intercept target solutions ? geometrical method ?

05-10-15, 06:36 AM

target speed 10
torpedo speed 28

10 + 28 = 38

So how does 38 shooting angle come close to the 20 you used? The only way you can make it fit together is by using a different formula.

You are wrong ....

speed target of 10 and torpedo of 28 knots = angle of shoot of 20 not 38 lol .....
angle of shoot of 38 will be for a target of 22 knots.

on my video i shoot the torpedo at 348. the torpedo arrive to the boat at 0 angle so. angle of shoot was 12 degres. 348 + 12 you arrive to the 0 degres. front of your u-boat so.

12 degres that means a target of 6 knots around. 5 to 6.

again.....

Distance is not taking in consideration will using my method.

shooting at 1km or 10 or more ...... is the same.

Until you got the real speed of target and know your speed of torpedo = you have a angle of shoot + being at a correct angle = 90 degres = you shoot the boat same as shown in my video.

>>>> Distance does not change the amout of lead, or the torpedo shooting angle if you will, but it does require your aiming to be more accuracte.

dont need to aim more accurate or else......

speed target + speed of torpedo + being at 90 D = angle of shoot ^^

you can miss your target if you did do a wrong calcul of the speed target nothing else.

example in that video ( when am was not using magui as a gui ) good one btw ^^

angle of shooting here is 20. you see i shoot the torpedo a 20 the boat get sunk at 0.

target speed was 10 knots.
this video is just an example, cause i never destroy neutral boat in my carrier ^^

you can see when the first part of the boat cross the 20 angle am not shooting yet the torpedo, i wait a little bit, cause that way i can destroy the exactly part of the boat that i want to destroy..

i explain my self bad, in the second video angle of shoot still 90 degres. what just change here is speed target, torpedo speed still 28 knots one.

05-10-15, 06:36 AM	#14
sENoZ Stowaway Posts: n/a Downloads: Uploads:	target speed 10 torpedo speed 28 10 + 28 = 38 So how does 38 shooting angle come close to the 20 you used? The only way you can make it fit together is by using a different formula. You are wrong .... speed target of 10 and torpedo of 28 knots = angle of shoot of 20 not 38 lol ..... angle of shoot of 38 will be for a target of 22 knots. on my video i shoot the torpedo at 348. the torpedo arrive to the boat at 0 angle so. angle of shoot was 12 degres. 348 + 12 you arrive to the 0 degres. front of your u-boat so. 12 degres that means a target of 6 knots around. 5 to 6. again..... Distance is not taking in consideration will using my method. shooting at 1km or 10 or more ...... is the same. Until you got the real speed of target and know your speed of torpedo = you have a angle of shoot + being at a correct angle = 90 degres = you shoot the boat same as shown in my video. >>>> Distance does not change the amout of lead, or the torpedo shooting angle if you will, but it does require your aiming to be more accuracte. dont need to aim more accurate or else...... speed target + speed of torpedo + being at 90 D = angle of shoot ^^ you can miss your target if you did do a wrong calcul of the speed target nothing else. example in that video ( when am was not using magui as a gui ) good one btw ^^ angle of shooting here is 20. you see i shoot the torpedo a 20 the boat get sunk at 0. target speed was 10 knots. this video is just an example, cause i never destroy neutral boat in my carrier ^^ you can see when the first part of the boat cross the 20 angle am not shooting yet the torpedo, i wait a little bit, cause that way i can destroy the exactly part of the boat that i want to destroy.. i explain my self bad, in the second video angle of shoot still 90 degres. what just change here is speed target, torpedo speed still 28 knots one. Last edited by sENoZ; 05-10-15 at 07:30 AM.