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fixed pluralisation and one complete rotation a turn e.g. 1 turn of the screw.
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Based on @TwoBitOperation's answer

and assuming his feet are 25 centimetercentimeters apart, if he walks $n$ circles with a radius of $r$ meters in a single direction, one feetfoot walks $2 \pi r n$ meters, and the other one $2 \pi (r + 0.25) n$. The difference, $\pi n / 2$ is 100 meters, so $n = 200 / \pi \approx 63.6$. The value of $r$ is then determined from $2 \pi r n = 3000$ so $2 r = 15$ and $r$ = 7.5 metermeters.

Whether these

63.6 roundsturns around e.g. a fountain or pond with a 15m diameter constitute a "normal walk" remains an open discussion.

Based on @TwoBitOperation's answer

and assuming his feet are 25 centimeter apart, if he walks $n$ circles with a radius of $r$ meters in a single direction, one feet walks $2 \pi r n$ meters, and the other one $2 \pi (r + 0.25) n$. The difference, $\pi n / 2$ is 100 meters, so $n = 200 / \pi \approx 63.6$. The value of $r$ is then determined from $2 \pi r n = 3000$ so $2 r = 15$ and $r$ = 7.5 meter.

Whether these

63.6 rounds around e.g. a fountain or pond with a 15m diameter constitute a "normal walk" remains an open discussion.

Based on @TwoBitOperation's answer

and assuming his feet are 25 centimeters apart, if he walks $n$ circles with a radius of $r$ meters in a single direction, one foot walks $2 \pi r n$ meters, and the other one $2 \pi (r + 0.25) n$. The difference, $\pi n / 2$ is 100 meters, so $n = 200 / \pi \approx 63.6$. The value of $r$ is then determined from $2 \pi r n = 3000$ so $2 r = 15$ and $r$ = 7.5 meters.

Whether these

63.6 turns around e.g. a fountain or pond with a 15m diameter constitute a "normal walk" remains an open discussion.

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Glorfindel
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Based on @TwoBitOperation's answer

and assuming his feet are 25 centimeter apart, if he walks $n$ circles with a radius of $r$ meters in a single direction, one feet walks $2 \pi r n$ meters, and the other one $2 \pi (r + 0.25) n$. The difference, $\pi n / 2$ is 100 meters, so $n = 200 / \pi \approx 63.6$. The value of $r$ is then determined from $2 \pi r n = 3000$ so $2 r = 15$ and $r$ = 7.5 meter.

Whether these

63.6 rounds around e.g. a fountain or pond with a 15m diameter constitute a "normal walk" remains an open discussion.

Based on @TwoBitOperation's answer

and assuming his feet are 25 centimeter apart, if he walks $n$ circles with a radius of $r$ meters in a single direction, one feet walks $2 \pi r n$ meters, and the other one $2 \pi (r + 0.25) n$. The difference, $\pi n / 2$ is 100 meters, so $n = 200 / \pi \approx 63.6$. The value of $r$ is then determined from $2 \pi r n = 3000$ so $2 r = 15$ and $r$ = 7.5 meter.

Based on @TwoBitOperation's answer

and assuming his feet are 25 centimeter apart, if he walks $n$ circles with a radius of $r$ meters in a single direction, one feet walks $2 \pi r n$ meters, and the other one $2 \pi (r + 0.25) n$. The difference, $\pi n / 2$ is 100 meters, so $n = 200 / \pi \approx 63.6$. The value of $r$ is then determined from $2 \pi r n = 3000$ so $2 r = 15$ and $r$ = 7.5 meter.

Whether these

63.6 rounds around e.g. a fountain or pond with a 15m diameter constitute a "normal walk" remains an open discussion.

Source Link
Glorfindel
  • 28.2k
  • 9
  • 98
  • 143

Based on @TwoBitOperation's answer

and assuming his feet are 25 centimeter apart, if he walks $n$ circles with a radius of $r$ meters in a single direction, one feet walks $2 \pi r n$ meters, and the other one $2 \pi (r + 0.25) n$. The difference, $\pi n / 2$ is 100 meters, so $n = 200 / \pi \approx 63.6$. The value of $r$ is then determined from $2 \pi r n = 3000$ so $2 r = 15$ and $r$ = 7.5 meter.