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Originally Posted by John Pancoast
Thanks for the great aid.........but question. You stated you noticed you and the target would arrive at the intercept point at the same time.
How do you know what time the target will get there ? His est speed/distance to point ?
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It's because this method solves for an intercept point using plots based on the same 'distance per kt of speed'. In essance you're solving:
"What course do I need to set so that the target and I arrive at the SAME POINT at the SAME TIME?"
Without going into the underlying trig, your intercept course will get you at YOUR speed to the same point the target will get at ITS speed in an identical amount of time - hence you will get there together. Plotted/calculated perfectly, you would actually collide.....:rotfl:
That's why the 'intercept speed x initial proportionality measure' circle is centred on the initial plot line for the target:
i.e. 1km per kt of speed of target = 6km line based on target's speed of 6kts
Hence, at an intercept speed of 12kts the 'proportionate circle' is a radius of 12 x 1 = 12kms, and this is centred on the point along the target plot 6km from initial plot point.
Hope that answered your question without making it more confusing!