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Old Aug 02, 2014, 06:21 PM
Lift is good
Rayven's Avatar
Olympic Penninsula, WA
Joined Dec 2001
758 Posts
Question
One meter vs 1.5 meter DLG launch difference?

Forgive me if this is a stupid question--but I'm not an aerodynamicist (though I am an over-educated carpenter), just an avid glider nut. I fly both a David Forbe's one meter "Sparrowhawk" and a 1.5 meter "Arrow", depending on the field I get to. The Arrow always launches to nearly twice the height of the little Sparrowhawk, but weighs twice as much. What is the fundamental principal at work here: lift generation? Inceased speed of launch because of the longer moment arm? Both?

I assume the larger wing generates more lift than the little one, but is it really enough to overcome the increased weight and drag? Does the 1.5 really launch that much faster?

Any ideas? Or even better, empirical answers?
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Old Aug 02, 2014, 11:52 PM
Thermal Naked!
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United States, AL, Mobile
Joined Dec 2007
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Weight (maybe mass?).. Throw a ping pong ball and a baseball. Which goes farther. I slept through physics class or I'd bombard you with lots of facts, figures and formulas, but it boils down to weight and mass type stuff.
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Old Aug 03, 2014, 12:36 AM
Lift is good
Rayven's Avatar
Olympic Penninsula, WA
Joined Dec 2001
758 Posts
That does remind of the whole momentum thing in physics: the more mass, the more momentum (assuming velocity is the same). So, basically, the 1.5 has more momentum than the 1.0 and therefore climbs for a longer period (and thus higher) until the momentum dissipates. Is it really as simple as that? And I suppose that if the weight of your DLG gets too heavy, then velocity declines so the launch isn't as high. Must be that sweet spot of around 9 ounces??
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Old Aug 03, 2014, 02:09 AM
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Mike4192's Avatar
United States, WA, Seattle
Joined Sep 2006
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There seem to be two things that vary between large and small DLGs: release velocity and aerodynamic drag.

Assuming you spin your body at the same speed releasing both a 1m and 1.5m DLG, the 1.5m DLG will probably be travelling just a little bit faster upon release. Remember, the tangential speed of a rotating object is rotation rate times radius.

Now simplify the idea by imagining we release the DLG with an initial velocity straight upwards. Naturally, we see the DLG go up and gradually slow down until it noses over. There are two forces at work slowing down the DLG: gravitational force and aerodynamic drag.

All we need to explain this is the one equation you hopefully remember from introductory physics:
sum of forces = mass*acceleration

What are our forces? Like we defined, gravitational force (mass*earth acceleration), and aerodynamic drag (0.5*rho*V^2*S*Cd, where rho is air density, V is velocity, S is planform wing area of the DLG, and Cd is a constant coefficient of drag (which we assume is roughly the same for similar DLGs because it is non dimensional)).

So if we write out the equation:
-0.5*rho*V^2*S*Cd - mass*(earth acceleration) = mass*(DLG acceleration)

And solve for DLG acceleration:
-0.5*rho*V^2*S*Cd/mass - earth acceleration = DLG acceleration

We see there is a tricky balance here. The acceleration from drag is proportional to velocity squared and wing area, and inversely proportional to mass. A 1.5m DLG must be just enough heavier than a 1m DLG that it's increased velocity upon launch and larger wing area doesn't increase it's drag too much. Even if it breaks even with a smaller DLG, it was still released with higher velocity.

Now think about a hypothetically larger DLG, say 2m wing span. Now it might be too big and released too fast that it's extra weight isn't enough to give it enough "uumph" to keep going.

Now that makes me think, maybe 1.5m isn't the optimal DLG size! Obviously 1.5m was just chosen because it is a reasonable size and a round number. There must be some wingspan that will yield a DLG that will launch higher than a 1.5m DLG.

As the textbooks say, I will leave that problem to the reader to work out for themselves.
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Old Aug 03, 2014, 05:09 AM
Looptastic!
sp00fman's Avatar
Enschede, Netherlands
Joined Nov 2009
2,342 Posts
Trying to grasp your equation... its early and coffee must sink in.
Attempting to rationalize things: i think launch height is simply the relationship between the launch speed and earths gravity.

An free falling object gains 10m/second of speed in Earths gravity. The other way around must be also true then, throwing something upwards must de-accelerate by 10m/s. Mass is not an factor here. We can only control the launch speed.

Attempting to find the correct formula, but a chart i found gives this data:

a launch speed of 106km/h gives 44m height before it stops, and that takes 3 seconds
at 140km/h (40m/s) it gets to 78 meter and it takes 4 seconds.

This is the data you would get in a vacuum. The arodynamic drag is the factor that restricts us from getting to the above mentioned heights.

The drag is loss of energy, and the more energy we give our plane in the launch (speed x mass), the less of a factor the arodynamic drag becomes. With more mass we need a bigger wing for our flying around purposes, and a big wing generating more drag. I dont know where the sweet spot is for weight/wing area.
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Old Aug 03, 2014, 05:27 AM
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Nashville, NC, USA
Joined Mar 1999
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You must impart energy into the plane.
KE = 1/2 x M x V^2
So, double the mass, same speed= twice the energy
So, same mass, double the speed=4 times the energy
So, heavier plane(more mass), longer wing span(more launch speed, assuming same rotational speed)= more height
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Old Aug 03, 2014, 05:36 AM
Looptastic!
sp00fman's Avatar
Enschede, Netherlands
Joined Nov 2009
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Hey, thanks for the correct energy equation. I was oversimplifying that. What would be fun is to make a chart or graph and put in the data of existing DLG planes, and see which one has the highest potential launch height. then take that plane and investigate why it is or isn't getting the predicted height.
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Old Aug 03, 2014, 06:05 PM
Will fly for food
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Bellevue WA,
Joined Dec 2003
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Quote:
Originally Posted by davecee View Post
You must impart energy into the plane.
KE = 1/2 x M x V^2
So, double the mass, same speed= twice the energy
So, same mass, double the speed=4 times the energy
So, heavier plane(more mass), longer wing span(more launch speed, assuming same rotational speed)= more height
Proof again that bigger fly's better
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Old Aug 03, 2014, 08:56 PM
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Thermaln2's Avatar
Reno Nevada
Joined Oct 2007
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Everyone is talking weight and speed, but that is not all that gives the height. If one can input speed, one has to look at what slows the plane down. Everyone has mentioned the drag of the plane, but there is more.

Additional items deal with launch technique and plane response. If we look at the Snipe it launches higher because of less drag, but it is a lot lighter than and even as light as the smaller planes. So cleanliness matter. Bigger planes fly better because they are more efficient, but so can smaller planes. Items that are not being considered is the launch "wag" of the plane as it responds to the throw. Each wag has been approximated to 10-15 ft of height loss. Smaller planes are harder to keep from wagging , especially when pone tries to put extra speed and effort into the launch. As the speed increases the wing and plane has a tendency to roll more and thus one cannot get the plane to climb as straight and true as a bigger plane. The tails are shorter so there is more wing weight at the tips. Launches also experience difficulties is the release of thw wingpeg. Many tend to hold onto the peg longer and release later in the swing.

I have seen the MiMi launch as high as bigger planes, even higher. So it can be done.

So even though one can impart more speed because it is easier to accelerate, other factor come into play.

Bigger planes, with longer wingspan just launch higher even at the same weight.
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Old Aug 03, 2014, 09:32 PM
I'd rather be Flying
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Nashville, NC, USA
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Quote:
Originally Posted by Thermaln2 View Post
I have seen the MiMi launch as high as bigger planes, even higher. So it can be done.


Bigger planes, with longer wingspan just launch higher even at the same weight.
So which is it?
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Old Aug 03, 2014, 11:26 PM
Lift is good
Rayven's Avatar
Olympic Penninsula, WA
Joined Dec 2001
758 Posts
How bout airfoil shapes? We give most modern DLG airfoils a little reflex to optimize for speed (I think)...so I guess that would enter the equation as well. Also: bigger might fly better, but the reality of trying to discus launch a much bigger wingspan than 1.5 gets problematic---unless you are really tall and lanky! So me at 5 ft 9 will probably never launch to the height my 6 ft 4 friend could achieve if he used a longer wingspan DLG. But even with us matched in glider size, his longer arm would likely impart more rotational speed. So there's yet another factor.
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Old Aug 04, 2014, 07:05 AM
I'd rather be Flying
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Nashville, NC, USA
Joined Mar 1999
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I'm not talking piloting or airframe design, I wouldn't presume to teach any of you anything about that. I'm just trying to explain (as I understand it) the energy transfer from the pilot to the DLG.

Let's assume all other things are equal, eg pilot skill, pilot arm length, rotational speed, airfoil, airframe stiffness and so on; and just consider 1 meter vs. 1.5 meter DLG as the only variable.

The longer wingspan plane will have it's center of mass further out from the pivot point. At the same rotational speed, the linear speed of the center of mass will be faster, thus have more kinetic energy to trade for launch height. A heavier plane will also have more energy at launch, assuming the same launch speed.

Energy imparted to the DLG goes up as the SQUARE of the speed increase. If we figure an arm length from the pivot point of 36 inches and add half of the wingspan and if we use the same rotational speed, the 1.5 meter DLG leaves the pilots hand 1.18 times faster and that translates into nearly 1.4 times the energy.

Now if we factor in a heavier plane, 10 ounces vs. 6 ounces, along with the faster speed, the energy we put into the plane at launch goes up 2.3 times. So, it looks to me that an average pilot with typical DLGs could easily transfer twice the energy into the 1.5 meter DLG compared to the 1 meter DLG.

Dave
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Old Aug 04, 2014, 09:06 AM
guamflyer - Avid Sloper
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GUAM.USA
Joined Dec 2009
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Quote:
Originally Posted by Mike4192 View Post
There seem to be two things that vary between large and small DLGs: release velocity and aerodynamic drag.

Assuming you spin your body at the same speed releasing both a 1m and 1.5m DLG, the 1.5m DLG will probably be travelling just a little bit faster upon release. Remember, the tangential speed of a rotating object is rotation rate times radius.

Now simplify the idea by imagining we release the DLG with an initial velocity straight upwards. Naturally, we see the DLG go up and gradually slow down until it noses over. There are two forces at work slowing down the DLG: gravitational force and aerodynamic drag.

All we need to explain this is the one equation you hopefully remember from introductory physics:
sum of forces = mass*acceleration

What are our forces? Like we defined, gravitational force (mass*earth acceleration), and aerodynamic drag (0.5*rho*V^2*S*Cd, where rho is air density, V is velocity, S is planform wing area of the DLG, and Cd is a constant coefficient of drag (which we assume is roughly the same for similar DLGs because it is non dimensional)).

So if we write out the equation:
-0.5*rho*V^2*S*Cd - mass*(earth acceleration) = mass*(DLG acceleration)

And solve for DLG acceleration:
-0.5*rho*V^2*S*Cd/mass - earth acceleration = DLG acceleration

We see there is a tricky balance here. The acceleration from drag is proportional to velocity squared and wing area, and inversely proportional to mass. A 1.5m DLG must be just enough heavier than a 1m DLG that it's increased velocity upon launch and larger wing area doesn't increase it's drag too much. Even if it breaks even with a smaller DLG, it was still released with higher velocity.

Now think about a hypothetically larger DLG, say 2m wing span. Now it might be too big and released too fast that it's extra weight isn't enough to give it enough "uumph" to keep going.

Now that makes me think, maybe 1.5m isn't the optimal DLG size! Obviously 1.5m was just chosen because it is a reasonable size and a round number. There must be some wingspan that will yield a DLG that will launch higher than a 1.5m DLG.

As the textbooks say, I will leave that problem to the reader to work out for themselves.
UUmm, I think even the reasonable thought of a particular wingspan being better suited for DLG launch , should be considered somewhat as important

SLOPE FAST - SOAR DEEP
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Old Aug 04, 2014, 09:10 AM
guamflyer - Avid Sloper
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theres gotta be a certain w.s. that would launch more consistently with most people..maybe...
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Old Aug 04, 2014, 09:13 AM
guamflyer - Avid Sloper
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contest flyers are gonna push da envelope anyways.. so how viable would the research be verses staying at the usual wingspans currently used
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