Carefully consider quantities of water ions
required for an effective particle stream.
Readily transformed into steam, superheated steam and then plasma, water molecules will become the ions needed to enter the ship's particle accelerator (PA) to gain near light speed velocity then exit from the spacecraft. (TE assumes "equivalent water molecules": H_{2}O broken into various ions, OH^{}, H^{} and O^{}.) Consequent momentum exchange imparts a slight velocity increase to the ship in the opposite direction.
To maintain gforce,
spaceships will need plenty of water. Life support requirements of an ample water supply are obvious; however, an even more compelling requirement is to sustain gforce throughout the flight. Fortunately, there is plenty of water in Earth's oceans for the first few gforce ships. For longer term, there are plenty of water bearing comets throughout interplanetary space to provision the fleet of gforce ships needed to service likely travel requirements.
Previous work causes us to conclude that gforce spaceship's Gross Weight (GW) is everdecreasing due to fuel consumption. In turn, decreasing GW results in decreased fuel requirements; thus, each day's fuel burn will be slightly less then previous day.  As fuel burns, gross weight decreases.
As vessel weight decreases, burn rate also decreases.
To consider ion quantity requirements,
TE constructs following three tables.
Table2: Any day: 86,400 Unique Seconds TE arbitrarily chooses Day 20 as an example; Day 20's fuel requirement is 222.91 mTs of water. Simple division approximates an average burn rate of .00258 mT (= 2,580 grams) per second.
Table3: Pulse requirements During each second, PA will create and expend many plasma packets. Each will contain small quantity of water but a large number of particles. TE arbitrarily assumes a Packet Repetition Frequency (PRF) of 10,000 per second. 
Table1: Everyday is different.
t  GW_{t}  F_{t}  ff_{tsec}  Δ_{t} 
days  metric Tonnes  mT/day  daily ave
gm/sec  daily ave
difference 
0  100,000 mT    n/a 
1   233.00  2,697 gm/sec  n/a 
2   232.46  2,690 gm/sec  6.90 gm/sec 
...   . . .  . . .  . . . 
19   223.45  2,586 gm/sec  6.044 gm/sec 
20   222.91  2,580 gm/sec  6.027 gm/sec 
21   222.39  2,574 gm/sec  6.010 gm/sec 
...     . . . 
39     5.76 gm/sec 
40     
Given    F_{t} 86,400 s/d  ff_{t1 } ff_{t} 
Δ_{t}= 0.233% / day %GW_{t} = (1Δ_{Day})^{t} TOGW = GW_{0}= 100,000 mT  TE work concludes that a vessel could maintain gforce acceleration if
exhaust particles achieve .866c
daily fuel consumption is .233% of its gross weight.
Table 1 arbitrarily assumes ship's initial gross weight (GW_{0}) to be 100,000 mTs. With fuel consumption of .233%/day, ship will consume 233 mT of water during day 1, and ship's gross weight will decrease by 233 mTs.
GW_{1 }= 99,767 mT Day 2's fuel consumption will be .233% of 99,767 mT which further decreases ship's gross weight and so on. Subsequent calculations (expedite with exponentials) eventually lead to fuel consumptions shown on Table 1 rows for midflight: days 19, 20, and 21. Since a metric Tonne (mT) equals 1,000,000 grams (gm), and one day equals 86,400 seconds, we can easily compute a daily per second average as shown in Table1a.
Table1a: Daily Per Second Average
t  F_{t}  ff_{tsec}  
day  grams/day  ave gm/s 
19  223,454,000  
20  222,912,000  
21  222,385,000  

*************
Total fuel requirement
for a 40 day, gforce mission is about 9,000 mTs of water.
(NOTE: One Olympic sized swimming pool contains about 3,000 mTs of water.)
*************
Assume each day's "average value" occurs exactly at midday.
Table1b. Three Days of MidFlight
t  sec  ff_{tsec}  Δ_{t} 
day of
flight  precise second
at midday  assume fuel flow
at precise sec  difference 
19  43,200^{th}sec   
20   2,580.00gm/sec  Δ_{20}=0.000 gms 
21   2,573.99gm/sec  Δ_{21}= 6.010gms 
Arbitrarily select Day 20 as a reference; set Δ_{20} to zero. Thus, burn rate for previous day (19) is a little greater (+), and burn rate for following day (21) is a little less ().
 Since vessel's GW and fuel burn rate gradually decrease throughout each day, common sense compels us to conclude the per second burn rate will range from slightly above the daily average to slightly below. TE artificially assumes each day's average per second fuel consumption value to be the "exact" value for the 43,200th second of the day (exactly halfway through the day's 86,400 seconds). This midday value will differ between succeeding middays and can be readily discerned. For example, Table1b shows the19th midday value to be 6.028 grams greater then value for 20th midday. In turn, the 21st midday value decreases by 6.01 grams. 
The name "Avogadro's Number" (N_{A}) is an honorary name attached to the calculated value of the number of atoms, molecules, etc. in a gram molecular weight of any chemical substance.
To determine number of particles in a given mass, use Avagadro's Number (N_{A }) approximately 6.0221415×10^{23}. N_{A }denotes quantity of molecules in one mole, material's atomic weight in grams. A standard mole is defined as the value of quantity of atoms in 12 grams of pure carbon12 (^{12}C), carbon's primary isotope with atomic weight, 12.
Any substance's mean molecular weight expressed in grams has same number of molecules. For example, the mean molecular weight of natural water is about 18.015; so, one mole of water is about 18.015 grams with 6.0221415×10^{23} molecules, same quantity as 12 grams (one mole) of Carbon12.
The mole proves to be a convenient way for chemists to express the amounts of reagents and products of chemical reactions.
For example, the chemical equation:
2 H_{2}+ O_{2} → 2 H_{2}O states that 2 mol of dihydrogen and 1 mol of dioxygen react to form 2 mols of water. Divide a mole by water's molecular weight to determine quantity of molecules in one gram of water.
Water: 1 gm = 6.0221415×10^{23} molecules / 18.015 1 gm = 0.334285×10^{23} molecules (of H_{2}O) If one gram of liquid water displaces one cubic centimeter (cc = cm^{3}); then,
33.428×10^{21} water molecules Rel. 1 cm^{3} Cube root both sides of this relation; then, TE presumes 3.2213×10^{7} molecules (of liquid water) take up 1 cm in length. Thus, average distance between liquid water molecules could be about:
10^{2} meter / 3.2213×10^{7} = 3.1×10^{10} m = .31 nanometers. 
Focus on Day 20 To more closely approximate per second fuel flows for Day 20, use different Δs for first and second half of the day. Table1c. Day 20: Start to Finish
t  sec  ff_{tsec}  Δ_{tsec} 
day_of
flight   assume fuel flow
at precise sec  difference 
20  00,001^{st}sec   
20   2,580.000gm/sec  Δ_{20Mid}=0.000 gms 
20   2,576.995gm/sec  Δ_{20PM}=3.005 gms 
Determine morning Δ by halving previous's day's difference. Δ_{20AM} = Δ_{19 }÷ 2=+3.014 gms Determine afternoon Δ by halving following day's difference. Δ_{20PM} = Δ_{21} ÷ 2 =3.005 gms 
For convenience, TE assumes linearity for determining precise and unique fuel consumption values for each of the 43,200 seconds in each half day. A linear model requires a per second decrement (let i represent quantity of elapsed seconds):
ff_{i}= FF_{Day}  i*Δ Decrement (Δ) can be expressed as mass or as quantity of potential particles (molecules not yet ionized).
At exactly midday of Day 20, 43,200th sec, vessel consumes 2,580 grams or 862.4549×10^{23} water molecules.
12 hours earlier, Day 20, first sec; fuel consumption is 3.014 grams more; 3.014 gms of water contain 1.007 534 99×10^{23} molecules. To determine per second decrement, divide by 43,200.
Δ = 2.332 257×10^{18} = .000 023 253×10^{23}
12 hours later, Day 20, 86,400th sec; fuel consumption is 3.005 grams less; 3.005 gms of water contain 1.004 524 92×10^{23} molecules. To determine per second decrement, divide by 43,200. Δ = 2.325 289×10^{18} = .000 023 253×10^{23} 
Table2. Day 20: 86,400 Unique Seconds.
t  ff_{sec}  N_{sec}  Remarks 
sec  estimated
grams  estimated
H_{2}O molecules  1 gram of H_{2}O contains 0.334 284 5×10^{23} molecules (or equivalent) 
1   863.461 878×10^{23}  First half day: 3.015 grams of H_{2}O contain 1.007 868 81×10^{23} molecules (or equivalent)
Determine average decrement per second, divide by 43,200.
Δ_{sec} = .000 023 33×10^{23} molecules 
2   863.461 854×10^{23} 
...   Δ≈.000 023 33×10^{23} 
43,199   862.454 033×10^{23}  Δ_{sec}= +.000 023 33×10^{23} 
43,200   862.454 010×10^{23}  Reference Point: Exact Midday: Δ= 0.0 
43,201   862.453 987×10^{23}  Δ_{sec}= .000 023 25×10^{23} 
...   Δ≈ 000 023 25×10^{23}  Second half day: 3.005 grams of H_{2}O contain 1.004 524 92×10^{23} molecules (or equivalent)
Determine average decrement per second, divide by 43,200.
Δ_{sec} = .000 023 25×10^{23} molecules 
86,399   861.449 508×10^{23} 
86,400  2,576.995 gm  861.449 485×10^{23} 
Given   N_{A}*ff_{sec} mole  
Assume day 20 of 40 day voyage. Fuel consumption is 222.91 mT for average fuel flow of 2,580 gm/sec 
*************
Avogadro, the Method In 1811, Avogadro first proposed his hypotheis: volume of a gas (at a given pressure and temperature) is proportional to the number of atoms or molecules regardless of the nature of the gas.
The term "Avogadro Number" was first used in a 1909 paper by Jean Baptiste Jean Perrin (18701942) entitled "Brownian Movement and Molecular Reality." (Translated from French to English in "Annales De Chimie et de Physique" by Fredric Soddy.)
In 1909, French physicist, Jean Perrin, proposed naming the constant in honor of Avogadro. Perrin: "The invariable number N is a universal constant, which may be appropriately designated 'Avogadro's Constant'." (now known as N_{A})
Perrin, won the 1926 Nobel Laureate in Physics for his work on the discontinuous structure of matter and the discovery of sedimentation equilibrium. A large part of his work determined the Avogadro constant by several different methods.
Determinating Avogadro's number required accurate and precise measurement of a single quantity on both the atomic and macroscopic scales.
In 1834, Michael Faraday's works on electrolysis contained value for electrical charge of a mole of electrons ("Faraday constant").
In 1910, Robert Millikan, an American physicist, measured the charge on a single electron. Divide the charge on a mole of electrons by the charge of a single electron to determine Avogadro's number. Since 1910, newer methods more accurately determined the values for Faraday's constant, the elementary charge, and N_{A}. 
 Thus far, TE has considered various water quantities required to gforce propel 100,000 mT vessel:
 For the entire trip, the 40 day requirement is about 10,000 metric Tonnes, enough water to fill a large pond or a small lake.
 For one day, the daily requirement will be about 250 mTs, a small swimming pool.
 Each day has 86,400 seconds. For each second, fuel consumption is a few liters, typical daily ration for drinking, cooking, etc.
 During each second, PA could eject 10,000 particle packets. Each packet could contain about a quarter gram of ionized water (1/4 gram liquid water is about 5 raindrops).
If the Packet Repetition Frequency (PRF) is 10,000 per second, the associated Period, P, is 100 µseconds. Thus, each packet has a 100 µsec window. If the particles travel at .866c (259,620,268 meters/sec), PRF's associated wavelength is about 26,000 m. Thus, a packet could travel 26 km in one packet period, more then sufficient to travel TE's projected PA guidepath of one km.
Lay literature states the PA particle beam is "pencil thin". Thus, TE arbitrarily assumes a circular cross section with max radius of one millimeter.
TE further assumes following packet dimensions: A= π r^{2} = 3.14 (.1 cm)^{2} = .0314 cm^{2} L = 10 m = 10 × 100 cm = 1,000 cm
V = A × L = .0314 cm^{2} × 1,000 cm = 31.4 cm^{3}
Since 30 cubic centimeters (cc) can contain 30 grams of liquid water, TE assumes 31.4 cc is enough volume for one quarter gram of water ions. 
TE assumes onboard PA uses "storage tubes" where particles will orbit at .866c for lengthy periods before being diverted into exhaust tube, final 1 km "guidepath" just prior to exit from spacecraft. This assumption justifies constant speed considerations reflected in diagram below. However, 25 extra kilometers leaves plenty of margin for adjustment. For example, if particles had to accelerate from zero to .866c within one km guidepath, this would reduce the average speed to .433 c, enough velocity to travel 13 km during the 100 µsec period; 12 km to spare.
Divide each second into 10,000 equal periods.
During each 100 µsec period, allocate one packet.
For each each 100 µsec period, the packet occupies the one km guidepath for only 4 µsecs; plenty of time to prepare for next packet.
At only one FLoating OPeration (FLOP) per nanosecond, each period has enough time for 100,000 FLOPs; TE assumes this to be sufficient computing power to make needed minor adjustments per packet.
Due to repulsion of like charged particles, plasma particles naturally tend to spread out into available volume. However,TE assumes that well designed PA uses superconducting magnets (sextupoles and quadrapoles) to focus each packet both longitudially and radially.
≈0  .008×10^{21}  0.181×10^{21}  1.17×10^{21}  2.94×10^{21}  2.94×10^{21}  1.17×10^{21}  0.181×10^{21}  .008×10^{21}  ≈ 0 
<<.001N_{P}  .001N_{P}  .021N_{P}  .136N_{P}  .341N_{P}  .341N_{P}  .136N_{P}  .021N_{P}  .001N_{P}  <<.001N_{P} 
N_{P} = 8.624 540 100×10^{21 }water molecules (ionized into particles) 

Longitudinal cross section shows particle density throughout the total packet length (10 meters). TE assumes normal distribution of particles throughout the 10m length; this results in a elongated bell curve with highest density at the mean (μ) length (5m) from start of packet. For convenience, TE assumes standard deviation (σ) of one meter; thus, 68.2% of particles are within one meter of μ; 95.4% of particles are within 2 meters of μ; and so on. 

Radial cross sections at various lengths along packet show particles filling up circular areas at various radii from the center. Radial focusing moves particles toward packet's center line. Particle density is greatest at the mean length of the packet, but particle density decreases as packet length differs from the mean. Thus, the lesser particle quantity is more affected by the magnetic radial focus which drives them closer to packet center line. 
As described above, TE assumes a packet length of 10 m (
packet volume > 30 cc) can easily contain .25 gms of water ions (
recall 1 gm of liquid water is about 1 cubic cm). However, this model leaves plenty of margin for adjustment, and the packet length can easily be increased as required. Other adjustable parameters include: PRF, packet size, guidepath length, cross sectional radius and no doubt others.
Table3a:Every day differs.
t  ff_{tsec}  P_{gm}  P_{par} 
days    
1   .270 gm  
...   . . .  . . . 
20   .258 gm  
...    
40    
Given   PRF  K * P_{gm} 
PRF= 10,000 packets per second Conversion Factor: K = 33.428×10^{21} (Qty of water molecules/gm) From Tables One and Two, determine average fuel flow per second for any day during the voyage. TE arbitrarily shows values for first, mid and last days of the 40 day trip.
For each day's 86,400 seconds, TE assumes a range of 86,400 decreasing, unique values. Thus, fuel flow values will range from slightly above average to slightly below. TE assumes each day's average fuel flow is the precise fuel flow for the 43,200th second of that day (precise midday).
For each precise midday, TE takes assumed value and determines average packet size by dividing by PRF (assume: 10,000 packets per second). Like values for seconds of the day, values for packets of the second will range from slightly above to slightly below.
TE assumes: At the precise midsecond, the 5,000th packet contains precisely that second's average particles per packet. 
Avogadro, the Man ************* Amedeo Carlo Avogadro was born in Turin in 1776 to a noble family of Piedmont, Italy. He dedicated himself to physics and mathematics (then called positive philosophy). In 1809, he started teaching at a liceo (high school) in Vercelli, where his family had property. In 1811, he published an article ("...Determining the Relative Masses of the Elementary Molecules of Bodies and the Proportions by Which They Enter These Combinations"), in a French journal, (Journal of Physics, Chemistry and Natural History); so, his major work "Avogadro's Hypothesis" was first written in French, even though Avogadro was Italian.
In 1820, he became Professor of Physics at the University of Turin. Due to regional politics, he lost his chair during 1823; however, Avogadro was recalled to the university in Turin in 1833, and he taught for another twenty years.
Though active in politics in younger life, Avogadro's private life was mostly sober and religious. He married Felicita Mazzé and had six children.
Professionally, Avogadro was very active; he held posts dealing with statistics, meteorology, and weights and measures (he introduced the metric system into Piedmont) and was a member of the Royal Superior Council on Public Instruction. 

TE uses different forms of Scientific Notation: a×10^{b} (operand, a, times power of ten)
Form  TE Example  Why this form? 
 1 mole = 6.0221415×10^{23} molecules  This normalized SN value is the the textbook definition of the mole, standard number of molecules in atomic weight (in grams) regardless of substance. 
Nonnormalized a < 1  1 gm (of H_{2}O)=1 mole/18.015
= 0.334285×10^{23} molecules  Nonnormalized value of water particles per gram readily compares with above value of particles per mole by keeping same value for exponent. 
Nonnormalized a > 1  typical ff_{sec} = 2,580 gm (of ionized H_{2}O)/sec
= 862.454 010×10^{23 }equivalent molecules/sec  Nonnormalized value of ionized water particles per second readily compares with above values of particles per mole and particles per gram. 
Change to Normalized  862.454010×10^{23}=8.62454010×10^{25 }  Equivalent value of particles per second can be expressed in Normalized SN by
decrease magnitude of operand (move decimal point left 2 digits) 
increase power of 10 (add 2 to exponent) 

Reduce Magnitude  8.624 540 10×10^{21 }particles per packet  At 10,000 packets per second, approximate particles per packet by dividing by 10^{4}. Due to nature of exponents, this is done by simply subtracting 4 from 25. 
Engineering Notation 10^{b}; b is mult, of 3  33.428×10^{21} molecules
(of liquid water)
displace 1 cm^{3} in volume  Returning to quantity of H_{2}O molecules in one gram, TE chooses to change from Scientific Notation (SN) to Engineering Notation (EN). 3.3428×10^{22} = 33.428×10^{21} 
Cube Root  3.2213×10^{7} molecules
(of liquid water)
take up 1 cm in length  Cube root volume to find cube's edge length in both cm and molecules. Cube root readily done when power of ten is multiple of 3. (33.428×10^{21})^{1/3} = 3.2213×10^{7} 
Negative Exponents  Intermolecular Distance = 3.1×10^{10} m  1 cm = 10^{2} m
10^{2} m/ 3.2213×10^{7} = 3.1×10^{10} = .31 nanometers. 

Table3b. Midday 20 of Gforce Flight
For each second of powered flight, average particles per packet is readily computed as shown in Table 2. TE assumes this value to also be actual value of packet which transits the guidepath during precise midsecond (500,000 µsec shown below as decimal ".500").
t  N_{P}  Δ_{sec}  Δ_{pkt} 
Decimal
Seconds   Per Second Diff
Particles/Second  Per Packet Diff
Particles/Packet 
43,199.500 sec   Δ_{sec}= +2.333×10^{15}  Δ_{pkt}= +2.333×10^{11} 
43,200.500 sec   Reference: Δ_{sec}= 0.0  Reference: Δ_{pkt}= 0.0 
43,201.500 sec   Δ_{sec}= 2.325×10^{15}  Δ_{pkt}= 2.325×10^{11} 
Midday    Δ_{sec} / 10^{4} 
Consider three seconds at the exact middle of 20th day of powered flight (41,199; 42,000; 42,001). Let exact middle of 42,000th second be the reference; then, compare values with one second before reference (midpoint of 41,199s) and with one second after reference point (midpoint of 42,001s).
 Table3c. Precise Middle Second of Day 20 First half second is slightly greater. First Half Second: packet size decreases by 1.1665×10^{15} particles.. Determine per packet decrement (divide by 5,000): Δ_{pkt} = Δ / 5,000 = 2.333×10^{11} particles/packet
P  Timing  N_{P}  Remarks 
Packet  µsec   Particles per Packet  SemiSecond Diff in Particles/Packet 
0001   000,100  8.624 540 22×10^{21}  Δ_{ }= +1.1665×10^{15} 
5,000   500,000  8.624 540 10×10^{21}  Reference: Δ= 0.0 
10,000   1,000,000  8.624 539 98×10^{21}  Δ_{ }= 1.1625×10^{15} 
Midday   Packet's
last µsec  See Table2  See Table2 
Last half second: Δ_{pkt} = 1.1625×10^{15}/5,000 = 2.325×10^{11} particles/packet Let exact middle of 42,000th second be the reference; then, compare values with 1/2 second before reference (start of 43,200s) and with 1/2 second after reference (end of 43,200s). 
Table3. 10,000 Unique Values for 10,000 Packets. Particle Quantity: First to Last Packet of Second 43,200
P  t  ff_{P}  N_{P}  Remarks 
Packet  µsec  estimated grams  estimated
H_{2}O molecules  
1  1100  .2583 gm  8.624 540 220 0000×10^{21}  First half second:
First to Mid: packet size decreases by 1.1665×10^{15} particles. Divide by 5,000 for per packet decrement: 2.333×10^{11} part. 
2  101200  ≈ same  8.624 540 219 7667×10^{21} 
...  ...  ...  ... 
4,999  499,801499,900  ≈ same  8.624 540 100 2333×10^{21}  Δ_{pkt} = .000 000 000 2333×10^{21} molecules 
5,000  499,901500,000  .2580 gm  8.624 540 100 0000×10^{21}  Reference Point: Exact Midday: Δ= 0.0 
5,001  500,001500,100  ≈ same  8.624 540 099 7675×10^{21}  Δ_{pkt} = .000 000 000 2325×10^{21} molecules 
 ...  ...  ...  Second half second: Mid to Last: packet size decreases by 1.1625×10^{15} particles. Divide by 5,000 for per packet decrement: 2.325×10^{11} part. 
9,999  999,801999,900  ≈ same  8.624 539 980 2325×10^{21} 
10,000  999,9011,000,000  .2577 gm  8.624 539 980 0000×10^{21} 
Given  1 packet passes through the PA every 100 µsec,
10,000 packets per second.  N_{A}*ff_{sec} mole  
Assume day 20 of 40 day voyage. Fuel consumption is 222.91 mT for average fuel flow of 2,580 gm/sec 
OTHER CONSIDERATIONS
Plasma Source. To continuously generate plasma, TE assumes significant PA design such that significant portion of particle stream diverts to impact a designated "pool" of water. At near light speed, this particle impact would impart sufficient energy to "superheat" the solid/liquid water and ionize the water molecules.
Real Efficiency. Hence, one of the many reasons for particle stream inefficiencies or less then100% conversion of plasma ions to propulsion particles. Thus, Thought Experiment arbitrarily assumed an efficiency of 70%; thus, best case scenario 30% of PA particles diverted to superheat water and create more plasma ions for more PA cycles.
Model Adjustments. Very likely that not all consumed fuel will exit (due to inherent inefficiencies); however, if some particles are put to other uses other then propulsion (whether by design or defect) can those consumed particles correctly be modeled as GW decrement. Smarter modelers will improve TE's current model; however, real data from actual flights will have the final word.
Particle Flow Precision. Finally, is it necessary or even possible to adjust particle quantities for individual packets??? Table 3 (above) approximates a theoretical per packet decrement of .23 parts per billion for subsequent packets. Even with greatly enhanced future technology, that sounds impractical if not impossible. A more practical method might be distribute weight scales throughout the vessel; these weight sensors could stream data into a centralized systems of servos/computers/etc to constantly make practical adjustments to PA's plasma flow. The objective: keep vessel's acceleration very close to gforce as perceived by humans.
Afterthought!
At 5 drops per packet of particles;
fuel flow is about 50,000 drops per second. 
2 Comments:
Thanks for sharing such beautiful information with us. I hope you will share some more information about ion bem. Please keep sharing.
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