Kites & kiting

[DFINITLYDISTRUBD]

There's a local that has something like that! For a few years he ran a "musical petting zoo" (name of his shop) and allowed anyone to come in and play any instrument they wanted to...not sure how he made his money. They were likely leather or rawhide on wood or just wood (like on a gurdy) as felt doesn't hold up to that kind of use very well (don't ask how I know). Was experimenting with leather as a replacement for horse hair...didn't work out as impregnating it with rosin is very difficult and the flex allowed caused all sorts of issues...even played with a rosined dowel , that worked but had too much rattle.

 

So you're saying I should convert my cello to a powered hurdy gurdy of sorts :P

http://www.youtube.c...h?v=FrRrK8kAK7c

Ironically I'm fairly certain there is no hurdy gurdy in the song...lol.

 

 

 

There is a product used by guitarists that uses a magnetic field to vibrate the strings called an E-Bow...expensive (but cheaper than a decent bow) and about the size of a capo...never seen one used, but I see they still sell them...they do have the drawback of requiring at least steel core strings, prefer synthetic (actually prefer gut but at close to to well over $300 a set, aint happening), the stranded steel core strings are pricey, solid steel cores are fairly cheap but sacrifice tone.

Edited September 14, 2013 by DFINITLYDISTRUBD

[Turtle]

...

So you're saying I should convert my cello to a powered hurdy gurdy of sorts :P

 

Yes. :P Let me know when you get enough kite & wind to get that cello airborne. Maybe just add wings and get a delta-cello. :smart:

 

So, with the pulleys cut deeper the rubber band works OK in a no-load test.

 

http://www.youtube.com/watch?v=xUiOLzHAaoc&feature=c4-overview&list=UUiDIgwwtUxDi7fxhELuBtxg

[Turtle]

Reporting in with results of first line test.. Epic fail.

 

When it wouldn't budge at 9v (the motor stalled) I bumped it up to 18v and the rubber band broke. With the c-clamps holding it together it weighs 2 lbs. Clearly will need intermediate pulleys to gear down.

 

 

Vid is just a fly-around; it was all over by then. Looks gnarly though, right?

 

 

http://www.youtube.com/watch?v=hFFWXU9U59w&feature=c4-overview&list=UUiDIgwwtUxDi7fxhELuBtxg

[DFINITLYDISTRUBD]

Hrrm! I was just coming here to retract my doubts about torque from the motor...spent the afternoon battling with trying to build then run a rock tumbler powered by a 12V compressor motor, They will run quite well on 9V given enough amperage....sadly my efforts to wrangle in the revs were fruitless and I destroyed the large diameter hairband I was using for a belt...get it not to slip on the motor it would spin up and fling the drum...so far attempt to drop the wattage motor stop. When I get it working I'll post the build in DIY over at CF....Well anyway good luck sir. If you can wrangle get the power where you need it and sort out the belt I'd wager she'll have all the grunt she needs to go up. I'm off to root through my secret stash of wall warts.

[DFINITLYDISTRUBD]

I imagine O-rings aren't cheap so won't be trying them any time soon.

They're pretty cheap for plumbing related ones, Paid less than a dollar for the one I use to lock my throttle on the bike (redneck cruise control).

 

I have a couple about the size you'd need based on the vid but they're cracked and dry rotted...will check prices...BTW thanx for reminding me I bet one would kinda sorta work for that rock tumbler I'm making. They're for sealing chlorinator and pump access covers.

 

Clearly will need intermediate pulleys to gear down.

I thought so...but then maybe not....this will sound off...but... I think you could improve performance by going to larger pulleys both at the motor and the windlass. Too small a pulley at the motor results in not enough contact surface for good gription, but of course a larger pulley at the motor then requires using a larger pulley at the windlass to keep the same reduction...but you knew that part already :P

[Turtle]

They're pretty cheap for plumbing related ones, Paid less than a dollar for the one I use to lock my throttle on the bike (redneck cruise control).

 

I have a couple about the size you'd need based on the vid but they're cracked and dry rotted...will check prices...BTW thanx for reminding me I bet one would kinda sorta work for that rock tumbler I'm making. They're for sealing chlorinator and pump access covers.

 

I borrowed a tumbler once and it used two 3/8" horizontal shafts in the base and the drum just sat on them. One shaft was driven by motor and it rolled the drum. (Do you have the grits by the way?) Will check O-ring prices at next opportunity. :thumbs_up Kite-powered rock tumbler anyone? :lol:

 

 

Need step-down pulleys

I thought so...but then maybe not....this will sound off...but... I think you could improve performance by going to larger pulleys both at the motor and the windlass. Too small a pulley at the motor results in not enough contact surface for good gription, but of course a larger pulley at the motor then requires using a larger pulley at the windlass to keep the same reduction...but you knew that part already :P

 

 

This is where kite application specifc issues arise. Larger pulley on drive-drum means increased wind-drag and may add to bobbling/swinging. Then again, since it would be parallel to wind direction it might steady things. :shrug: Weight of 2 big(ger) pulleys vs. 4 smaller to arrive at same result sounds like a problem in linear programming, though that's not to say I have sensitive enough scales for collecting data or that I remember how to use matrices to solve such an equation system.

 

Since motor stalled with the rubber-band-belt the small pulley wasn't gription deficient, but of course that may be different with O-ring or hair-tie belts.

 

Raining now and last night they said we may have seen the last 80º day of the year. No joy.

[DFINITLYDISTRUBD]

Found source of a sizeable fee "O" ring...bucket lids! requires a fair amount of tension to not slip...but too much pulls the bucket off the casters...no grit yet....threw sand water and rocks in just to prove to myself it would run, had the lid off to be sure things were being tumbled...was quite messy...will be researching grits now that it runs.http://www.youtube.c...h?v=EdA4pTiSuE8

Edited September 16, 2013 by DFINITLYDISTRUBD

[Turtle]

I have 2 new pulleys @ 2:1 reduction. Gonna be a bit while I work out the engineering particulars of attachment.

 

Meantime, went searching for someone with a working robot already. Found this guy who dances all around my concept. One of his machines is very similar to my layout with the wound-drive-drum, but it's not self-powered; he also has a propeller driven rig. Trey kewl

 

:coffee_n_pc:

 

 

http://www.youtube.com/watch?v=jqtRe1DeY60

 

 

http://www.youtube.com/watch?v=_oSjFhqfM5k

 

 

 

 

PS Here's some folks spreading seeds from a kite. Ya don't say?!

Home page:FirstKite

 

http://www.youtube.com/watch?v=htMJ0AlAgzc

Edited September 16, 2013 by Turtle

[DFINITLYDISTRUBD]

Wow, they're both way coolz! Love the simplicity of the first ant it give me an idear! Skip the belt, replace with string on a direct drive spool with a small spindle that pulls string off of a larger spool fitted to a smaller traction spool that drives the unit up the line...basically what the guy in the first vid has only not manual...and the line stays with the traveler.

[Turtle]

For the time being, stick a fork in me.

 

Rants & raves.

 

First and foremost my participation in this forum is as a writer and the scientific subjects I choose more-or-less secondary. I judge the success of my writing partly by the number of responses -thanks again responders- as well as by the numbers of readers. I judge the readership by watching the Views display, and since I picked up this thread in July after a 3 year hiatus it has been getting on average 100 views per day. Views were just over 12,000 in July and now stand at just over 18,000. Thank you so much Dear Tender Readers for your interest and support. :thumbs_up Since any-and-all views of threads here at the forum contribute to its overall success, I would like to thank the staff for their help and encouragement but alas they have given none so I return the favor. :thumbs_do

 

A number of factors contributed to my putting things on hold. A mainstay of my shopwork, a high-speed rotary tool, self-destructed while I was machining the last pulleys. School is back in session so my flying field is off-limits. Fall has fallen and here in the Pacific Northwest that means rain in one form or another. Lastly, I have been neglecting domestic duties and responsibilities and I need to get with the program.

 

I will keep an eye on the thread and make an effort to answer any questions or comments as they come. Kites are and will remain the oldest scientific venture into aeronautics on our humble little rock and I encourage you one-and-all to investigate it first hand and just go fly a kite. Thanks again for your readership.

[Turtle]

While my building/flying season is done, I left Double-D hanging where I got stumped on calculating the height of one of his flights. I/we asked for help, but none was forthcoming and we went on. I/we are asking for help again. My start is in post #243, including links to the NASA pages I was using for reference info and equations. Double--D's flight pics and info start in the post previous, and my efforts to a solution follow from there in a more-or-less tangled form. Besides working out his specific solution, I'd like to be able to do these calculations on my own for next season.

 

Little help?

[LaurieAG]

Besides working out his specific solution, I'd like to be able to do these calculations on my own for next season.

 

Hi Turtle,

 

Unfortunately the Lift equation and the Drag equations for the method shown both have coefficients that must be determined experimentally so a protractor and a fudge factor might be the way to go once you get the kite up again :) You could also record the different heights you get from different wind speeds (and line lengths) to get a rough and ready reckoner chart from wind speed.

 

http://www.grc.nasa.gov/WWW/K-12/airplane/lifteq.html

http://www.grc.nasa.gov/WWW/K-12/airplane/drageq.html

 

The Control Line Equations page has a link at the bottom that takes a different tack to calculating the height. The only problem is that the kite must be flying for the height to be calculated.

 

With a little more mathematical knowledge you can calculate the altitude at which the kite is flying. In this case, you can use trigonometry, because it is based on a line of sight which is a straight line.

 

http://www.grc.nasa.gov/WWW/K-12/airplane/kitehigh.html

[Turtle]

Hi Turtle,

 

Unfortunately the Lift equation and the Drag equations for the method shown both have coefficients that must be determined experimentally so a protractor and a fudge factor might be the way to go once you get the kite up again :) You could also record the different heights you get from different wind speeds (and line lengths) to get a rough and ready reckoner chart from wind speed.

 

http://www.grc.nasa....ane/lifteq.html

http://www.grc.nasa....ane/drageq.html

 

The Control Line Equations page has a link at the bottom that takes a different tack to calculating the height. The only problem is that the kite must be flying for the height to be calculated.

 

 

 

http://www.grc.nasa....e/kitehigh.html

 

Thanks for responding Laurie. Alas this was not my kite, but Dfndstrb's and doing another such flight is doubtful now that Fall is here. I got around the difficulties in lift & drag coefficients by using NASA's modeler which fortunately has his type of kite programmed. (material, size, weight). The program does not do the altitude calc for 5000yds of line out though, so I took their output needed and tried to do the calc myself. [using this page: Control Line Equations] I'll find the specific posts to refer you to, and in the mean time here is the modeler's info. Note you need Java to use the modeler, and I'll give that link again.

 

I have forgotten most my trig and don't know if I'm using my calculator correct or what range of answers I should expect in the intermediate calculations. Thanks again for helping. :)

 

modeler output:

g= line-weight = 0.001173333 oz/ft

W = weight = 1.773oz for plastic kite and 1/4" birch sticks

D = drag = 11.52oz

L = lift = 25.624oz

T= tension = 21.432oz

V = velocity = 9mph = 13.2 fps

A = kite area = 582"sq

p =

E = elevation above sea level = 800 feet

a = angle of attack = 20 deg = 0.34906585 radians

NASA Kite Modeler 1.5a beta

 

 

Edit: PS Can you check if I did the calcs right here in post #252? Thanks.

 

Alrighty thens. Let's see how well I screw up finding C1. :lol: Anybody???

 

g = line-weight = 1.1 lbs = 17.6oz

W = weight = 2 oz

D = drag = 11.52oz

L = lift = 25.624oz

 

C1 = sinh^-1[(L - g - W) / D ]

= sinh^-1[(25.624 - 17.6 - 2) / 11.52]

= sinh^-1[6.024/11.52]

= sinh^-1[0.5229166]

= 0.5016146144159724309580791536331

 

Oui? :thumbs_up No!? :thumbs_do

Edited September 26, 2013 by Turtle

[LaurieAG]

BTW, the Kite Modeler output is a bit strange as g should equal the total line weight and p should equal g/s i.e. weight/line length.

 

The complete equation for height y is a bit of a monster but could be simplified.

 

y = - D/p * COSH(SINH^-1((L-g-W)/D)) + D/p * COSH(p/D * [sINH^-1((L-W)/D)) - SINH^-1((L-g-W)/D))] + SINH^-1((L-g-W)/D)

 

where sinh^-1 indicates the inverse hyperbolic sine of the expression. That just means that the sinh(C1) is equal to (L - g - W) / D. There are tables of inverse hyperbolic sines, just like there are tables of inverse sines in trigonometry.

 

http://en.wikipedia.org/wiki/Inverse_hyperbolic_sine

 

In mathematics, the inverse hyperbolic functions provide a hyperbolic angle corresponding to a given value of a hyperbolic function. The size of the hyperbolic angle is equal to the area of the corresponding hyperbolic sector of the hyperbola x y = 1, while the area of a circular sector of the unit circle is one-half the corresponding circular angle. Some authors have called inverse hyperbolic functions "area functions" to realize the hyperbolic angles.

The abbreviations arcsinh, arccosh, etc., are commonly used, even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area.[1][2][3] Other authors prefer to use the notation argsinh, argcosh, argtanh, and so on. In computer science this is often shortened to asinh. The notation sinh−1(x), cosh−1(x), etc., is also used, despite the fact that care must be taken to avoid misinterpretations of the superscript −1 as a power as opposed to a shorthand for inverse (e.g., cosh−1(x) versus cosh(x)−1).

[Turtle]

BTW, the Kite Modeler output is a bit strange as g should equal the total line weight and p should equal g/s i.e. weight/line length.

 

The complete equation for height y is a bit of a monster but could be simplified.

 

y = - D/p * COSH(SINH^-1((L-g-W)/D)) + D/p * COSH(p/D * [sINH^-1((L-W)/D)) - SINH^-1((L-g-W)/D))] + SINH^-1((L-g-W)/D)

 

 

 

http://en.wikipedia....hyperbolic_sine

 

 

 

I just tried to run the modeler again to double check and it won't let me set the angle. :rant:

 

As to the rest on the trig, did I do the C1 calculation correct, or not? To get it I did the algebraic calculation, put the result in the calculator, pushed Inv key and then pushed sinh key. Is that correct? Where's a physics/math expert when we need one? :rolleyes:

[phillip1882]

import math
g= 0.001173333
W = 1.773 #oz for plastic kite and 1/4" birch sticks
D = 11.52
L = 25.624
p = 14.694

y = (- D/p * math.cosh(math.asinh((L-g-W)/D))
+ D/p * math.cosh(p/D * (math.asinh((L-W)/D) 
-math.asinh((L-g-W)/D))) + math.asinh((L-g-W)/D))
print y

outputs 0.456106520742

and i can confirm your result for C1

[Turtle]

import math
g= 0.001173333
W = 1.773 #oz for plastic kite and 1/4" birch sticks
D = 11.52
L = 25.624
p = 14.694

y = (- D/p * math.cosh(math.asinh((L-g-W)/D))
+ D/p * math.cosh(p/D * (math.asinh((L-W)/D) 
-math.asinh((L-g-W)/D))) + math.asinh((L-g-W)/D))
print y

outputs 0.456106520742

and i can confirm your result for C1

 

Thank you thank you! Now for using C1 to find X =XK I think.

 

...

Thennnnn...also if I got C1 correct I use it to calculate XK.

 

XK = (D / p) * [sinh^-1((L - W) / D) - C1]

= (11.52/ 0.0011733) * [sinh^-1((25.624 - 2) / 11.52) - 0.5016146144159724309580791536331]

= 9818.4607517259013040143185885963 * sinh^-1(2.0506944444444444444444444444444 - 0.5016146144159724309580791536331)

= 9818.4607517259013040143185885963 * sinh^-1(1.5490798300284720134863652908113)

= 9818.4607517259013040143185885963 *1.2216830548152906878629509132376

= 11995.047124752534496020791375179

 

:coffee_n_pc:

 

 

 

So that last result, i.e. 11995.047124752534496020791375179, I expected to be the final answer, that is it should be the altitude of the kite in feet. Did I do that right then?

Edited September 26, 2013 by Turtle