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Making a box move left to right over and over

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Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.

7 replies to this topic

#1Anddos  Members

Posted 17 August 2013 - 01:40 PM

Ok i am getting stuck with the maths on how to make a box move left to right over and over none stop , i understand you change the position by the velocity but i dont understand how to get the if conditions right so it actually moves none stop,here is what i have

bool hitleft = false;
bool hitright = false;
if(hitleft == false)
{
velocity = 0.01f;
Pos.x = Pos.x + velocity;

if(Pos.x == 2.0f)
{
hitleft = true;
}
}
if(hitright == false)
{
velocity = -0.01f;
Pos.x = Pos.x + velocity;
if(Pos.x == -2.0f)
{
hitright = true;
}
}

d3ddev->Clear(0, NULL, D3DCLEAR_TARGET, D3DCOLOR_XRGB(0, 0, 0), 1.0f, 0);

d3ddev->Clear(0, NULL, D3DCLEAR_ZBUFFER, D3DCOLOR_XRGB(0, 0, 0), 1.0f, 0);

d3ddev->BeginScene(); D3DXMatrixTranslation(&matTrans,Pos.x+velocity,Pos.y,Pos.z);

d3ddev->SetTransform(D3DTS_WORLD, &(matTrans));

box->DrawSubset(0); d3ddev->EndScene();

d3ddev->Present(NULL, NULL, NULL, NULL);

Edited by Anddos, 17 August 2013 - 01:55 PM.

#2Medo3337  Members

Posted 17 August 2013 - 01:55 PM

Do you mean rotating the box?

If so, you can use D3DXMatrixRotationYawPitchRoll():

http://msdn.microsoft.com/en-us/library/windows/desktop/bb205361(v=vs.85).aspx

#3belfegor  Members

Posted 17 August 2013 - 02:56 PM

static D3DXVECTOR3 dir(1.0f, 0.0f, 0.0f);
pos = pos + dir * moveSpeed * dt;

if(pos.x > maxX)
{
dir.x = -1.0f
}
else if(pos.x < minX)
{
dir.x = 1.0f
}


Edited by belfegor, 17 August 2013 - 02:56 PM.

#4Tom KQT  Members

Posted 17 August 2013 - 03:35 PM

To fix your original code a bit:

// once:
float velocity = 0.01f;

// ----------------------------------------------------------------

// each frame:
if (Pos.x >= 2.0f || Pos.x <= -2.0f) {
velocity = -velocity;
}
Pos.x = Pos.x + velocity;

d3ddev->Clear(0, NULL, D3DCLEAR_TARGET, D3DCOLOR_XRGB(0, 0, 0), 1.0f, 0);
d3ddev->Clear(0, NULL, D3DCLEAR_ZBUFFER, D3DCOLOR_XRGB(0, 0, 0), 1.0f, 0);
d3ddev->BeginScene(); D3DXMatrixTranslation(&matTrans, Pos.x, Pos.y, Pos.z);
d3ddev->SetTransform(D3DTS_WORLD, &(matTrans));
box->DrawSubset(0); d3ddev->EndScene();
d3ddev->Present(NULL, NULL, NULL, NULL);


It's much shorter

• Never use == with float variables. Your condition if(Pos.x == 2.0f) can actually never be true, because even if you think the value is 2.0f, it can be something like 2.0000001f or 1.9999998f. It's the typical floating point precision problem. In your case, you must use <= and >= to check whether the position isn't at OR BEHIND the border.
• The code should work, but unless you have a fixed timestep, the box will move at different speed on different computers. And even if you have fixed timestep, you're doing it wrong, because you in fact cannot fix the timestep with 100 % certainty. The right way to do it isn't pos = pos + velocity, but pos = pos + velocity * dt, where dt is the time difference between two frames. If you do it this way, then suddenly velocity isn't a magic number 0.01f without obvious meaning, but it is a real physical velocity in [your_length_units / s]. Dt is a small number, so your new velocity won't be 0.01f anymore, it will be something higher. If your working are is -2 to +2 and if you wanted the box to move from left to right in for example 5 seconds, then velocity would be 0.8f and the box would travel for 5 seconds regardless the CPU and anything.

Edited by Tom KQT, 17 August 2013 - 03:40 PM.

#5Anddos  Members

Posted 18 August 2013 - 12:00 AM

static D3DXVECTOR3 dir(1.0f, 0.0f, 0.0f);
pos = pos + dir * moveSpeed * dt;

if(pos.x > maxX)
{
dir.x = -1.0f
}
else if(pos.x < minX)
{
dir.x = 1.0f
}


That works perfectly, i wouldnt of thought it would be such simple code

#6Tom KQT  Members

Posted 18 August 2013 - 03:11 AM

static D3DXVECTOR3 dir(1.0f, 0.0f, 0.0f);
pos = pos + dir * moveSpeed * dt;

if(pos.x > maxX)
{
dir.x = -1.0f
}
else if(pos.x < minX)
{
dir.x = 1.0f
}


That works perfectly, i wouldnt of thought it would be such simple code

It can be even simpler with less redundant data:

static D3DXVECTOR3 dir(1.0f, 0.0f, 0.0f);
pos = pos + dir * moveSpeed * dt;

if(pos.x > maxX || pos.x < minX)
{
dir.x = -dir.x;
}

This way you are not repeating the number 1.0f (for the velocity value) three times in the code.

Btw, the code has one problem. Pos.x can go behind the maxX and minX values, because you modify it, then check whether it is behind the threshold and if it is you change the velocity. But position already is behind and you render it this way.

Edited by Tom KQT, 18 August 2013 - 03:15 AM.

#7Anddos  Members

Posted 18 August 2013 - 09:09 AM

How would you check floating point numbers for accurate values such as 1.0f,2.0f etc , is there anyway or do you have to use < > all the time?

#8Tom KQT  Members

Posted 18 August 2013 - 09:44 AM

If you need to compare a float variable with some specific number (of course you cannot use < or > in all cases), you need to use some tolerance. Simply said - you don't ask whether it is exactly 2.0f, but whether it is close enough to 2.0f.

if (a == 2.0f)


You write something like:

if (abs(a - 2.0f) < 0.00001f)


It's good to make a function for this, it also helps for readability, then it can be for example:

if (IsEqual(a, 2.0f))


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