Calculating maximum speed of Bungee Launch

[Texas Buzzard]

I am building a bungee launcher for my ducted fan powered Jet Fighters.
The motive power is a 24' length of 1/4th" dia. Latex tube from Lowes.

When I stretch this tubing to 2 1/2 times it's original length the tension is close to 4 pounds. My plane has a weight of 1.5 pounds.

I know that during the launch the tension will decrease as the plane moves.
I am interested in the maximum speed on release of ring from plane.
I estimate the average force to be about 3 pounds for the initial acceleration
Remember this force and the plane are acting near the horizontal:so we are not concerned with gravity. Almost all of the force is available to cause acceleration of the plane.

I have solved for final velocity ...Will one of you Check My Work?

Using 3 pounds for the accerating force .......

F/m=a OR F/g = mass ; 1.5 lb/32 ft/sec^2 = 0.0468 slugs for mass of plane

Again F= m x a or a = F/m ; 3 slug ft/sec^2 divided by 0.0468 slugs = 64 ft/sec^2 for the acceleration
.................................................. .....
acceleration = 64 ft/sec^2 ; This is twice the acceleration of gravity.

We know that a rate of 88 ft/sec = 60 mph ; so 64 ft/sec^2/88 mph = X/60 mph..... X = 43 mph
for final velocity ( yes, this is an estimate, for time is not known.)

The Jets stall at well over 20 mph... maybe 25 mph. So if we are over estimating the speed of the launch by 25% that gives 0.75 x 43 mph 0r 32 mph.

Will someone please check my work. You will get a prize - a pic of my Jet. Thanks.

[JetPlaneFlyer]

Here's what I calculate:

(BYW...I find it easier working in metric)

First of all the distance travelled between bungee fully stretched and relaxed is 36' (11m)

If the force is 4 lb at fully stretched and zero when relaxed, assuming the force builds up in a linear fashion, then the average force is 2lb (8.9N)

avg. acceleration (a) = avg. force / mass = 8.9N/0.68Kg = 13.09m/s^2

Assuming the starting velocity is zero then:

V = sq. root (2.a.d)

where:
V = final velocity (m/s)
a = avg. acceleration (m/s^2
d = dist travelled (m)

v = sq root (2 x 13.09 x 11) = 16.97m/s ( 38mph )

So 38mph is the figure I get ... I guess the thrust of the motor could increase the velocity, there again, hysteresis in the latex may work against it?

[BMatthews]

The assumption here is that the bungee has a linear buildup of tension. The way to test that would be to start at 0 stretch and then go up in increments of something like 1 meter and measure the pull at each point. Then graph the pull to force chart and see just what it really is. It's likely that it'll have some curve to it that sort of mimics a discharge curve of a battery or it may well be linear up to a point and then rise sharply at the end. In either case you can do a pretty good extimate of putting in a flat line that cuts the curve in such a way that the areas above and below are about equal. It ain't perfect but it'll give you a flat line to work with and from there go with JPF's method.

Although really it would be good to also try doing this by cutting that line into 4 segments and calculate for each segment to see if it matches.

One thing that'll be hard to guage is the drag from any sort of launch rail and also the air drag during the last 1/2 of the launch where speed is high enough that this is an issue. But if the model is up to flying speed at some point around the 1/2 way mark then it really doesn't matter much does it...

[Al M]

Rubber does not return energy with the same curve as it stored energy either. You will have to rely on good luck to get accurate results.

[ghoti]

There is always the empirecal method: Get a doppler speed measuring device and see how fast your rubber will launch a 1.5 pound stone, maybe? That would be without prop power added, of course. LOL Bill

[JetPlaneFlyer]

Rubber does not return energy with the same curve as it stored energy either.
I know, it's called hysteresis and i mentioned it in my original reply ;)

[Tom Harper]

ghoti,

Not a bad idea. You could use a weighted fuselage (with rudder), in place of your 'stone', and an Eagle Tree data logger. The data logger will take 10 samples per second. It would still be approximate due to start up issues and delays in the system, but it would yield useful engineering data.

The weather has been too bad for flying. Good time to take data.

[Texas Buzzard]

BMatthews is giving a description of an experiment that is done in Physics Classes Actually he is discussing Hooks Law ....go see Hook's Law in a text book or INTERNET.

For my purposes I don't need an exact figure (mph). I just wanted a "working first approximation" assuring the launch would give a sped above stall speed.

Thanks BMatthews
>>>>>>>>>>>>>>>MXYZ<>><>>>...........

BMatthews is on the right track in his first 6 sentencesThe assumption here is that the bungee has a linear buildup of tension. The way to test that would be to start at 0 stretch and then go up in increments of something like 1 meter and measure the pull at each point. Then graph the pull to force chart and see just what it really is. It's likely that it'll have some curve to it that sort of mimics a discharge curve of a battery or it may well be linear up to a point and then rise sharply at the end. In either case you can do a pretty good extimate of putting in a flat line that cuts the curve in such a way that the areas above and below are about equal. It ain't perfect but it'll give you a flat line to work with and from there go with JPF's method.

Although really it would be good to also try doing this by cutting that line into 4 segments and calculate for each segment to see if it matches.

One thing that'll be hard to guage is the drag from any sort of launch rail and also the air drag during the last 1/2 of the launch where speed is high enough that this is an issue. But if the model is up to flying speed at some point around the 1/2 way mark then it really doesn't matter much does it...

[MarkusN]

One thing that is ignored up to now: The weight of the bungee itself is not neglectable. When you stretch it and let it go it does not accelerate with infinite acceleration. That "empty" acceleration is an upper limit.
No time to dive into that in detail right now, but I assume that you have to add 1/3 to 1/2 of the bungee's weight to the mass accelerated.

[Texas Buzzard]

Here's what I calculate:

(BYW...I find it easier working in metric)

First of all the distance travelled between bungee fully stretched and relaxed is 36' (11m)

If the force is 4 lb at fully stretched and zero when relaxed, assuming the force builds up in a linear fashion, then the average force is 2lb (8.9N)

avg. acceleration (a) = avg. force / mass = 8.9N/0.68Kg = 13.09m/s^2

Assuming the starting velocity is zero then:

V = sq. root (2.a.d)

where:
V = final velocity (m/s)
a = avg. acceleration (m/s^2
d = dist travelled (m)

v = sq root (2 x 13.09 x 11) = 16.97m/s ( 38mph )

So 38mph is the figure I get ... I guess the thrust of the motor could increase the velocity, there again, hysteresis in the latex may work against it?
>>>>>>>>>>>>>>><<<<<<<<<<<<<<<
I found my error. When I converted m/sec to ft/sec I was in error.

THANK YOU FOR CONFIRMING MY WORK. Now I get 38 mph too.

So we know that upon launch the Park Jet will be flying well above the stall speed of an estimated 20 to 24 mph. (Heavy)

The friction and aerodynamic drag are not significant for my purposes. Yes the super theoritical people will mention aerodynamic drag, a -force due to friction of guide rails % and these off physical properties of the Latex tube. All I wanted was to find the aprox. velocity at moment of release from the tow ring. Thanks to ALL of you who contributed.

JetPlaneFlier - Thanks again

[mnowell129]

Alternately you can just work with energy.
The total energy in the bungee is 72 lb-ft (36 feet * 2 lb average).
If all the energy in the bungee is converted to kinetic energy in the jet 72 ft-lb = 1/2 m * v^2.
v = sqrt(144 (ft-lb) / mass (slug))
v = sqrt( 144 (ft-lb) / w (lb)/g (ft/sec*sec))
v = sqrt( 144 )/ (1.5 /32.2) (ft*ft)/(sec*sec))
v = sqrt(3091) ft/sec
v = 55.6 ft/sec
v = 37.9 mph

As has been addressed already some of the energy goes into the bungee (this could be estimated by how much shorter the bungee is than 24 feet after launch) and drag losses in the airplane.

[Radioguy]

Bruce:

It's been about 35 years since we first met and your brainpower still impresses me!

I'm on vacation now, and yesterday we passed an old-style steam laundry and Seagulls were thermalling above the shop..... just the same way we used to at the park in the early '70's. That was a good group of folks.

Keep up the good work Bruce.

I'll be flying a Convair 580 on fire suppression operations and be in charge of taking a Convair cockpit and turning it into a simulator ... doing the computer interfacing, etc.

All the best,
Lee

[Texas Buzzard]

Alternately you can just work with energy.
The total energy in the bungee is 72 lb-ft (36 feet * 2 lb average).
If all the energy in the bungee is converted to kinetic energy in the jet 72 ft-lb = 1/2 m * v^2.
v = sqrt(144 (ft-lb) / mass (slug))
v = sqrt( 144 (ft-lb) / w (lb)/g (ft/sec*sec))
v = sqrt( 144 )/ (1.5 /32.2) (ft*ft)/(sec*sec))
v = sqrt(3091) ft/sec
v = 55.6 ft/sec
v = 37.9 mph

As has been addressed already some of the energy goes into the bungee (this could be estimated by how much shorter the bungee is than 24 feet after launch) and drag losses in the airplane.
................................................
mnowell, great...this solution was deep down in my mind in the beginning but just couldn't bubble out. I like your approach. There are more way than one to skin a cat. Sure we ignore friction and other small degrading factors. We wanted a first approximation value ------ You got it, so did JetPlaneFlier. I staggered into agreeing with both of you. Good stuff.

So with a stall speed of about 22 to 25 mph of the Jet, we can launch with confidence.

[N9DP]

Is no one concerned about V^^2 drag?

Good landings,

Dennis

[N9DP]

Is no one concerned about V^^2 drag?

Good landings,

Dennis

Never mind. A full-size Mooney 201 has an equivalent flat-plane area of 3 sq ft. At 38 mph the flat-plane drag is 3.7 lbs/sq ft.
Dennis

[ghoti]

Dennis, That number is hard to believe when I hold my flat hand out of a car window at 50 mph. But what do I know? bill

[N9DP]

Dennis, That number is hard to believe when I hold my flat hand out of a car window at 50 mph. But what do I know? bill

Which number are you having a problem with? Flat-plate drag is the simplest geometry for calculating drag.

Lbs/sqft = MPH^^2 x 0.00256.

If we take your hand as 4.5 x 8 inches, at 50 mph the force is 1.6 lbs. If you extend your arm instead of just sticking out your hand, the total drag force is roughly 2.5 lbs, given the cylindrical cross section of your arm. The intersection drag at the window/airflow interface is quite large, but I don’t know how to calculate that.

A friend of mine constructed a fixture to hold his sailplane with a fish scale arrangement that measures drag force. He has his wife drive the car down a back road while he holds the fixture with sailplane in the breeze and logs drag force as a function of speed. I want to make a similar gadget to mount on the roof of my jeep and extend through the sunroof. Science!

Good landings,

Dennis

[JetPlaneFlyer]

Is no one concerned about V^^2 drag?

Good landings,

Dennis

I dont think it's a problem because the model is fitted with an EDF unit that will provide thrust which would be far greater than drag at the speeds calculated... So in reality the launch speed will be underestimated because net thrust is not considered.

If it were a glider then drag should be factored in.

As noted earlier in the thread the weight of the bungee itself (or about 1/3 of it) should also really be added to the model's weight.