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Absolute maximum slinging length (Read 5349 times)
Apex-apoc
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Re: Absolute maximum slinging length
Reply #45 - Aug 11th, 2017 at 7:47pm
 
Parmenion wrote on Aug 11th, 2017 at 4:20am:
rpm is round per minute
(1 round/second=60rpm)


I said that already - and McNamaras number of rounds are easily to count while viewing the video and remember the rhythm of a "second-to-second-time" like a "tick-tack-tick-tack ...". I am musically too, you know?!

So I have seen and count that McNamara turns even more rounds than 3 "rps" - perhaps up to 3,2. Half a round more or less doesn't matter, because already 2,5 rps are far more than could be done by performing pirouettes. And what you can do in the last half round of release dosen't matter too and in no way, because the same can be done too in each other style of slinging.

Parmenion wrote on Aug 11th, 2017 at 4:20am:
let's not compare this to slinging.


Why not? You did it compare with discus throwers and other "athlets"!

Parmenion wrote on Aug 11th, 2017 at 4:20am:
as i said earlier you need to reach maximum speed only at the moment of release and not for a longer time


I said that already too.

Parmenion wrote on Aug 11th, 2017 at 4:20am:
if a way of throwing has more leverage and bigger acceleration than other styles then if it takes a little time to perform it, it doesn't matter.


Bigger acceleration comes not from longer slings or more leverage ... or perhaps I didn't have understood the point of this statement now.

Parmenion wrote on Aug 11th, 2017 at 4:20am:
don't get too consumed with theory,math and physics unless you can make a real research


Sure - this I already learned at school 35 years ago, but "Newtons math & physics" and Keplers third and second law of course is verified by facts of expirience already long.

So what?

Parmenion wrote on Aug 11th, 2017 at 4:20am:
... even then it will be useless if it isn't supported by experimental data.so the question should be how many of us can achieve, what jax and timpa achieve, but with shorter slings and/or more conventional styles 


My expirience in slinging includes more than 27 years of practice and minimum 100.000 throws (no joke!), although while performing allways the same style. And now sorry for "shocking" you with incredible news, but im able to throw a steelball (d = 26 mm; 72,3 g) MUCH FURTHER (!) than Yurek and did it already in 1998, 1999, 2000, 2001, ... several times.

In 2009 for the first time I threw a ball of tungsten* (type "heavy metall alloy" - dens. 18,5 g/cm³) with diameter 20 mm (= 77 g) about 716 m after throwing it about several times about 680 m. In 2011 I threw nearly the same distance (670 - 708 m) once more, respectivley several times.

Even with simple natural stones (100 -130 g) I achieve 400 m EASYLY and ABSOLUTLY (for) SURE and can land / place it nevertheless all within a rectangle of 15 x 30 meters. That means, the number of stones or balls that I threw could also found again for sure and easily - maximum 3 % of all my shots get lost.

Since 2008 I own 5 balls of tungsten* (red lacquered), threw them about 80 times for more then 600 meters and lost only one until today. Some of them i had to seek for more than 15 hours (in three or four days), but most of all I found again within 3 or 5 hours, because most of them I get placed right there where I mean it to place (means in that case whthin a rectangle of 25 x 50 m - dry mown lawn or a dry mown field - which I couldn't see while slinging. I had to throw it over a row of trees and bushes, what feels (felt) like a couple of "blind shots" - nevetheless I could place it relatively accurat in the intended "target-field").

So believe me: How much the range depends on rotation speed (or frequence) and elliptical "extension" (while release) do I know very well. My "theory" as you said is very good verified by "own expirience from self-made experiments".



______________________________________________________________________________

* "Tungsten" in german means "Wolfram" - in Reinstform hat das eine Dichte von 19,2 g/cm³, ist dann aber ein ziemlich sprödes Metall (bricht fast wie Glas). Gegen Schlag widerstandsfähiger ist jedoch die sog. "Schwermetall-legierung", die dann aber eine Dichte von nur noch 18,5 g/cm³ aufweist (3 - 5 % Eisen und Nickel sind dann mit untergemischt). Das ist dann aber immer noch mehr als doppelt so "schwer" als etwa Chromstahlkugeln (7,8 g/cm³) und auch noch deutlich schwerer als Blei (11,3 g/cm³).

Eine Kugel Wolfram mit D = 20 mm kommt heute in etwa auf 35 bis 40 Euro - das hängt stark davon ab, über welche Kanäle man sie bezieht, denn nicht jeder Betrieb, der mit Halbzeug aus Wolfram handelt, ist zugleich dazu in der Lage (oder willens), daraus auch Kugeln zu drehen, während "Sinterware" in genau dieser Legierung nur äußerst schwer aufzutreiben ist. Ein Rundstab mit D = 20 mm und L = 300 - 330 mm kostete um 2008 etwa 245 Euro (den aktuellsten Preis kenne ich gerade nicht).

Kugeln aus Wolfram-carbid findet man online schon wesentlich leichter (sind auch noch deutlich billiger), aber dessen Dichte beträgt dann auch nur noch 15,3 g/cm³.

Sofern man Wolfram nicht gerade stundenlang im Mund lutscht, ist dieses "Schwermetall" im übrigen auch vollkommen ungiftig und sehr korrosionsbeständig - rein optisch von Edelstahl (Nirosta) praktisch nicht zu unterscheiden - genauso "silbrig", gut polierbar und ebenso "hart" - nur eben schon beinahe dreimal so schwer.
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« Last Edit: Aug 15th, 2017 at 6:12am by Apex-apoc »  
 
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Re: Absolute maximum slinging length
Reply #46 - Aug 12th, 2017 at 3:04am
 
Sorry Apex.  Ich kenne sie nicht . Ich kenne auch keine Videos von ihnen. Ich schieße seit früher Kindheit. Ich lernte in dieser Zeit sehr viele slinger kennen. Nur ganz wenige waren wirkliche slinger. Die allermeisten waren einfach nur angeber. Sind sie ein angeber? Ich möchte das wirklich gerne wissen.  Wo immer sie in Deutschland auch wohnen. Sicherlich sind sie erreichbar für mich. Bei einem gemeinsamen Treffen werden wir sehen. Was 100000 Schüsse erbracht haben. Was halten sie davon? Facta locuuntur. Damit bin ich bisher immer gut gefahren.
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Re: Absolute maximum slinging length
Reply #47 - Aug 12th, 2017 at 4:12am
 
Apex-apoc wrote on Aug 11th, 2017 at 7:47pm:
Even with simple natural stones (100 -130 g) I achieve 400 m EASYLY and ABSOLUTLY (for) SURE and can land / place it nevertheless all within a rectangle of 15 x 30 meters.


if that's true, then thank you for insisting on your technique.
Please make a gift to the slinging community and share your knowledge.


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Re: Absolute maximum slinging length
Reply #48 - Aug 12th, 2017 at 4:14am
 
700m I'm calling your bluff on this one.
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Re: Absolute maximum slinging length
Reply #49 - Aug 12th, 2017 at 4:30am
 
This is the drawing I mentioned to load up here. It should imagine the physikal process of slinging long ranges.

The small black & white circle in the middle of the slinging means the slinging hand (right-handed) and its moving while slinging.

The grey figure besides the middle of the larger red circle means the slinger itself (sorry - his right arm isnt drawn).

"Extension" means the length of track of the slinging-circle while release (respectivly in the last half round, right befor release). From this length of extension results the most important amount of "speed up" while release.

If I said already, the "top speed" becomes twice if the relation between small and large axis (of ellipse) is like 1:2 (or 2:1 = 2). If this relation (quotient) is higher - perhaps 1:3 or 1:2,5 so the "speed up" is higher too.

Because a natural man isn't able to rotate a 1,2 m long sling much higher than 3 rps (= 180 rpm) what is 15 - 20 m/s only, he always have to try to perform the extension as wide as possible.



Of course I never could give them a technical "measuring", but only this "drawn physic" can explain how really measured top speeds (35 - 60 m/s) can come up of those slow rotations.

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Re: Absolute maximum slinging length
Reply #50 - Aug 12th, 2017 at 5:24am
 
On base of this drawn process and "principle" I could build up the following three tables:

For better compareability each table shows only one fix amount of rotation speed (rps). Therefore this condition is written in RED types. If all other conditions are the same you can compare the top speeds from one rotation speed with top speeds of an other rotation speed.

And surely: Top speeds far above 65 m/s are very difficult to perform, because to rotate a 1,2 m long sling with more then 2,5 rps (and widest extension) is very difficult too.
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Re: Absolute maximum slinging length
Reply #51 - Aug 12th, 2017 at 6:22am
 
Jaegoor wrote on Aug 12th, 2017 at 3:04am:
Die allermeisten waren einfach nur angeber.


Tja gut - das tut mir natürlich leid für dich. Aber glaube mir - ich selbst habe noch überhaupt keinen "Slinger" kennengelernt / getroffen, und habe auch nicht vor, mir zu aller notwendigen "action" auch noch das Drehen von mehr oder minder tauglichen Beweis-Videos aufzuhalsen.

Für ein Treffen im Raum Nürnberg darfst du mich natürlich gerne mal besuchen - nach "Augenzeugen", die  für ein solches Unterfangen auch die Zeit und Nerven haben, suche ich ohnehin noch. Nur vor dem kommenden Frühjahr (2018) käme mir das sicherlich noch äußerst ungelegen, da ich derzeit auch noch einen ganzen Arsch voll anderer "Baustellen" aufgerissen habe.

Zum Werfen und "Bullets-suchen" kam ich mitllerweile schon seit 18 Monaten nicht mehr, und ich fürchte, das selbe "Problem" wird sich auch noch eine paar weitere Monate hinziehen. Aber keine Sorge - ich laufe nicht weg, und werde mich auch oder vor allem hier noch des öfteren melden.

Außerdem habe ich so ein bisschen den Eindruck, dass du lediglich die Vorteile höherer Materialdichten unterschätzt, denn 716 m mit Wolframkugeln ist gerade mal soviel wie die Hälfte (358 m) mit einem gleichschweren Kieselstein. Der Luftwiderstand - übrigens die einzige Kraft, die der schlußendlich erzielten Schußkraft entgegenwirkt - liegt bei gleichschweren Wolframkugeln um ziemlich genau 50% niedriger als bei natürlichen Steinen (2,4 - 2,7 g/cm³). Gegenüber Würfen mit banalen "Findlingen" verdoppelt sich die Wurfweite also schon allein dadurch beinahe "ganz automatisch".

Ein "Weltwunder" sind diese 700 m also noch immer keines, und ich werfe eigentlich nur deshalb nicht noch weiter, weil derart kostspielige Geschosse einen so armen "Hirten" wie mich beim Beschleunigen ganz schön zu "hemmen" vermögen. Ich fürchte halt, wenn ich so bis an die 750 m weit werfen würde, die teuren Kugeln ganz sicher los zu sein. Die mitunter viele Stunden andauernde Suche danach ist schon so nervenaufreibend genug (es ist ja eben nicht nur der materielle Preis der Kugeln, sondern auch die Action, dieselben neu aufzutreiben).


Was hingegen die erwähnten 100.000 Würfe betrifft, wären genau nur diese in einem Zeitraum von 27 Jahren ebenfalls nicht mehr als ein Klacks, denn 250 Würfe werfe ich schon allein an einem einzigen Tag. In "Hoch-zeiten" (Frühjahr bis Spätherbst) bin oder war (!) ich aber regelmäßig bis zu zwei mal pro Woche auf der Piste.

Wären es tatsählich immer zwei Trainigs-gänge pro Woche gewesen, hätten bis dato eigentlich schon über 400.000 Würfe absolviert werden müssen - die "Winterpausen" (und / oder sonstige Ausfälle) bereits rausgerechnet:

27 Jahre x 35 Wochen pro Jahr = 945 Wochen

Multipliziert mit 500 Würfen pro Woche (weil zwei Trainingsgänge pro Woche) = 472.500 Würfe.

Lass nun mal zum Füße-baumeln-lassen nur die Hälfte aller Würfe weg, und zieh meinetwegen auch noch 5 oder 7 Jahre des "totalen Faulenzens" ab. Was ist dann nichtsdestotrotz noch übrig?
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Re: Absolute maximum slinging length
Reply #52 - Aug 12th, 2017 at 9:30am
 
Really interesting stuff Apex. I am not a distance slinger so I have no idea if 700 meters is possible even with tungsten but you've definitely done your homework. Thanks for posting all this.
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Re: Absolute maximum slinging length
Reply #53 - Aug 12th, 2017 at 12:18pm
 
@Apex-apoc
since i can't see what the rest of us are doing differently when slinging for range or accuracy what is your tip to achieve distances like you did?
(consistently)

is your form/technique depicted well by any video on youtube?

what sling(length, materials,etc) did you use for 400m stone throw?
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Re: Absolute maximum slinging length
Reply #54 - Aug 12th, 2017 at 12:30pm
 
also what is "Factor by Extention 2 to 3m" and how is it calculated?
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Re: Absolute maximum slinging length
Reply #55 - Aug 12th, 2017 at 3:04pm
 
Apex-apoc wrote on Aug 11th, 2017 at 7:47pm:
Parmenion wrote on Aug 11th, 2017 at 4:20am:
rpm is round per minute
(1 round/second=60rpm)


I said that already - and McNamaras number of rounds are easily to count while viewing the video and remember the rhythm of a "second-to-second-time" like a "tick-tack-tick-tack ...". I am musically too, you know?!

So I have seen and count that McNamara turns even more rounds than 3 "rps" - perhaps up to 3,2. Half a round more or less doesn't matter, because already 2,5 rps are far more than could be done by performing pirouettes. And what you can do in the last half round of release dosen't matter too and in no way, because the same can be done too in each other style of slinging.

Parmenion wrote on Aug 11th, 2017 at 4:20am:
let's not compare this to slinging.


Why not? You did it compare with discus throwers and other "athlets"!

Parmenion wrote on Aug 11th, 2017 at 4:20am:
as i said earlier you need to reach maximum speed only at the moment of release and not for a longer time


I said that already too.

Parmenion wrote on Aug 11th, 2017 at 4:20am:
if a way of throwing has more leverage and bigger acceleration than other styles then if it takes a little time to perform it, it doesn't matter.


Bigger acceleration comes not from longer slings or more leverage ... or perhaps I didn't have understood the point of this statement now.

Parmenion wrote on Aug 11th, 2017 at 4:20am:
don't get too consumed with theory,math and physics unless you can make a real research


Sure - this I already learned at school 35 years ago, but "Newtons math & physics" and Keplers third and second law of course is verified by facts of expirience already long.

So what?

Parmenion wrote on Aug 11th, 2017 at 4:20am:
... even then it will be useless if it isn't supported by experimental data.so the question should be how many of us can achieve, what jax and timpa achieve, but with shorter slings and/or more conventional styles 


My expirience in slinging includes more than 27 years of practice and minimum 100.000 throws (no joke!), although while performing allways the same style. And now sorry for "shocking" you with incredible news, but im able to throw a steelball (d = 26 mm; 72,3 g) MUCH FURTHER (!) than Yurek and did it already in 1998, 1999, 2000, 2001, ... several times.

In 2009 for the first time I threw a ball of tungsten* (type "heavy metall alloy" - dens. 18,5 g/cm³) with diameter 20 mm (= 77 g) about 716 m after throwing it about several times about 680 m. In 2011 I threw nearly the same distance (670 - 708 m) once more, respectivley several times.

Even with simple natural stones (100 -130 g) I achieve 400 m EASYLY and ABSOLUTLY (for) SURE and can land / place it nevertheless all within a rectangle of 15 x 30 meters. That means, the number of stones or balls that I threw could also found again for sure and easily - maximum 3 % of all my shots get lost.

Since 2008 I own 5 balls of tungsten* (red lacquered), threw them about 80 times for more then 600 meters and lost only one until today. Some of them i had to seek for more than 15 hours (in three or four days), but most of all I found again within 3 or 5 hours, because most of them I get placed right there where I mean it to place (means in that case whthin a rectangle of 25 x 50 m - dry mown lawn or a dry mown field - which I couldn't see while slinging. I had to throw it over a row of trees and bushes, what feels (felt) like a couple of "blind shots" - nevetheless I could place it relatively accurat in the intended "target-field").

So believe me: How much the range depends on rotation speed (or frequence) and elliptical "extension" (while release) do I know very well. My "theory" as you said is very good verified by "own expirience from self-made experiments".



______________________________________________________________________________

* "Tungsten" in german means "Wolfram" - in Reinstform hat das eine Dichte von 19,2 g/cm³, ist dann aber ein ziemlich sprödes Metall (bricht fast wie Glas). Gegen Schlag widerstandsfähiger ist jedoch die sog. "Schwermetall-legierung", die dann aber eine Dichte von nur noch 18,5 g/cm³ aufweist (3 - 5 % Eisen und Nickel sind dann mit untergemischt). Das ist dann aber immer noch mehr als doppelt so "schwer" als etwa Chromstahlkugeln (7,8 g/cm³) und auch noch deutlich schwerer als Blei (11,3 g/cm³).

Eine Kugel Wolfram mit D = 20 mm kommt heute in etwa auf 25 bis 35 Euro - das hängt stark davon ab, über welche Kanäle man sie bezieht, denn nicht jeder Betrieb, der mit Halbzeug aus Wolfram handelt, ist zugleich dazu in der Lage (oder willens), daraus auch Kugeln zu drehen, während "Sinterware" in genau dieser Legierung nur äußerst schwer aufzutreiben ist. Ein Rundstab mit D = 20 mm und L = 300 - 330 mm kostete um 2008 etwa 210 Euro (den aktuellsten Preis kenne ich gerade nicht).

Kugeln aus Wolfram-carbid findet man online schon wesentlich leichter (sind auch noch deutlich billiger), aber dessen Dichte beträgt dann auch nur noch 15,3 g/cm³.

Sofern man Wolfram nicht gerade stundenlang im Mund lutscht, ist dieses "Schwermetall" im übrigen auch vollkommen ungiftig und sehr korrosionsbeständig - rein optisch von Edelstahl (Nirosta) praktisch nicht zu unterscheiden - genauso "silbrig", gut polierbar und ebenso "hart" - nur eben schon beinahe dreimal so schwer.

What length sling and what style are you using to achieve said distance?
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Re: Absolute maximum slinging length
Reply #56 - Aug 12th, 2017 at 4:12pm
 
If David Engvall world record is 477 meters, then your over 700 meters is really good.
I congratulate you!
With you would be a pleasure to throw a competition of length. Even though I would lose a lot  Sad.
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Re: Absolute maximum slinging length
Reply #57 - Aug 13th, 2017 at 11:36am
 
Parmenion wrote on Aug 12th, 2017 at 12:30pm:
also what is "Factor by Extention 2 to 3m" and how is it calculated?


When the extension is 2 m, than the FACTOR (F) is an other as if the extension is 3 m. In the branches are written the factors only - one for extension = 2m and one for extension = 3m. With this factors are to multiplicate the "wind-up-speeds" (= what is written under "2,0 Rd./s" or "2,5 Rd./s" or "3,0 Rd./s"). For example (as in the drawing):

3 x 19 m/s = 57 m/s

"3" is the "factor" (= relation between small and large axis of ellipse)

"57 m/s" is what is called "Top speed" in the tables.

I wrote this tables for comparison! You can see how the "Top Speeds" change when the length of sling, the rotation speed and / or the "extension" changes.

What "Extension" means is shown in the drawing. That is not the same as the "factor", because the shape of ellipse depends on "Circumference" (respektively "diameter") and "Extension" at the same time, but both of it can change independently of each-other.

All this "dependings" and "relations" I can't explain better - not in english  Embarrassed - because partially they are "co-relations" and damned difficult to bring into a few words (of foreign language). Therefore I made the drawing and right this you have to "study" and understand  Cheesy.

In the tables (branches) I have written two different "Top speeds" (the lowest & the highest), because in front of it I have written the lowest and highest factor (which depends on "extension").

Important to know: In the branch "Factor by Extension ..." I wrote the factors only - no "extensions" !

The "extensions" are allways "2m" and / or "3m" (respectively a range in between).



But if you have understood the drawn relations / dependings completely, then let me know what is a better expression as: "Factor by Extension 2 to 3 m". But this expression must be as short as the small place in a branch.

Should I have written there "Factor" only?    Undecided
How should I call that what is now called "extension"? Do you know an better expression?
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Re: Absolute maximum slinging length
Reply #58 - Aug 13th, 2017 at 1:56pm
 
Blowgunman123 wrote on Aug 12th, 2017 at 3:04pm:
What length sling and what style are you using to achieve said distance?


The small balls of tungsten (and steel) I threw with slings which were especially made for very small an middle weighted ammo. Its "pouch" was a small piece of Y shaped leather (retention cord was linked a the tripe or "foot" of this Y). A crotched release cord of very thin (fine braided) yarn of polyester (2 x 1,2 mm) was linked at each "arm" of that Y.

This type of sling hasn't any "release resistance", because for release the release cord must not become accelerated or full opened / evolved. In the short moment of release the ball slips through the splitted / crotched release cord. So this cord in relaese moves only along 3 or 4 cm what is practically done within "0,01 seconds".

Other types of slings waste energy for accelerate the release cord, for "sonic boom" and / or to rotate the stone. Right this energy should be safed for better ranges.

Engvall in principle does the same, because his "hook" (for his arrow) had only to tilt by a very small move too. The same advantage easily can be given to a rock sling by crotching or splitting its release cord (and the release side of pouch).

My retention cord at this sling was made of a doubled nylon line (what is a little bit like "wire" and taken for mowing grass - D = 1,4 mm but doubled). This nylon line comes relatively "stiff", but it helps to provide the sling of embroiling the cords while winding up such a small pouch.

Length of sling was 1,32 m (1,26 or 1,27 m in other cases) and my throwing style always is helicopter (more or less over head or more "side-arm").


But another conditions are important too:

For good flowing practise you need two fileds of lawn in a distance of 400 - 500 m and always to wait until they are mown ... and always to wait on dry weather, because the practise is not to absolve with balls of tungsten (their number is to low because their price to high), but with some more heavy (and cheaper) steelballs, which would sunk to deep in / under the ground while impact if this ground still is too humid /wettish / moist.

And you need to increase the range in a lot of small steps only, because you get not the right feed-back if the thrown stones or balls couldnt be found anymore. Ammo always must be found again and for this they always must be thrown relativley accurate.

At last you always have to know and control the "exact" range BEFORE (!) the ball is thrown. The ball (always) have to land right there where you ment it to land. Othewise you can't found it or measure its distance anymore.

To throw a lot of stones straight into the woods or bushes trains only your power, skills and technique. But "to achieve distance" is a little bit different, because range or distance at last is a very clear "number" of yards or meter(s). Therefore it doesnt help to throw the balls or stones only for "somewhere". Somewhere is nowhere! Much more you have to know the measure of distance and right for this the exact position of all thrown balls ... and for this to find it all again.

But if the seeking for "hidden" balls tooks to much time, than the practise never comes in a good flow. Therefore you should increase your range only simultanously to your accuracy.

I mean, at first you have to know the place and distance where the balls SHOULD BE placed. For example in a rectangle of 10 x 20 m (the longer side in direction of throwing) which lays in a distance of exactly 300 m. Then you throw always exact that distance only until you are able to place all balls "absolutely correkt" (means: "within the rectangle"). No ball must be lost.

First when you are able to do this for sure, then you can try the next distance (350 m) ... and so on. Never increase this distance if you still loose steelballs. All balls have to be placed right or throwen not at all.



And importend as well: Don't make much more than a hundred "shots" within one month when throwig so strong, and never throw so strong if your "rest" (pause) was longer than 14 days. To absolve a lot of "easy" shots in between (minimum all week) is very importent to prevent "Epicondylitis ulnaris humeri (a type of "tennis ellbow). If necessary throw a couple of stones with pure hands only, but always keep on throwing.

If you get a tennis ellbow for longer than a month than the practice is absolutly "over" for more than a year, because its healing is damned slowly. On the other side it is absolutly normal to have always a little bit "tennis ellbow", but that must be a kind of tennis ellbow which subsides within three days always.

So always have a very attentively eye on this little pains in your ellbow.
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« Last Edit: Aug 14th, 2017 at 1:58am by Apex-apoc »  
 
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Re: Absolute maximum slinging length
Reply #59 - Aug 13th, 2017 at 4:17pm
 
Apex-apoc wrote on Aug 13th, 2017 at 11:36am:
All this "dependings" and "relations" I can't explain better - not in english  Embarrassed - because partially they are "co-relations" and damned difficult to bring into a few words (of foreign language). Therefore I made the drawing and right this you have to "study" and understand  Cheesy.


use your language i'll try to understand  through google translate. Also what principles of physics are used to explain velocity multiplication in your theory?
The trajectory(in the image you uploaded) doesn't explain where/how/how much force  is applied by the hand during the throw . So it's of little use without the theory.


For 700m in vacuum you need 82m/s at 45 degrees
for 680m 81m/s.
Since you throw very dense projectiles we can do the hypothesis that it is like you throw it in a vacuum.

82m/s in your tables above is achieved with  at least 1,4m sling, 3 rounds per second, extension 3m.
I have to note 3m extension is as much difficult as doing more than 3 rounds per second with long sling.

About your model I have understood these:
1. the longer the sling the more difficult it is to turn it around fast
2. the higher the rounds/sec the smaller the extension  (biomechanical limits... the higher the velocity of an object the more difficult it gets to apply force to it)
you have to do 3m extension and 3 rounds/sec , two difficult things at a time …

so either
a. your model is wrong, and there's more to it...
b. you are lying
c. you do more than 3m extension
ci. you do more than 3 rounds/sec
d. “instant” release slings have a tremendous advantage over simple slings
e.you are HULK
f. it is pretty easy to achieve the mentioned distance/speed but not  many of us can spot the landing, and do measurements ,so didn't notice this miracle . You are accurate and consistent ,you noticed....

which of the above is true?



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