pounds, vertical dive terminal velocity would increase to 138.1 mph. If 10 pound airplane is extremely clean, and capable of 120 mph straight and level, thrust at 70% prop efficiency would be 4.375 pounds. Vertical dive terminal velocity would increase to 181.4 mph, neglecting the increase expected from the higher Reynolds number.
The height needed to attain this high a terminal velocity would be probably be over 5,000 feet.
Height needed to attain a terminal velocity of 115.5 mph would be about 2,000 feet, do-able with good pilot eyesight.
Consider a fairly clean airplane that can achieve 100 mph (146.7 feet/sec.) straight and level, with a 2 hp engine. Note that one hp = 550 foot-pounds per second.
At 70% propeller efficiency, thrust = (2x550x0.7)/146.7 = 5.25 pounds.
Getting back to my proposal for measuring drag by finding terminal velocity in a vertical dive, with a dead engine, when drag equals weight: If the airplane weighs 7 pounds, it would clearly go faster in a vertical dive than at full power, straight and level.
Its actual speed in vertical dive, neglecting the drag of the stationary prop, would be the square root of (7/5.25), multiplied by the maximum powered speed, straight and level of 100 mph, which equals 115.5 mph.
If the airplane weight is increased from 7 to 10 pounds, vertical dive terminal velocity would increase to 138.1 mph. If 10 pound airplane is extremely clean, and capable of 120 mph straight and level, thrust at 70% prop efficiency would be 4.375 pounds. Vertical dive terminal velocity would increase to 181.4 mph, neglecting the increase expected from the higher Reynolds number.
The height needed to attain this high a terminal velocity would be probably be over 5,000 feet.
Height needed to attain a terminal velocity of 115.5 mph would be about 2,000 feet, do-able with good pilot eyesight.
04-11-2009 | 08:40 AM
#19
Senior Member
RE: Measuring Total Plane Drag
I made a math mistake in preceding post, and inadvertently submitted while checking calculations.
Correct dive height to attain 115.5 mph terminal velocity, neglecting air resistance would be about 445 feet, and probably about 600 feet with air resistance.
Correct dive height needed to attain 181.4 mph without air resistance would be about 1100 feet, and probably about 1500 feet with air resistance factored in.
Both appear do-able if the airplane hangs together and doesn't turn itself into a dangerous missile.
Correct dive height to attain 115.5 mph terminal velocity, neglecting air resistance would be about 445 feet, and probably about 600 feet with air resistance.
Correct dive height needed to attain 181.4 mph without air resistance would be about 1100 feet, and probably about 1500 feet with air resistance factored in.
Both appear do-able if the airplane hangs together and doesn't turn itself into a dangerous missile.
04-11-2009 | 01:38 PM
#20
RE: Measuring Total Plane Drag
ORIGINAL: pimmnz
By measuring the drag Seriously, the only practical way of doing this is to find a university handy that has an aeronautical thread and a tunnel big enough to put the whole model in. Knowing the tunnel corrections they can the measure all the parameters at various airspeeds and come up with Cd and Cl for the model. They can do lots of other things, stall speed and stall aoa etc. Could be a useful trade off here, it gives the students a practical set of problems, and you all you need to know about your model, and the cost of doing this, which could otherwise be horrendous, may be within some sort of bargaining reach. It may mean you lose the model for a month or two, but the results could be well worth the time. Based on what you find, the answers could be extrapolated across a wide range of models, and answer quite a few questions.
Evan, WB #12.
By measuring the drag
Seriously, the only practical way of doing this is to find a university handy that has an aeronautical thread and a tunnel big enough to put the whole model in. Knowing the tunnel corrections they can the measure all the parameters at various airspeeds and come up with Cd and Cl for the model. They can do lots of other things, stall speed and stall aoa etc. Could be a useful trade off here, it gives the students a practical set of problems, and you all you need to know about your model, and the cost of doing this, which could otherwise be horrendous, may be within some sort of bargaining reach. It may mean you lose the model for a month or two, but the results could be well worth the time. Based on what you find, the answers could be extrapolated across a wide range of models, and answer quite a few questions.Evan, WB #12.
Thanks for the knock to my noggin!
04-11-2009 | 02:37 PM
#21
RE: Measuring Total Plane Drag
ORIGINAL: HighPlains
You can determine quite a bit by compairing speed in level flight to speed in a dive if you have use of a Radar gun and have spotters reading the dive angle.
You can determine quite a bit by compairing speed in level flight to speed in a dive if you have use of a Radar gun and have spotters reading the dive angle.
04-11-2009 | 02:48 PM
#22
Senior Member
RE: Measuring Total Plane Drag
ORIGINAL: Rotaryphile
Correct dive height to attain 115.5 mph terminal velocity, neglecting air resistance would be about 445 feet, and probably about 600 feet with air resistance.
Correct dive height needed to attain 181.4 mph without air resistance would be about 1100 feet, and probably about 1500 feet with air resistance factored in.
Both appear do-able if the airplane hangs together and doesn't turn itself into a dangerous missile.
Correct dive height to attain 115.5 mph terminal velocity, neglecting air resistance would be about 445 feet, and probably about 600 feet with air resistance.
Correct dive height needed to attain 181.4 mph without air resistance would be about 1100 feet, and probably about 1500 feet with air resistance factored in.
Both appear do-able if the airplane hangs together and doesn't turn itself into a dangerous missile.
Have you an explanation for consideration of density altitude? I've been recording it since the 70s whenever I do any kind of testing, especially prop tests. Most would be amazed exactly how much it differs, even from day to day. And not many would realize how much it affects our model performance. It's probably the one most important unknown that makes a great deal of our testing of little real value.
As an aside, nowadays it's actually quite easy to take very accurate readings of it. There are handheld "weather stations" quite reasonably priced that do quite an excellent job of recording just about everything.
04-11-2009 | 03:13 PM
#23
RE: Measuring Total Plane Drag
ORIGINAL: Rotaryphile
Measuring static thrust is easy - I just use a fish scale tied to the tail. Thrust in flight is normally lower than static thrust, and needs a wind tunnel for accurate measurement.
Consider a fairly clean airplane that can achieve 100 mph (146.7 feet/sec.) straight and level, with a 2 hp engine. Note that one hp = 550 foot-pounds per second.
At 70% propeller efficiency, thrust = (2x550x0.7)/146.7 = 5.25 pounds.
Getting back to my proposal for measuring drag by finding terminal velocity in a vertical dive, with a dead engine, when drag equals weight: If the airplane weighs 7 pounds, it would clearly go faster in a vertical dive than at full power, straight and level.
Its actual speed in vertical dive, neglecting the drag of the stationary prop, would be the square root of (7/5.25), multiplied by the maximum powered speed, straight and level of 100 mph, which equals 115.5 mph.
If the airplane weight is increased from 7 to 10 pounds, vertical dive terminal velocity would increase to 138.1 mph. If 10 pound airplane is extremely clean, and capable of 120 mph straight and level, thrust at 70% prop efficiency would be 4.375 pounds. Vertical dive terminal velocity would increase to 181.4 mph, neglecting the increase expected from the higher Reynolds number.
The height needed to attain this high a terminal velocity would be probably be over 5,000 feet.
Height needed to attain a terminal velocity of 115.5 mph would be about 2,000 feet, do-able with good pilot eyesight.
Measuring static thrust is easy - I just use a fish scale tied to the tail. Thrust in flight is normally lower than static thrust, and needs a wind tunnel for accurate measurement.
Consider a fairly clean airplane that can achieve 100 mph (146.7 feet/sec.) straight and level, with a 2 hp engine. Note that one hp = 550 foot-pounds per second.
At 70% propeller efficiency, thrust = (2x550x0.7)/146.7 = 5.25 pounds.
Getting back to my proposal for measuring drag by finding terminal velocity in a vertical dive, with a dead engine, when drag equals weight: If the airplane weighs 7 pounds, it would clearly go faster in a vertical dive than at full power, straight and level.
Its actual speed in vertical dive, neglecting the drag of the stationary prop, would be the square root of (7/5.25), multiplied by the maximum powered speed, straight and level of 100 mph, which equals 115.5 mph.
If the airplane weight is increased from 7 to 10 pounds, vertical dive terminal velocity would increase to 138.1 mph. If 10 pound airplane is extremely clean, and capable of 120 mph straight and level, thrust at 70% prop efficiency would be 4.375 pounds. Vertical dive terminal velocity would increase to 181.4 mph, neglecting the increase expected from the higher Reynolds number.
The height needed to attain this high a terminal velocity would be probably be over 5,000 feet.
Height needed to attain a terminal velocity of 115.5 mph would be about 2,000 feet, do-able with good pilot eyesight.
However, since this is a brainstorming thread I'll throw this convoluted modification out there and see if it lands. A parachute opens and a jumper quickly reaches TV in a short distance because of the extremely high drag force. What if we deployed a drag chute that the planes engine could overcome well enough to stay airborn in level flight.
So we find the airspeed at max power level flight with and without the chute deployed. Now we do a Terminal Dive with the chute deployed and record that speed(the TV). We can calculate the drag force with chute deployed (and it's Cd) from the TV. now we can calculate the in-flight thrust from the engine in level flight with the chute deployed. Next, assuming the thrust doesn't change with speed, we can calculate the drag force (and it's Cd) of the clean plane without the chute deployed!
The sticky part is whether the motor can keep the plane flying level with a big enough chute to slow a terminal dive in a reasonable distance. As a bonus, this method would also eliminate the danger of overspeeding the airframe. The chute could be released from the plane high enough to build momentum and glide in like a normal deadstick.
04-11-2009 | 05:26 PM
#24
RE: Measuring Total Plane Drag
CC
I hope you are able to get access to a wind tunnel, I am very interested to see the results of the test.
Do you by chance have access to an unmodified Duellist twin to compare the data with? I am intrigued to see how much of a difference that reduce frontal area, improve the nose and nacelle transition can make.
I would love to take something like the "Velocity" and create a dual tractor powered version.
Also as I have an real interest in the particular area that you are dealing with I am curious to the results such efforts make .
I would love to see more instrumentation for our models become available, to allow for us to obtain more data.
As far as what you threw out...
Sounds a bit like towing a banner to me...
I hope you are able to get access to a wind tunnel, I am very interested to see the results of the test.
Do you by chance have access to an unmodified Duellist twin to compare the data with? I am intrigued to see how much of a difference that reduce frontal area, improve the nose and nacelle transition can make.
I would love to take something like the "Velocity" and create a dual tractor powered version.
Also as I have an real interest in the particular area that you are dealing with I am curious to the results such efforts make .
I would love to see more instrumentation for our models become available, to allow for us to obtain more data.
As far as what you threw out...
Sounds a bit like towing a banner to me...
04-11-2009 | 08:05 PM
#25
My Feedback: (1)
RE: Measuring Total Plane Drag
CC,
Read though the following:
http://www.rcuniverse.com/forum/m_68..._1/key_/tm.htm
Do you have a link to some more information?
http://www.rcuniverse.com/forum/m_68..._1/key_/tm.htm
