OT: Looking for a good physics site

[Suicide Junkie]

It should probably be noted that constant acceleration is a basic premise of these formulae; to get a join-fleet, your acceleration will have to vary over time unless you are already in the same location (in which case, A1 = A2 for all time).

[Jack Simth]

Yeah - they're the dumb-fire (accelerates in a straight line in whatever direction you pointed it - doesn't make decisions in-flight) missile targeting solution (goal: HIT THE TARGET HARD) for a target that isn't evading (e.g., another dumb-fire missile). If the target is evading, or you can track target mid-flight, you want something a little more organic (such as a simple "accelerate in direction of current target's location")

[narf poit chez BOOM]

I was thinking I could probably get a matching intercpt by treating the targets' velocity and acceleration as 0, by adding them (Or perhaps subtracting them) from or to the enemies' velocity and acceleration.

Or something like that.

(1/2) * A * T^2 + V * T + P = D

A = 5
T = 3
P = 50
V = 15
D = 117.5

(1/2) * A / D^2 + V / D + P = T
1/2 * 5 / 117.5^2 + 15 / 117.5 + 50 = 3?
1/2 * 0.042 + 0.127 + 50 = 3?
0.021 + 0.127 + 50 = 3? // It was at this point that my suspicions became rather definite...
50.148 != 3

(1/2) * A * T^2 + V * T + P = D

(1/2) * A * D^2 + V * D + P = T
1/2 * 5 * 117.5^2 + 15 * 117.5 + 50 = 3?
2.5 * 13,806.25 + 1,762.5 + 50 = 3? // Yeah, didn't think so.
34,515.625 + 1,812.5 != 3
36,328.125 != 3 // Curiousity.

(1/2) * A * D^0.5 + V / D + P = T?
1/2 * 5 * 117.5^0.5 + 15 / 117.5 + 50 = 3?
2.5 * 10.84 + 0.128 + 50 = 3? // Erm, no...
77.228 != 3.

(1/2) * A * T^2 + V * T + P = D

(1/2) * A / D + V / D + P = T^2 + T
1/2 * 5 / 117.5 + 15 / 117.5 + 50 = 12?
1/2 * 0.043 + 0.128 + 50 = 12?
0.0215 + 0.128 + 50 = 12? // Er, no...
50.1495 != 12

Ok, my brain is frying...

[narf poit chez BOOM]

(1/2) * A * T^2 + V * T + P = D

A = 5
T = 3
P = 50
V = 15
D = 117.5

T^2 + T = (1/2) * A + V + P?
3^2 + 3 = 1/2 * 5 + 15 + 50?
9 + 3 = 2.5 + 15 + 50? // Erm, no...
12 = 17.5 + 50?
12 != 67.5

T^2 * T = (1/2) * A + V + P?
3^2 * 3 = 1/2 * 5 + 15 + 50?
9 * 3 = 2.5 + 15 + 50? // Fiddle, fiddle...
27 = 17.5 + 50?
27 != 67.5

T^2 * T = (1/2) * A + V + P + D?
3^2 * 3 = 1/2 * 5 + 15 + 50 + 117.5?
9 * 3 = 2.5 + 15 + 50 + 117.5? // ...Lute?...
27 = 17.5 + 50 + 117.5?
27 != 67.5 + 117.5?

T^2 * T = ((1/2) * A + V) * 2 + P + D?
3^2 * 3 = (1/2 * 5 + 15) * 2 + 50 + 117.5?
9 * 3 = (2.5 + 15) * 2 + 50 + 117.5? // ...Mumble...
27 = 17.5 * 2 + 50 + 117.5?
27 = 35 + 50 + 117.5?
27 != 202.5

T = (((1/2) * A + V) * 2 + P + D) / ((1/2) * A + V + P)?
3 = ((1/2 * 5 + 15) * 2 + 50 + 117.5) / (1/2 * 5 + 15 + 50)?
3 = ((2.5 + 15) * 2 + 50 + 117.5?) / (2.5 + 15 + 50)?// ...Hmm...
3 = (17.5 * 2 + 50 + 117.5) / (17.5 + 50)?
3 = (35 + 50 + 117.5) / (17.5 + 50)?
3 = 202.5 / 67.5
3 = 3!

A = 8
T = 4
P = 35
V = 25
(1/2) * A * T^2 + V * T + P = D

1/2 * 8 * 4^2 + 25 * 4 + 35 = D
4 * 16 + 100 + 35 = D
64 + 135 = D
199 = D

A = 8
T = 4
P = 35
V = 25
D = 199

T = (((1/2) * A + V) * 2 + P + D) / ((1/2) * A + V + P)?
4 = ((1/2 * 8 + 25) * 2 + 35 + 199) / (1/2 * 8 + 25 + 35)?
4 = ((4 + 25) * 2 + 234) / (4 + 60)?
4 = (29 * 2 + 234) / 64?
4 = (58 + 234) / 64?
4 = 292 / 64?
4 != 4.5625 // Nope...Error check...Still nope. So close, yet so far.

[Fyron]

Why are you randomly rewriting the equation? The way to solve a quadratic equation has already been posted..

Why are you starting with a complete set of values, including time and distance, that you plug in to verify the refactored equations? Where are they coming from?

[narf poit chez BOOM]

If you thought that was random, you need to look again. I was de-rusting my algebra skills. I know I wasn't using the method that was posted; that was not the point. I was working off the theory that if 'T' is used to solve the equation, then some permutation of the equation could be used to solve for T.

I did not start with a distance value. I calculated it. I use a complete set of values for the simple reason that I can use them to see if the refactored equation is the correct one.

[Fyron]

Getting T^3 as the only factor with T out of a quadratic polynomial looks entirely random to me.

[narf poit chez BOOM]

@Fyron: Hence why I need to de-rust my algebra skills. It's the circle of maaaaath...

@Jack: I did try google. More than once.

[AngleWyrm]

narf poit chez BOOM said:I was working off the theory that if 'T' is used to solve the equation, then some permutation of the equation could be used to solve for T.

Given: D = (1/2) * A * T^2 + V * T + P

T = (-V +/- Sqrt(2AD-2AP+V^2) )/A

A = 5
T = 3
P = 50
V = 15
D = 117.5

3=(-15 +/- Sqrt(2x5x117.5 - 2x5x50 + 15^2))/5
3=(-15 +/- Sqrt(1175 - 500 + 225))/5
3=(-15 +/- Sqrt(900))/5
3=( -15 +/- 30 )/5
3=( 15 or -45 )/5
3=( 3 or -9)

Also:
V = ((D-P)/T) - AT/2
P = D - (T(AT+2V))/2
A = -( (2(-D+P+TV))/T^2 )

--
Mathematica: The new calculator

[narf poit chez BOOM]

Thanks!