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### #ActualGuyWithBeard

Posted 15 May 2013 - 08:19 AM

Thanks, I'll have to look through that. I am using the coefficients for several things. For example, to get the x at a given t I can just do:

float Curve::QuadraticBezierSegment::getXAtT( float t ) const
{
// A*t^2 + B*t + C
return ( m_AX * t * t ) + ( m_BX * t ) + m_CX;
}


There might be some other way to do that too, I am not sure. Also, if I take the derivative of the formula and plug in the coefficients I can solve it and get local extrema, ie. the bounds of the curve. Again, I am not a math person, and there might be other ways to do this. If some come to mind, please feel free to share

### #2GuyWithBeard

Posted 15 May 2013 - 08:18 AM

Thanks, I'll have to look through that. I am using the coefficients for several things. For example, to get the t at a given t I can just do:

float Curve::QuadraticBezierSegment::getXAtT( float t ) const
{
// A*t^2 + B*t + C
return ( m_AX * t * t ) + ( m_BX * t ) + m_CX;
}


There might be some other way to do that too, I am not sure. Also, if I take the derivative of the formula and plug in the coefficients I can solve it and get local extrema, ie. the bounds of the curve. Again, I am not a math person, and there might be other ways to do this. If some come to mind, please feel free to share

### #1GuyWithBeard

Posted 15 May 2013 - 08:18 AM

Thanks, I'll have to look through that. I am using the coefficient for several things. For example, to get the t at a given t I can just do:

float Curve::QuadraticBezierSegment::getXAtT( float t ) const
{
// A*t^2 + B*t + C
return ( m_AX * t * t ) + ( m_BX * t ) + m_CX;
}


There might be some other way to do that too, I am not sure. Also, if I take the derivative of the formula and plug in the coefficients I can solve it and get local extrema, ie. the bounds of the curve. Again, I am not a math person, and there might be other ways to do this. If some come to mind, please feel free to share

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