Как сбит MH17: гражданское расследование

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Математическое моделирование траектории падения центроплана #2

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На картинке в клеточку запредельные скорости, этого неможет быть.

Если сбросить балванку весом 250 тонн с высоты 10000 метров  без крылошек под углом к горизонту 70 градусов. то максимальная скорость у земли будет 460 м/сек. http://forumupload.ru/uploads/0016/23/c6/51/804874.png
http://forumupload.ru/uploads/0016/23/c6/51/804874.png

Предыдущая часть темы: Математическое моделирование траектории падения центроплана

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oper написал(а):

Возьми поэкспериментируй, возьми массу 5 тонн, свой мидель и найди Сх, все эти параметры введи в свою программу и получи результат и его запомни.


Возьми линейку из свой жопы и вставь её себе обратно, но только не вдоль заднего прохода, а поперёк.  :D

Как тебе такой эксперимент Дебил конченный.  :D

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oper  требует от DSB запилить ему дверной проём передней правой двери (1R),  вырванный внутренним взрывом.   :D
     
         

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Тетя, а ты значит программой Усчена моделируешь?

Тетя написала:
У тебя карта крушения с 5-ти киллометровым радиусом перед глазами. ВСЕ ОБЛОМКИ перед тобой.
Я хочу что бы сначала ТЫ СКАЗАЛ - За пределами 5-ти киллометровой зоны ЕСТЬ МАССИВНЫЕ ФРАГМЕНТЫ MH17  ???

Как ты СЧИТАЕШЬ, есть они там или нет ?

http://forumupload.ru/uploads/0016/23/c6/51/936576.png

Ты же про это мог узнать только с помощью программы Усчена. Там где у тебя твой кокпит начал падать (проекция на землю) туда они и прилетят. Ну как, что-то нашел? :D

Отредактировано oper (2021-10-23 22:20:07)

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oper написал(а):

Risto_Koivula,_Tampere написал(а):

    v´(t)/v²(t) = constant = c/m, when no extra forces influence.  c =ρAC/2. The dimension c is that of ρA = kg/m³xm² = kg/m. The mass dimension comes here from the mass of the gas or fluid.

Вот сейчас стало понятно, но вначале исправляем вашу оплошность, это не верно  mdv/dt = mv´(t) = - ρCv²(t)/2 = - cv²(t) должно быть так mdv/dt = mv´(t) = - ρCАv²(t)/2 = - cv²(t).

Тогда c =ρAC/2 и равенство - c/m = constant = -2g/vo² верно и у обоих выражений размерность m⁻¹.

You are right that there were both typing and copy-paste errors in the text. I have corected what have found.

Хорошо тогда такой вопрос, вы взяли перегрузку 2.0 g и 2.6 g из отчета DSB:

    The fixed Emergency Locator Transmitter was automatically activated by a
    longitudinal deceleration of between at least 2.0 g and 2.6 g.

    Its signal was first
    detected between 13.20:35 and 13.20:36 (15.20:35 - 15.20:36 CET). System logic
    means that the ELT was activated between about 13.20:05 and 13.20:06
    (15.20:05 - 15.20:06 CET).

но это технические характеристики ELT,

These are "usual drag decelerations" , not overloadings from headwind etc. like in this link.:

http://mh17.forum.camp/viewtopic.php?id=39#p1945

https://scontent.fqlf1-1.fna.fbcdn.net/ … e=6199416C--

" Перегрузки >2g могут быть и при штормовой турбулентности. Вектор хаотичен, так почему бы ему не быть продольным?

И еще вопрос, выясняли тип замков, удерживающих переднюю стойку? Если завязаны на электричество, могли отключится когда пропало питание и тогда вполне возможный вариант

    Альберт Валиев написал(а):

    по какой-то причине передняя стойка сходит с замков и вываливается, по причине чего начинает отламываться кокпит - в кокпите в этот момент большая перегрузка. В таком случае момент срабатывания ELT = момент отрыва кокпита. "

These are also easily calculated from Newton´s drag law. Drag overloadinds from headwind and usual loading in normal use can be same magnitude due to the scond power on speed in equqtions.

во время взрыва ракеты неизвестно сколько было g, могло быть и 3g, и 4g, и 5g.

The accelerations during explosion are not here useful. Explosion as such did not wake up ELT of longitudinal deceleration (acceleration), it was activated from drag deleleration when when cockpit was disconnected from fuselage. Despite deceleration 2 ... 2.6 g, speeds where measured. We can compare them with theoretical functions:

Time, s    Measured  v, m/s  v´(0) = 2.0 g  v(t) = 3125 m/(t + 12.5s)  v´(0) = 2.6 g  v(t) =  v(t) = 2403m /(t + 9.6s)   v´(0) = 2.3g v(t) = 2717,4m/(t + 10.87s)

   0                 250                                    250                                                           250                                                                                          250,0

   1                 230                                    231                                                           226                                                                                          229,8

   2                 211                                    215                                                           207                                                                                           211,1

Вам же неизвестно сколько было g вы просто взяли паспортные данные ELT.

Отредактировано oper (Вчера 18:25:16)

According to ELT data the acceleration was -2.3 g (and deceleration 2.3 g.

Falling speed for -2.3 g   

w(t) = p⁻¹tanh(gpt) = 164,8m tanh(t/16.48s),

and falling "height" (from the top):

h(t) = k⁻¹ln cosh(gpt) = 2717,4m ln cosh(t/16.4s)   

Falling time T: => 2717,4m (T/16.4s - ln2) = 10000m   => T = 16.4s (10000/2717,4 + 0.693) = 71.7 s.

Horizontal translation during the fall:

D = 2717,4m ln(1 + T/10,87s) = 5510m.

This is best what can be reached based on informations of DSB / Bellingcat "report".  But instead of this, there is bullshit in it.

Отредактировано Risto_Koivula,_Tampere (2021-10-29 18:54:00)

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Alex написал(а):

Risto_Koivula,_Tampere написал(а):

    That "DSB strike point" is no radar poit at all, but it is a calcuated by the radar ekstrapolation for following signal of the airplane. Almaz-Antey say the former point is the pointi where the airlane was oserved to be "in disintegraion". The distance of radar points is 10 sconds which means, taht on earth it was 2.5 km, because the plane was flying 250m/s. the real strike point is two poits, 5 m backwards from "DSB strike point".

http://forumupload.ru/uploads/0016/23/c6/61/230097.jpg

The cockpit "flew" to the right from the fuselage, not to the left. The direction of the route is somewhat wrong.

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Risto_Koivula,_Tampere написал(а):

The cockpit "flew" to the right from the fuselage, not to the left. The direction of the route is somewhat wrong.

I am trying to understand exactly where (according to your ideas) 2 events occurred.
1. Rocket hit the plane.
2. Cockpit department.

Can you specify the coordinates for Google Earth Pro so I can move them to the map?

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Risto_Koivula,_Tampere написал(а):

Exlosion as such did not wake up ELT longitudinal deceleration (acceleration), it was activated from drag deleleration when when cockpit was disconnected from fuselage. Despite deceleration 2 ... 2.6 g, speedswhere measured. We can compare them with theoretical functions:

Если это действительно так, то ваш метод гениален!!!

Фактически по  c/m можно приблизительно определить баллистический коэффициент BC=m/Cx*A и используя эту величину, решив уравнения численным методом, построить траекторию падения кокпита и тем самым найти координаты отрыва кокпита от МН17, а также начальное направление (курс) падения кокпита.

Как вы определили или как вы вывели,  что c/m =2g/vo²?

Давайте вашим методом при перегрузки 2g найдем баллистический коэффициент.

c/m=2g/vo²=20/250²=0,00032
ρAC/(2m)=0,00032

На высоте 10000 метров ρ=0,4 (примерно) тогда AC/(2m)=0,00032/0,4=0,0008
AC/m=0,0008*2=0,0016
BC=1/(AC/m)=m/(AC)=1/0,0016=625
Итого при 2g получаем баллистический коэффициент ВС=625

При перегрузки 2,6g
c/m=0,000416
AC/m=0,00208
BC=1/0,00208=480,77
BC=480,77

Правильно?

Отредактировано oper (2021-10-24 17:05:15)

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oper написал(а):

Если это действительно так, то ваш метод гениален!!!

Фактически по  c/m можно приблизительно определить баллистический коэффициент BC=m/Cx*A и используя эту величину, решив уравнения численным методом, построить траекторию падения кокпита и тем самым найти координаты отрыва кокпита от МН17, а также начальное направление (курс) падения кокпита.

BC/ρ = m/c = k⁻¹ 

Как вы определили или как вы вывели,  что c/m =2g/vo²?

I have explained already two times. But I do it still once and korrect the values.

Математическое моделирование траектории падения центроплана #2

" According to Newton´s (first and) second law and his drag law, for a body moving in a dragging fluid

v´(t)/v²(t) = constant = v´(0)/v²(0) = 2.3g/vo² = c/m  = 23m⁻¹/250² = (2717,4m)⁻¹, when no extra forces influence. 

c =ρAC/2. The dimension c is that of ρA = kg/m³xm² = kg/m. The mass dimension comes here from the mass of the gas or fluid. 

(The density of air ρ depeds on height: 1n 10000m it is about 0.5kg/m³, on sea level it is about 1.0kg/m³. We use now the former. If we use the latter for the horizontal fall, it means longer fall time T and longer fall flight D.)

When we know a (trajectory) constant in one point of a trajectory, we know it in every point.

In this case we know at the disconnection point t = 0 both the speed v(0) = 250 m/s and the acceleration v´(0) = -2.3g = -23m/s²

A verification:

v´(t)/v²(t) = - c/m = constant = -2g/vo² = -20ms⁻²/(250ms⁻¹)² = - (2717,4m)⁻¹ = -k "

Давайте вашим методом при перегрузки 2g найдем баллистический коэффициент.

c/m=2g/vo²=20/250²=0,00032

ρAC/(2m)=0,00032

На высоте 10000 метров ρ=0,4 (примерно) тогда AC/(2m)=0,00032/0,4=0,0008
AC/m=0,0008*2=0,0016
BC=1/(AC/m)=m/(AC)=1/0,0016=625
Итого при 2g получаем баллистический коэффициент ВС=625

In this case BC = ρk⁻¹ = 0.4kg/m³ x 2717,4m = 1087 kg/m²

При перегрузки 2,6g
c/m=0,000416
AC/m=0,00208
BC=1/0,00208=480,77
BC=480,77

Правильно?

No. In my presentation ρ is included in c and c/m.

I have understood "overloadings" (перегрузки) mean extra drag forces from hard headwinters in high speeds like here. They can be same magnitude as as normal drag forces in usual use, for instance -2g, due to the second power on speed according to air in drag equation.

Horizontal overloadings in this sense can be calcutated from formula

mv´(t) + c (v(t) - vo)² + F, where F is external force from motors etc. Headwinter (vstretsnij veter) is sign minus. V(t )is  according to inertial frame of reference (not according to the air).

Отредактировано Risto_Koivula,_Tampere (2021-10-26 03:14:45)

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Risto_Koivula,_Tampere написал(а):

v´(t)/v²(t) = constant = v´(0)/v²(0) = 2.3g/vo²

Спасибо! Теперь понято!

Risto_Koivula,_Tampere написал(а):

No. In my presentation ρ is included in c and c/m.

c/m=2g/vo²=k=20/250²=0,00032
BC = ρ/k=0.5/0,00032=1562,5.
Так?

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Alex написал(а):

Risto_Koivula,_Tampere написал(а):

    The cockpit "flew" to the right from the fuselage, not to the left. The direction of the route is somewhat wrong.

I am trying to understand exactly where (according to your ideas) 2 events occurred.
1. Rocket hit the plane.
2. Cockpit department.

Can you specify the coordinates for Google Earth Pro so I can move them to the map?

Alex написал(а):

https://scontent.fqlf1-1.fna.fbcdn.net/v/t1.6435-9/247574802_6295542997187030_6261580368922463063_n.jpg?_nc_cat=108&ccb=1-5&_nc_sid=730e14&_nc_ohc=xktwITsXFxsAX894D6i&_nc_ht=scontent.fqlf1-1.fna&oh=554dcb15f81512bfcc7636007cdb4c4d&oe=61981F62

That "DSB strike point" is no radar poit at all, but it is a calcuated by the radar ekstrapolation for following signal of the airplane. Almaz-Antey say the former point is the pointi where the airlane was oserved to be "in disintegraion". The distance of radar points is 10 sconds which means, taht on earth it was 2.5 km, because the plane was flying 250m/s. the real strike point is two poits, 5 m backwards from "DSB strike point".

https://www.youtube.com/watch?v=1tKmoygQudI

In Almaz´s analysis of radar and other points, this flap refers to the first pinpointed radar point, where the disintegration of the plane "has started".

http://forumupload.ru/uploads/0016/23/c6/63/t456734.jpg

This is 10 seconds later than the flight data acquisition unit stopped working due to energy supply breaking.

This means that the strike point has been about the FORMER radar point, which is about 12 km right west from Hrabove map denotation (the administrative center).

But Bellingcat, DSB and JIT say that the strike had been the FOLLOWING from the first pinpointed "radar point", whici is NO RADAR POINT AT ALL as there is pinted too, but the radar point were it was remarked that the plane has disappeared from radar.

Bellingcat, DSB and JIT try to "move the strike point" 5 km closer to Snizhne where BELLINGCAT "analysed from warmth
tracks in US satellite observation" (which no "third part hss not seen) to have been "the shooting place" of a BUK missile.

http://forumupload.ru/uploads/0016/23/c6/63/t269407.png

http://forumupload.ru/uploads/0016/23/c6/63/t550350.png

Отредактировано Risto_Koivula,_Tampere (2021-12-21 06:57:09)

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oper написал(а):

Risto_Koivula,_Tampere написал(а):

    v´(t)/v²(t) = constant = v´(0)/v²(0) = 2.3g/vo²

Спасибо! Теперь понято!
Risto_Koivula,_Tampere написал(а):

    No. In my presentation ρ is included in c and c/m.

c/m=2g/vo²=k=20/250²=0,00032
BC = ρ/k=0.5/0,00032=1562,5.

Так?

Yes. But we need about 5 significant numbers, because lose to each other big numbers are substracted in calculations.

In horizontal translation is the high-air value ρ = 0,5 impotatnt, because high speeds are in the beginning.

But for the fall the highest speeds are at the end, near eath surface. That is why we can try what a double c/m =k = 4.6g/vo² = a²g/vo² = (1359m)⁻¹ means at fall.

p = sqrt⁻¹(k/g) = a/vo =(116,6m/s)⁻¹

w(t) = p⁻¹ tanh(gpt) = 116.6ms⁻¹tanh (t /11.66s) , ja

h(t) = int (w(t))dt = g⁻¹p⁻²ln cosh(gpt) = k⁻¹ln cosh(gpt) = 1359m ln cosh(t/11.66s)

For "larger" x: ln cosh x > x - sqrt2 = x - 0,693.

1359m (t/11.66s - 0.693) = 10000m => t /11.66s = 7.35 + 0.693= 3.893 => 8.05, from which the fall time

T = 94s.

from which using now horizontally upper air deceleration:

D = 2717.4m(ln(1 + 94s/10.87s) =6159m

Отредактировано Risto_Koivula,_Tampere (2021-10-29 22:08:32)

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Here are measures, but also lies.

http://graphics.wsj.com/mh17-crash-map/

Map of a Tragedy: How MH17 Came Apart Over Ukraine

The Wall Street Journal cataloged and mapped some of the debris of Malaysia Airlines Flight 17, which fell across three Ukrainian villages after the Boeing 777 was brought down on July 17, allegedly by a surface-to-air missile.

Read the related story »

Reporting by Jon Ostrower in Washington, D.C., Alexander Kolyandr, Margaret Coker and Paul Sonne in Ukraine

Interactive by Stuart A. Thompson / The Wall Street Journal

http://graphics.wsj.com/mh17-crash-map/img/villagemap.jpg
Renée Rigdon/The Wall Street Journal

First and worst lie: It was not USA that said the missile was shot from Snizne, but it was BELLINGCAT.

What USA recognized was "old fashioned Soviet BUK system ("Gadfly, Gainful") tracking (radar) radioation. They did and do not where the telar was with less than 100 km accuracy. Aand Bellingcat does not know anything

https://www.ibtimes.com/biden-mh17-blow … rm-1632062

" U.S. officials have said ouU.S. officials have said outright that Malaysia Airlines Flight MH17 was shot down with a surface-to-air missile, but are unsure of the missile’s origin.

Vice President Joe Biden said during an appearance in Detroit that it appears MH17 was “shot down, not an accident. Blown out of the sky,” and other American officials confirmed that a missile was fired at the Boeing 777, but are unsure who shot the missile.

A senior U.S. official told CNN that just before MH17 crashed, a radar system noted a surface-to-air missile system tracking an aircraft, and a second system saw a heat signature. By analyzing the trajectory of the missile, investigators can try to pinpoint the source of the attack.

... "

The crash of Malaysia Airlines Flight 17 has been extraordinarily public. Investigators of aircraft accidents typically cordon off the scene immediately after a crash, and tightly control access for media and others to avoid disturbing the debris.

Because Flight 17 fell in a war zone —across farm fields and homes near where pro-Russia separatists are fighting the Ukraine government — control of the wreckage has been chaotic, and the comparatively sanitized photos typically released by aviation safety officials have been replaced by thousands of images distributed across wire services and social media.

Mapping that wreckage is intended to illustrate how Flight 17 may have come down and the geographical challenges of conducting an investigation in a war zone.

Hrabove

http://graphics.wsj.com/mh17-crash-map/img/map-top-flat.jpg

1. Vertical Tail

The jet's vertical tail with the Malaysia Airlines logo lies just down the road from the aft fuselage section to which it was once attached.

2. Aft Fuselage and Horizontal Tail Structure

One of the most intact pieces of Flight 17 is a portion of the aft fuselage of the Boeing 777. This rear part, which attaches to the rest of the jet's body behind the passenger cabin, landed upside down next to a road in Hrabove, near what was left of the connection between the jet’s horizontal tail, a wing on the tail that points the nose. One of the carbon-fiber-composite horizontal tails was severed and found roughly 1,200 feet away in a grazing field.

3. Center Structure

http://graphics.wsj.com/mh17-crash-map/img/main-debris.png
http://graphics.wsj.com/mh17-crash-map/img/final/HRABOVE_12.jpg

The single largest and most concentrated debris field is along a roadside in Hrabove, more than five miles east of the western edge of the documented debris field. The underside of the wings is visible in the wreckage, identifiable by access holes used by manufacturing and maintenance crews to work inside the fuel tanks. In the charred debris field are parts of both of the jet's Rolls-Royce engines and a major structure called the center wing box that connects both wings to the fuselage and the well that holds the main landing gear. The landing gear was tucked inside the jet when it was cruising at 33,000 feet.

4. Right Rear Door Frame

A cattle pasture in Hrabove is littered with pieces of the jet's fuselage skin. A part of the right side of the rear passenger cabin at the last set of the jet’s doors sits mangled, its door missing. The skin shows the identity of the Boeing 777, with the Malaysian registration 9M-MRD.

5. Left Wingtip

The remains of the left wingtip, with visible signs of what appear to be shrapnel or puncture damage, sits severed from the rest of the wings, which fell and burned more than a half mile away.

6. Horizontal Tail

The carbon-fiber-composite horizontal tail landed a few feet away from the remains of the wingtip near a small pond. Partially covered in mud, the tail was pulled from the nearby pond, according to one photographer.

7. Cargo Door Frame

Scattered pieces of the rear fuselage, like this one around the aft cargo bay, litter a field in Hrabove a half-mile from the largest concentration of debris containing the charred remains of the jet's center structure and wings.

8. Crew Rest Bunks

On long-range flights like those connecting Southeast Asia to Europe, cabin crews use rest bunks to work in shifts. Malaysia Airlines installed its berths underneath the floor of the 777 at the front of the rear cargo bay. The under-floor bunks and their contents landed in a field about 1,800 feet from the main debris site in Hrabove.

Rozsypne

http://graphics.wsj.com/mh17-crash-map/img/map-2-full.jpg

9. Cockpit and Lower Nose

The Boeing 777’s cockpit and lower nose section landed in a sunflower field in Rozsypne, approximately 1.6 miles from the cargo floor section to which it was once attached and nearly four miles from Hrabove. The remains of the flight deck, its floor, displays and pilot controls were strewn nearby. Personal effects from the passengers and crew litter the site. The jet’s avionics components and wiring from the under-floor electronics bay were crushed into the nose landing gear, which became separated while it was tucked up into the forward wheel well.

http://graphics.wsj.com/mh17-crash-map/img/cockpit-lower-nose.png
http://graphics.wsj.com/mh17-crash-map/img/final/452428560.jpg

Petropavlivka

http://graphics.wsj.com/mh17-crash-map/img/map-3-full.jpg

10. Lower Forward Cargo Floor

The jet’s lower forward cargo floor landed nearly five miles from the main debris field in Hrabove on a side road in the town of Petropavlivka. Local residents collected debris littered across the neighborhood and deposited it on this section of the aircraft that was once part of the forward cargo bay. Bundles of wires, air ducting, the nose landing-gear bay door and even demolished drink carts now sit on top of the silver rollers that make up the lower belly of the Boeing 777.

http://graphics.wsj.com/mh17-crash-map/img/lower-fuselage.png http://graphics.wsj.com/mh17-crash-map/img/final/PROBE_GRAPHIC_2.jpg

Отредактировано Risto_Koivula,_Tampere (2021-10-25 08:13:04)

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Risto_Koivula,_Tampere написал(а):

Yes.

Значит у кокпита МН17 при расчете вашим методом, путем ввода продольной перегрузки 2,0 g, BC=1562,5.

Давайте сравним ВС двух кокпитов.

В отчете  "IN-FLIGHT BREAKUP OVER THE TAIWAN STRAIT
NORTHEAST OF MAKUNG, PENGHU ISLAND
CHINA AIRLINES FLIGHT CI611
BOEING 747-200, B-18255"
https://www.skybrary.aero/bookshelf/books/1349.pdf
http://forumupload.ru/uploads/0016/23/c6/51/29582.png
http://forumupload.ru/uploads/0016/23/c6/51/431429.png
http://forumupload.ru/uploads/0016/23/c6/51/842472.png

у кокпита CI611 BC=45.

По вашей методике определим перегрузку, зная что BC = ρ/k. можно найти k и соответственно перегрузку.
k=0,5/45=0,01111111
теперь найдем перегрузку a=0,01111111*250²=694,44 или примерно 69,4 g

При отрыве кокпита у МН17  продольная перегрузка 2 g, а при отрыве кокпита у CI611 продольная перегрузка 69,4 g.

Как объясните такую разницу?

Я считаю. что просто вводить 2,0 - 2,6 g неверно, надо знать величину продольной перегрузки. которое создается при отрыве кокпита МН17 и ее вводить.

Эту величину продольной перегрузки на МН17 вы не знаете!    ELT сработал от 2,0 - 2,6 g, продольная перегрузка на МН17 была больше 2,0 - 2,6 g,  но какая на самом деле была продольная перегрузка на МН17 при отрыве кокпита неизвестно.

Отредактировано oper (2021-10-25 17:43:59)

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Ралив написал(а):

1 яблоко х 1 яблоко  = яблоко квадратное = 1^2

Я, например, считаю, что в результате мгновенного отрыва кокпита кратковременная продольная перегрузка МН17 была 39 g.

С помощью метода Risto_Koivula,_Tampere определим дальность падения копита.

mdv/dt = mv´(t) = - ρACv²(t)/2 = - cv²(t) = the drag force

v´(t)/v²(t) = - c/m = constant = -39g/vo² = -390ms⁻²/(250ms⁻¹)² = - (160.26m)⁻¹ = -k

Integrating implicitely:

-v⁻¹(t) = -k(t - to), where to is integation constant, which gives

v(t) = k⁻¹(t - to)⁻¹

v(0) = 250ms⁻¹ = vo = -k⁻¹to⁻¹, which gives

to = -k⁻¹vo⁻¹ = - vo/(2g) = - 250ms⁻¹(390ms⁻²)⁻¹ = - 0,64s.

The distance d(t) at time t:

d(t) = int v(t)dt = k⁻¹[ln (t - to) - ln (0 - to)] = k⁻¹ ln (1 - t/to) = 160.26m ln (1 + t/0,64s)

At the fall the same c/m = k = (160.26m)⁻¹ can be used. (It can be regarded to be constant in trajectory direction.)

Now v(0) = 0, and the acceleration downwards dv(t)/dt = -g.

Now

m dv(t)dt = mv´(t) = mg - cv²(t),

v´(t)/(v²(t) - mg/c) = c/m = k, which is taken from the former task.

Let´s define still one coefficient:

k/g = p² = c/(mg), p = sqrt⁻¹(1602.6) m⁻¹s = 40.03⁻¹m⁻¹s

(m in two meanings: the mass and meter...)

v´(t)/(v²(t) -p⁻²) = gp²:

- p artanh (p v(t)) = (t - t1) gp²,

where t1 = 0 from v(0) = 0.

v(t) = p⁻¹ tanh(gpt) = 40.03ms⁻¹tanh (t /4.003s)

The fall height (from the top)

h(t) = int (v(t))dt = g⁻¹p⁻²ln cosh (gpt) = k⁻¹ln cosh(gpt) = 160.26m ln cosh(t/4.003s)

For "larger" x: ln cosh x > x - sqrt2 = x - 0,693.

160.26m (t/4.003s - 0.693) = 10000m => t /4.003s = 3.2 0.693= 3.893 =>

T = 15.58s IS THA FALL TIME.

In this time the cockpit goes in horisontal direktion distance

d(T) = 160.26m ln (1 + 15.58s/0,64s) = 518.05 m

Почти такое же расстояние падения кокпита получено при компьютерном моделировании с использованием локуса 10000-250-118-0.   
http://forumupload.ru/uploads/0016/23/c6/51/570490.png

Отредактировано oper (2021-10-25 22:20:12)

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oper
Не спрашивали у этого Куала-Лумпур, где у него изменение плотности воздуха учитывается?

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oper
(Бороду-то я сбривал, куда умище девать?)

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uschen написал(а):

Не спрашивали у этого Куала-Лумпур, где у него изменение плотности воздуха учитывается?

Он берет плотность воздуха 0,5 , но я думаю лучше брать среднюю величину. Интересный метод. Сколько же было g при отрыве кокпита? Кратковременная перегрузка, наверное доли секунд, кокпит отвалился и удар воздушной массы по дыре.

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oper написал(а):

Он берет плотность воздуха 0,5 , но я думаю лучше брать среднюю величину. Интересный метод. Сколько же было g при отрыве кокпита? Кратковременная перегрузка, наверное доли секунд, кокпит отвалился и удар воздушной массы по дыре.

А почему не 5, или 25? Стволовой это, думаю.

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uschen написал(а):

А почему не 5, или 25? Стволовой это, думаю.

Нет, это не Стволовой! Это профессиональный математик.

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oper написал(а):

Risto_Koivula,_Tampere написал(а):

    Yes.

Значит у кокпита МН17 при расчете вашим методом, путем ввода продольной перегрузки 2,0 g, BC=1562,5.

I have no "Method".

I have basic equation and I have initial data measured from a falling body in the air. The horizontal acceleration is v´(0) = -2.3g and the horizontal speed v(0) = 250m/s.

In the vertical direction i have speed w(0) = 0 and acceleration w´(0) = -g.

The trajectory can be calculated from these exactly (= without loosing or addind information in "method"), in spite of the somewhat deceleranting effect of higher air density lower.

We need BC only for that correction. BC = 1359 at the top and 2717 at the bottom. k = c/m = (2717m)⁻¹ is according to the former. 

This gives results which Almaz has also got in its simulations.

Давайте сравним ВС двух кокпитов.

В отчете  "IN-FLIGHT BREAKUP OVER THE TAIWAN STRAIT
NORTHEAST OF MAKUNG, PENGHU ISLAND
CHINA AIRLINES FLIGHT CI611
BOEING 747-200, B-18255"

Cockpit and cockpit... Here as "the cokpit" is regarded all material interconnected to fall with the loose from the fuselage cockpit. The details on that paggage are not needed.

This "cockpit" was about 10000 kg.

http://mh17.forum.camp/viewtopic.php?id=39#p1945

" Альберт Валиев написал (а):

по какой-то причине передняя стойка сходит с замков и вываливается, по причине чего начинает отламываться кокпит - в кокпите в этот момент большая перегрузка. В таком случае момент срабатывания ELT = момент отрыва кокпита. "

This is not "peregruzka", this is usual deceleration from drag to loose "cocpit package".

Cockpit gets loose in transversal direction form bolts etc. (замки) of a flange (стойка) connecting it to fuselage, like a sewer tube breaks at its rusty flange bolts. The exploding on the missile does not get rise to significant longitudinal acceleration to the the top of cockpit at moment.

http://forumupload.ru/uploads/0016/23/c6/63/t19622.png

Boeing787 fuselage and cockpit flanges. Another left below.

https://www.skybrary.aero/bookshelf/books/1349.pdf
http://forumupload.ru/uploads/0016/23/c6/51/29582.png
http://forumupload.ru/uploads/0016/23/c6/51/431429.png
http://forumupload.ru/uploads/0016/23/c6/51/842472.png

у кокпита CI611 BC=45.

No interest...

По вашей методике определим перегрузку, зная что BC = ρ/k. можно найти k и соответственно перегрузку.

This in not перегрузка but basid drag deceleration. Don´t use таблицы перегрузoк here.

Give headwind (vstrechnyi veter), and I will calculate you перегрузка from it for MH17 "cockpit" in disintegration moment.

Don´t try to confuse with empty words.

k=0,5/45=0,01111111

теперь найдем перегрузку a=0,01111111*250²=694,44 или примерно 69,4 g

При отрыве кокпита у МН17  продольная перегрузка 2 g, а при отрыве кокпита у CI611 продольная перегрузка 69,4 g.

Как объясните такую разницу?

You write bullshit.  My coefficient k is measured for MH17´s loose cockpit. It has nothind to do with other cockpits.

Я считаю. что просто вводить 2,0 - 2,6 g неверно, надо знать величину продольной перегрузки.

-2.3 g. It was measured in two ways: directly and through speeds. And it is usual deceleration of free in the air under drag moving body. It is not any accidental impact or oveloading factor. And it gives very good results in the poit vof you forming the debris deviation as it is.

которое создается при отрыве кокпита МН17 и ее вводить.

Missile explosion accelerates cockpit in taransversal direction, from left to right, and maybe also somewhat downwards.

Эту величину продольной перегрузки на МН17 вы не знаете!

I dont´t need "such". I know the bodies speed and acceleration free in the air. The only interesting thing is WHAT THEY ARE - not that, how they were formed!

ELT сработал от 2,0 - 2,6 g, продольная перегрузка на МН17 была больше 2,0 - 2,6 g,  но какая на самом деле была продольная перегрузка на МН17 при отрыве кокпита неизвестно.

You mean "the (part of) drag to the cockpit just before it had got disconnected". It is unknown, and useless!

Отредактировано Risto_Koivula,_Tampere (2021-10-29 08:10:17)

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oper написал(а):

Сегодня 01:40:44

oper написал(а):

Ралив написал(а):

    1 яблоко х 1 яблоко  = яблоко квадратное = 1^2

Я, например, считаю, что в результате мгновенного отрыва кокпита кратковременная продольная перегрузка МН17 была 39 g.

For this magnificient drag coefficient is presupposed for MH17 10 tons´ (10000kg) "cockpit package" a transversal to the trajectory  effective cross-section (projection) area

CA = 2 x 39 x 10ms⁻² x 2.0 kg⁻¹m³ x 10000kg x 250⁻²m⁻²s² = 250m², which presupposes an equivalent plate with diameter 17.8m.

For g = 2.3 g the equivalent plate area is 2 x 2.3 x 10 x 2.0 x 10000 x 250⁻² = 14,72 m² and equivalent diameter 4.32m, which is what it is at MH17.

https://en.wikipedia.org/wiki/Boeing_777

С помощью метода Risto_Koivula,_Tampere определим дальность падения копита.

    mdv/dt = mv´(t) = - ρACv²(t)/2 = - cv²(t) = the drag force

    v´(t)/v²(t) = - c/m = constant = -39g/vo² = -390ms⁻²/(250ms⁻¹)² = - (160.26m)⁻¹ = -k

Where and when has this "overloading deceleration" been MEASURED?

To create that deceleration you would need for MH17 10000kg "cockpit package" an externmal force of 390MP (megaponds), which is towrds the cockpit a double of the support force, if MH17 were standing straightforward  on earth supporting only to its cockpit.

I remark, that you understand nothing about anything.

mdv/dt = mv´(t) = - ρACv²(t)/2 = - cv²(t) = the drag force

v´(t)/v²(t) = - c/m = constant = -39g/vo² = -390ms⁻²/(250ms⁻¹)² = - (160.26m)⁻¹ = -k

Integrating implicitely:

-v⁻¹(t) = -k(t - to), where to is integation constant, which gives

v(t) = k⁻¹(t - to)⁻¹

v(0) = 250ms⁻¹ = vo = -k⁻¹to⁻¹, which gives

to = -k⁻¹vo⁻¹ = - vo/(2g) = - 250ms⁻¹(390ms⁻²)⁻¹ = - 0,64s.

The distance d(t) at time t:

d(t) = int v(t)dt = k⁻¹[ln (t - to) - ln (0 - to)] = k⁻¹ ln (1 - t/to) = 160.26m ln (1 + t/0,64s)

At the fall the same c/m = k = (160.26m)⁻¹ can be used. (It can be regarded to be constant in trajectory direction.)

Now v(0) = 0, and the acceleration downwards dv(t)/dt = -g.

Now

m dv(t)dt = mv´(t) = mg - cv²(t),

v´(t)/(v²(t) - mg/c) = c/m = k, which is taken from the former task.

Let´s define still one coefficient:

k/g = p² = c/(mg), p = sqrt⁻¹(1602.6) m⁻¹s = 40.03⁻¹m⁻¹s

(m in two meanings: the mass and meter...)

v´(t)/(v²(t) -p⁻²) = gp²:

- p artanh (p v(t)) = (t - t1) gp²,

where t1 = 0 from v(0) = 0.

v(t) = p⁻¹ tanh(gpt) = 40.03ms⁻¹tanh (t /4.003s)

The fall height (from the top)

h(t) = int (v(t))dt = g⁻¹p⁻²ln cosh (gpt) = k⁻¹ln cosh(gpt) = 160.26m ln cosh(t/4.003s)

For "larger" x: ln cosh x > x - sqrt2 = x - 0,693.

160.26m (t/4.003s - 0.693) = 10000m => t /4.003s = 3.2 0.693= 3.893 =>

T = 15.58s IS THA FALL TIME.T = 15.58s IS THE FALL TIME.

You have calculated here wrong:

160.26m (t/4.003s - 0.693) = 10000m => t /4.003s = 10000/160.26 + 0.693 = 63,09 =>

T = 252,6s IS THe FALL TIME.

Of course a body with magnificiant drag coefficient falls slowly.

You had a good try (but bad "result").

In this time the cockpit goes in horisontal direktion distance

d(T) = 160.26m ln (1 + 15.58s/0,64s) = 518.05 m

d(T) = 160.26m ln (1 + 252.6/0.64s) = 958m

Почти такое же расстояние падения кокпита получено при компьютерном моделировании с использованием локуса 10000-250-118-0.

Your children´s "mechanics" computer e-play computes "potatoes".

That is what Bellingcat, DSB and JIT say, and they may also lie, that "Almaz says"...

Отредактировано Risto_Koivula,_Tampere (2021-10-27 07:06:02)

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Alex написал(а):

Risto_Koivula,_Tampere написал(а):

http://forumupload.ru/uploads/0016/23/c6/61/857847.jpg

             

http://forumupload.ru/uploads/0016/23/c6/61/27114.jpg

https://youtu.be/acd_67jrXxM

Отредактировано Alex (2021-10-22 01:07:05)

http://forumupload.ru/uploads/0016/23/c6/63/t393036.jpg

This strike point, точка пoражения, means that possible missile flight from Pervomaike (DSB, Bellingcat) would have been abot 25 km.

http://forumupload.ru/uploads/0016/23/c6/63/t236306.png

http://forumupload.ru/uploads/0016/23/c6/63/t630967.png

In some Bellingcat/DSB flaps there is a "turbo-free fall": the plane ia said to fall FASTER THAN THE FREE FALL TIME FROM 10000m: 45 s.

http://forumupload.ru/uploads/0016/23/c6/63/t15334.png

That is not the real strike poin, but "DSB_Strike_Point".

Here are the flight lines of Boing jumbojets during ne week fefore and includind 14.7.

https://www.nytimes.com/interactive/201 … pe=PAYWALL

" Before Crash, Some Airlines Avoided Ukraine

Published July 18, 2014

A survey of flights to Asia from Europe in the last week found that some airlines had been flying over eastern Ukraine and some had been avoiding the area.

Source: Flight path data from flightradar24.com

http://forumupload.ru/uploads/0016/23/c6/63/t559408.png

Отредактировано Risto_Koivula,_Tampere (2022-01-23 14:29:01)

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Risto_Koivula,_Tampere написал(а):

Here are the flight lines of Boing jumbojets during ne week fefore and includind 14.7.

I'm showing you flight MH17 for July 17th.

http://forumupload.ru/uploads/0016/23/c6/61/186489.jpg

"Corridor" Boeing according to the Ministry of Defense of the Russian Federation and the FlightRadar24.com service.
Data overlapping. 

Besides that
         
   
http://forumupload.ru/uploads/0016/23/c6/61/296747.png     
           
   
http://forumupload.ru/uploads/0016/23/c6/61/56980.jpg
     
     
http://forumupload.ru/uploads/0016/23/c6/61/345123.jpg

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Risto_Koivula,_Tampere написал(а):

Where and when has this "overloading deceleration" been MEASURED?

А вы где взяли "overloading deceleration"? ELT не измеряет "overloading deceleration", он срабатывает при достижении перегрузки 2 - 2,6 g, а какое на самом деле было "overloading deceleration" неизвестно.
 

Risto_Koivula,_Tampere написал(а):

This "cockpit" was about 10000 kg.

Масса "сockpit" неизвестна!
"Сockpit" может иметь массу 5000 kg, а может быть 25000 kg. В ваших расчетах не учитывается масса , у вас независимо от массы "cockpit" падает на одно и тоже расстояние, которое равно  5851 m.

Вы не учитываете массу (m), cross sectional area (A), drag coefficient (Cd).

Ваши расчеты по определению дальности падения неверны!!!

Отредактировано oper (2021-10-26 15:33:29)

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oper написал(а):

Risto_Koivula,_Tampere написал(а):

    Where and when has this "overloading deceleration" been MEASURED?

А вы где взяли "overloading deceleration"? ELT не измеряет "overloading deceleration", он срабатывает при достижении перегрузки 2 - 2,6 g, а какое на самом деле было "overloading deceleration" неизвестно.

How do you say in English "перегрузкa g"?  Перегрузкa is "overloading, transloadling". g is a measure of acceleration, ускоре́ние. ELT doesn´t and can´t  make differense with accelerations ackording to their causes. Why do you keep harping перегрузкa insted of ускоре́ние?

Risto_Koivula,_Tampere написал(а):

    This "cockpit" was about 10000 kg.

Масса "сockpit" неизвестна!

Unknown, and useless, becaude we have better masured information about the trejectory.

"Сockpit" может иметь массу 5000 kg, а может быть 25000 kg. В ваших расчетах не учитывается масса , у вас независимо от массы "cockpit" падает на одно и тоже расстояние, которое равно  5851 m.

Вы не учитываете массу (m), cross sectional area (A), drag coefficient (Cd).

I need only ratio ρACd/m/2 = c/m = k  = v´(0)/v²(0), not necessarily separation of the factors.

Ваши расчеты по определению дальности падения неверны!!!

They are right enough to prove that Bellincat´s, DSB´s and JIT´s "calculations" and "observations" are bullshit.

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Risto_Koivula,_Tampere написал(а):

Risto_Koivula,_Tampere написал(а):

    This "cockpit" was about 10000 kg.

Масса "сockpit" неизвестна!

Unknown, and useless, becaude we have better masured information about the trejectory.

У вас кокпит с любой массой  падает в одно и тоже место, на расстояние 5851 m. Этого не может быть.

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Ралив написал(а):

Масса разная и соответственно площадь разная, а результат стабилен.

А если один и тот же кокпит, но другой ELT, который включается  при 1 g, то кокпит должен вместо 5851 метров улететь на 10000 метров?
А если взять ускорение 0,001 g  то кокпит улетит на 100 000 метров. :D

Отредактировано oper (2021-10-26 23:55:08)

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oper написал(а):

Risto_Koivula,_Tampere написал(а):

    Risto_Koivula,_Tampere написал(а):

        This "cockpit" was about 10000 kg.

    Масса "сockpit" неизвестна!

    Unknown, and useless, becaude we have better masured information about the trejectory.

У вас кокпит с любой массой  падает в одно и тоже место, на расстояние 5851 m. Этого не может быть.

If the mass is in same ratio with the dragging transversal to trajectory surface.

The free fall parabola is almost in a square 10km x 10km.

The drag decelerates both falling and horizontal transition in close to each other magnitudes.

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oper написал(а):

Ралив написал(а):

    Масса разная и соответственно площадь разная, а результат стабилен.

А если один и тот же кокпит, но другой ELT, который включается  при 1 g, то кокпит должен вместо 5851 метров улететь на 10000 метров?
А если взять ускорение 0,001 g  то кокпит улетит на 100 000 метров. :D

No. The ELT can be regulated to "wake up" in several accelerations/decelerations, but it measures real acceleration according to inertial frame of reference.

It is not good idea to speak about перегрузки (in sense of this table, for instance) with ELT: it measures переускорение.

http://forumupload.ru/uploads/0016/23/c6/63/t339672.jpg

This kind of overloadings are measured with static means in the constructions. The cockpit can be under severe перегрузкa, for instance the external loads can be double to usual (2g + 2g for instance) whitout the ELT "remarking" anything, if the motors are regulized to keep the speed stable in the winds.

Отредактировано Risto_Koivula,_Tampere (2021-10-27 07:07:22)

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oper написал(а):

Нет, это не Стволовой! Это профессиональный математик.

Да? А Вы спрашивали, это действительно он?
Я исхожу из того, что за эти годы подобный бред был слышен только от Стволового.

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