求解点在线段上的位置进度值（0.0-1.0范围）

No problem, let's figure out how to compute that progress value (let's call it t) you need. This value will sit between 0.0 and 1.0 if the point lies directly on the segment, and outside that range if it's beyond either end of the segment.

Core Mathematical Idea #

The line segment from (x1, y1) to (x2, y2) can be represented with a parametric equation:

P(t) = (x1 + t*(x2 - x1), y1 + t*(y2 - y1))

Where:

• t = 0.0 corresponds exactly to the start point (x1, y1)
• t = 1.0 corresponds exactly to the end point (x2, y2)
• Values of t < 0.0 mean the point is beyond the start, and t > 1.0 mean it's beyond the end.

To find t for your target point (x3, y3), we use vector projections (this works even if the point isn't perfectly on the line segment—we'll get the t value for the closest point on the line to your target):

Step-by-Step Calculation #

• Compute vectors:
  • Vector from start to end: dx = x2 - x1, dy = y2 - y1
  • Vector from start to target: vx = x3 - x1, vy = y3 - y1
• Calculate dot product and squared length:
  • Dot product of the two vectors: dot = vx*dx + vy*dy
  • Squared length of the segment vector: len_sq = dx*dx + dy*dy
• Compute t:
  • If the segment is a single point (len_sq = 0), we can define t = 0.0 (or handle this edge case based on your needs—like checking if the target equals the point)
  • Otherwise: t = dot / len_sq

Example Code (Python) #

Here's a reusable function that implements this logic:

def calculate_progress(x1, y1, x2, y2, x3, y3):
    dx = x2 - x1
    dy = y2 - y1
    vx = x3 - x1
    vy = y3 - y1
    
    len_sq = dx * dx + dy * dy
    
    # Handle case where segment is a single point
    if len_sq == 0:
        # Return 0.0 if target matches the point, else NaN
        return 0.0 if (vx == 0 and vy == 0) else float('nan')
    
    dot_product = vx * dx + vy * dy
    t = dot_product / len_sq
    
    return t

How It Works #

• If your target point is exactly on the segment, t will land cleanly between 0.0 and 1.0. For example, the midpoint will return 0.5.
• If the point is beyond (x2, y2), you'll get a value like 1.5 (meaning it's half the segment length past the end).
• If it's beyond (x1, y1), you'll get a negative value like -1.1.
• If the point isn't on the line at all, t corresponds to the closest point on the segment's infinite line to your target.

Edge Cases to Note #

• Horizontal segments: The function still works—dy will be 0, so the dot product simplifies to vx*dx, and len_sq is dx*dx, so t = (x3 - x1)/(x2 - x1) as you'd expect.
• Vertical segments: Similarly, dx is 0, so t = (y3 - y1)/(y2 - y1).
• Zero-length segments: The function checks for this and returns 0.0 if the target matches the point, or nan if not (you can adjust this behavior if needed).

内容的提问来源于stack exchange，提问作者user3024235