Vector me This

[Chipcrap]

OK this is a VERY basic question. A vector is a line on a graph representing something (Velocity, Displacement, ect...). Which is the best way to define a Vector:

1. Vector1 = (x1,y1,x2,y2) or

2. Vector1 = x1,y1, degree, magnitude


I working on a basic physics sim, so which would be easier? Or are both fine? Or are both wrong?

[Shrog]

Both are fine from a design principle point of view, but neither are syntactically correct for VB.

How are you planning to do it in VB?

Shrog

[kedaman]

not really, a vector is a one value per dimension measurement, therefore a line can either be a position vector and a offset vector or two position vectors. The second is more commonly used for lines, but since you use velocity, which is derived from displacement, you should use an offset vector.

[Chipcrap]

Kedamin- Using your example, say I had Rock at XY Location 0,16 (Y Being Verticle Axis). I want to find out the rocks location in Realtime using Gravity. At time 0 the rock is 16 meters high. So could I use the following 2 vectors to determine rocks position using a calculation based on time elapsed?:

Dim Rock as Vector
Rock.X=0
Rock.Y=16

Dim Gravity As Vector
Gravity.X = 0?
Gravity.Y = -9.8

Do those two vectors look right? I'm not so sure about the Gravity. THANKS!

[kedaman]

yes, if the first is suppose to be a displacement and the second, gravity. gravity though is acceleration and derived from velocity, so displacement is velocity*time+0.5*acceleration*time^2

[SurfDemon]

Just to throw in my two pennies, you could have a class Vectore which handled bot the 2 point vector and the 1 point, angle/magnitude.

Changing one set of parameters would automatically change the other.

e.g.

Vector1.x1=0
.y1=0
.x2=1
.y2=0

Reading the .angle for this would give 90 (if working in degrees) and .magnitude would give 1

Changing .angle to 0 would change .x2=0 and .y2=1 (assuming 0,0 is bottom left).

I don't know if this helps matters,

SD

[Chipcrap]

OK So I have:

Private Type Vector
x As Double
y As Double
End Type

Private Type xyObject
Displacement As Vector
Velocity As Vector
End Type

Dim Rock As xyObject
Rock.Displacement.x = 0
Rock.Displacement.y = 16
Rock.Velocity.x = 0
Rock.Velocity.y = 0
Dim Gravity As Double
Dim Time As Double

CurrentVel = Rock.Velocity.y + (Gravity * Time)
CurrentY = Rock.Displacement.y + ((CurrentVel + Rock.Velocity.y) / 2 * Time)

It works, but I'm not sure if it is right. Do these equations look right?

[kedaman]

Are you operating in 2 dimensions at all? because otherways drop the vector concept and work with scalars

[Chipcrap]

LOL! Yes-eventually---I want to simulate an object falling straight down first, and then when I get that right I'm going to add an x direction, and then a Z. So do those look right?

[kedaman]

well not really, youre a bit incosistent, having scalar results, are you just doing an analysis or is this a simulation?

[Chipcrap]

It's a simulation

[kedaman]

then you have a loop or a timer procedure in which you
1. add acceleration to velocity
2. add velocity to displacement
and vector addition is done by adding each component .x+.x and .y+.y and so on

[Chipcrap]

Like so:

Do Until Rock.Displacement.y < -0.01
Time = modPhysics.GetTime(False) 'This tells how much time has passed since last loop.

Rock.Velocity.y = Rock.Velocity.y + (Gravity * Time)
Rock.Displacement.y = Rock.Displacement.y + ((2 * Rock.Velocity.y + Gravity * Time) / 2 * Time)

Loop

I got those EQ's out of a physics book and they seem to work. I just don't get why you have to add Gravity*Time to the Rocks Velocity in both EQ's. Is this what you mean?

[kedaman]

more like

Rock.Velocity.y = Rock.Velocity.y + Gravity
Rock.Displacement.y = Rock.Displacement.y + Rock.Velocity.y

assuming one time unit per loop, where time unit is the measurement for velocity mt^-1 and acceleration mt^-2

[SurfDemon]

Originally posted by kedaman
Are you operating in 2 dimensions at all? because otherways drop the vector concept and work with scalars

Mmmmmmmm, yes, but wouldn't this be a contrarian approach?

SD

[kedaman]

maybe sD .p