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I have the Sherlock fake suicide worked out...oh yes


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Well its Ms Reggie and myself got it.

Moffat hinted thus:True devotees will spot them (clues). THROUGH THEIR TEARS!!!


What does Sherlock do that he never does....cry.

Why does he cry...to create tears.

You can see his tears fall...thus helping those down below to move the soft landing into place directly below him.


Got it!

Ah well a gust of wind would also affect his fall which is the part of the reason for the teardrop, to make allowances for wind.

In a roundabout way Sherlock admits to the 'fake' tears with his speech to Watson on being a fake.


He also needs to have Watson stand far away so that the gunman cannot see Watson and the prepared landing at the same time.

Throwing his mobile down sends the message 'Im jumping now'(he doesn't want it in his pocket when he lands).

Exactly! And with wind-shift surely...

http://t3.gstatic.com/images?q=tbn:ANd9GcT64bvZTta0cZAwnBcrlt7VjloO4sbwGF6F0tHTj6JEJMWIcAHNIg + http://t1.gstatic.com/images?q=tbn:ANd9GcRevJMOgBclkCzIpoykCNV1Ku05MY3g65QaZ227z-RUt49h63PDpw > http://i.dailymail.co.uk/i/pix/2012/01/17/article-2087814-0F8035B200000578-692_634x396.jpg



OR



A spherical teardrop of mass m, radius r, vertical velocity v (positive downward), and

mass density ρd ≈ 1 g cm−3 satisfies

m =

4

3

πρdr3 (1)

dm

dt = πρmr2|v| (2)

m

dv

dt = mg −

C

2

πρar2v|v| − v

dm

dt

(3)

where the terms on the right-hand side of equation (3) give the weight, air drag, and mist drag

acting on the growing drop.

When the air drag is written as above, the drag coefficient C = C® is a function only of

the Reynolds number [6, 7] R = 2r|v|/ν, where ν is the kinematic viscosity of air. Figure 1

shows C® measured for falling liquid drops (data points, [8]), together with the small-R

theoretical Stokes result C = 24/R (broken line, [6]) and a new empirical result (full line),

C = 12R−1/2 (4)

valid over the intermediate range 10 < R < 1000. Such simple algebraic results allow us

to avoid complicated numerical results for C® obtained from the nonlinear Navier?Stokes

equations [7]. Inserting equation (4) into equation (3) and setting dm/dt = dv/dt = 0 yields

the mist-free terminal or ?settling? speed of drops of radius r with 10 < R < 1000:

v0 = σr (5)

There's a great post here with a bunch of interesting theories, but have a look at the screenshot at the very bottom. The figure on the stretcher looks pretty clearly like Mycroft, not Sherlock...


http://bloodredorion.tumblr.com/post/16159755295/how-did-sherlock-survive

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