Second Sik Path of Wing of Milky-way.

First Sika Path of Wing of Milky-way. 

It is a Wing in which the sun is present. and we are doing count of stars of Wing. Now width of Sika path is 9400 Time-year. therefore width of wing is also 9400 Time-year. and length is also 9400 Time-year of Sika-path. 

therefore W x L = total stars of a layer. or 9400 x 9400 = 88360000 stars of upper layer. and thickness is 1000 time-year of Sika-Path. therefore stars are 88360000000 into the Sika-path. or 88360000 x 1000Ty thickness = 88360000000 stars into the Sika-Path. Or, there is a first upper layer of Sika-Path. In which stars are 88360000. because there is one Time-year distance between two stars. 

and it is also a first path of Milky-way. and 9400 Time year is width of first path. therefore first path width is 9400Ty. and maths says 2 + 2 = 4 and 2 x 2 = 4. therefore width + length = total amount. or width x length = total amount. So width 9400Ty x 9400Ty length = 88360000 Time year. or 88360000 number of stars. and it is first layer. Now wing. and thickness is 1000 Time Year. therefore 1000 layers are present of 88360000 stars. So 88360000 x 1000 thickness = 88360000000 stars into the Sika-Path. 

Second Sik Path of Wing of Milky-way. 

It is second path of Milky way. of which name is Sik. and length is 9400 Time-year with 9300 Time-year width. because the width is happening short slowly. and Thickness is 966 time year of the second Sik Path. because the thickness happening short slowly. and there is a left side in the photograph. 

and left side is a beginning. where stars will have entered from the last Sika path. and left side contain on one Time-year distance. or one Time year distance of stars is into the left side. and one is a right side. which is a end of second Sik path. and here are two Time-year distance between two stars. or after the sik path will have start the third path. 

Now for number of stars into the Sik path. So Length x Width x Thickness is equal to total stars of Path. therefore 9400 x 9300 x 966 = 8.4447720000, stars into the second Sik path. Now this second Sik path is contain on one Time-year distance between in two stars beginning. and at the last two time-year distance. 

but we shall use 1.5 average gap of each star. because there is one time-year distance and two time-year distance at the last between two stars. So we have 8.4447720000 total number of stars. and we are doing solve through the gap between two stars. So 8.4447720000 / 1.5 average gap = 5.6298480000 total number of stars into the second Sik path.  

First Sika path and second Sik path. 

So Today we have travelled of two path of one wing of Milky-way. of which distance was 9400 + 9400 = 18800 Time-year. and 88360000000 Stars are first Sika path. and 5.6298480000 stars of second Sik path. then 1.44658480000 total stars of both path. or 88360000000 + 5.6298480000 = 1.44658480000 stars of Sika path and Sik path. and It was details of a wing in which the sun is existent.