Convert centimeters to degrees












0














I am working on an autonomous robot. This robot can recognize an object and pick it up. I have a pickup algorithm already.
$$x=Dcos(θ) quad text{and} quad y=Dsin(θ),$$ where $D$ is the length of my robot arm and $θ$ is the angle that it moved in radians.



This code gives an understanding of where the arm is on the Cartesian plane. I have an object detection system and using ultrasonic sensor I want it to stop $2 ,text{cm}$ before the object. If the robot is located at the origin, $(0, 0)$, and the object is $2 ,text{cm}$ from me, I want to know its $(x, y)$ coordinates. How do I convert from centimeters to degrees so I can know the object's $x$ and $y$ values?










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  • Maybe I don't get your question, but you can use the fact that a radian has a length equal to the radius, in your case, the length of the robot arm, D1.
    – Patricio
    Dec 1 at 4:15










  • I get your point but the length of the arm can be smaller
    – Kyrylo Kalashnikov
    Dec 1 at 4:18










  • OK, I guess that's the beauty of radians. Whatever the length of the arm (whether it is fully extended or not), the length of the radian in cm is equal to that of the radius. A fully extended arm, say 100 cms long, moved 1 radian each will travel precisely 100 cms. If the arm is contract to just 10 cms long, it will travel by just 10 cms.
    – Patricio
    Dec 1 at 4:26
















0














I am working on an autonomous robot. This robot can recognize an object and pick it up. I have a pickup algorithm already.
$$x=Dcos(θ) quad text{and} quad y=Dsin(θ),$$ where $D$ is the length of my robot arm and $θ$ is the angle that it moved in radians.



This code gives an understanding of where the arm is on the Cartesian plane. I have an object detection system and using ultrasonic sensor I want it to stop $2 ,text{cm}$ before the object. If the robot is located at the origin, $(0, 0)$, and the object is $2 ,text{cm}$ from me, I want to know its $(x, y)$ coordinates. How do I convert from centimeters to degrees so I can know the object's $x$ and $y$ values?










share|cite|improve this question
























  • Maybe I don't get your question, but you can use the fact that a radian has a length equal to the radius, in your case, the length of the robot arm, D1.
    – Patricio
    Dec 1 at 4:15










  • I get your point but the length of the arm can be smaller
    – Kyrylo Kalashnikov
    Dec 1 at 4:18










  • OK, I guess that's the beauty of radians. Whatever the length of the arm (whether it is fully extended or not), the length of the radian in cm is equal to that of the radius. A fully extended arm, say 100 cms long, moved 1 radian each will travel precisely 100 cms. If the arm is contract to just 10 cms long, it will travel by just 10 cms.
    – Patricio
    Dec 1 at 4:26














0












0








0







I am working on an autonomous robot. This robot can recognize an object and pick it up. I have a pickup algorithm already.
$$x=Dcos(θ) quad text{and} quad y=Dsin(θ),$$ where $D$ is the length of my robot arm and $θ$ is the angle that it moved in radians.



This code gives an understanding of where the arm is on the Cartesian plane. I have an object detection system and using ultrasonic sensor I want it to stop $2 ,text{cm}$ before the object. If the robot is located at the origin, $(0, 0)$, and the object is $2 ,text{cm}$ from me, I want to know its $(x, y)$ coordinates. How do I convert from centimeters to degrees so I can know the object's $x$ and $y$ values?










share|cite|improve this question















I am working on an autonomous robot. This robot can recognize an object and pick it up. I have a pickup algorithm already.
$$x=Dcos(θ) quad text{and} quad y=Dsin(θ),$$ where $D$ is the length of my robot arm and $θ$ is the angle that it moved in radians.



This code gives an understanding of where the arm is on the Cartesian plane. I have an object detection system and using ultrasonic sensor I want it to stop $2 ,text{cm}$ before the object. If the robot is located at the origin, $(0, 0)$, and the object is $2 ,text{cm}$ from me, I want to know its $(x, y)$ coordinates. How do I convert from centimeters to degrees so I can know the object's $x$ and $y$ values?







algebra-precalculus trigonometry






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edited Dec 1 at 5:52









Saad

19.7k92252




19.7k92252










asked Dec 1 at 4:05









Kyrylo Kalashnikov

32




32












  • Maybe I don't get your question, but you can use the fact that a radian has a length equal to the radius, in your case, the length of the robot arm, D1.
    – Patricio
    Dec 1 at 4:15










  • I get your point but the length of the arm can be smaller
    – Kyrylo Kalashnikov
    Dec 1 at 4:18










  • OK, I guess that's the beauty of radians. Whatever the length of the arm (whether it is fully extended or not), the length of the radian in cm is equal to that of the radius. A fully extended arm, say 100 cms long, moved 1 radian each will travel precisely 100 cms. If the arm is contract to just 10 cms long, it will travel by just 10 cms.
    – Patricio
    Dec 1 at 4:26


















  • Maybe I don't get your question, but you can use the fact that a radian has a length equal to the radius, in your case, the length of the robot arm, D1.
    – Patricio
    Dec 1 at 4:15










  • I get your point but the length of the arm can be smaller
    – Kyrylo Kalashnikov
    Dec 1 at 4:18










  • OK, I guess that's the beauty of radians. Whatever the length of the arm (whether it is fully extended or not), the length of the radian in cm is equal to that of the radius. A fully extended arm, say 100 cms long, moved 1 radian each will travel precisely 100 cms. If the arm is contract to just 10 cms long, it will travel by just 10 cms.
    – Patricio
    Dec 1 at 4:26
















Maybe I don't get your question, but you can use the fact that a radian has a length equal to the radius, in your case, the length of the robot arm, D1.
– Patricio
Dec 1 at 4:15




Maybe I don't get your question, but you can use the fact that a radian has a length equal to the radius, in your case, the length of the robot arm, D1.
– Patricio
Dec 1 at 4:15












I get your point but the length of the arm can be smaller
– Kyrylo Kalashnikov
Dec 1 at 4:18




I get your point but the length of the arm can be smaller
– Kyrylo Kalashnikov
Dec 1 at 4:18












OK, I guess that's the beauty of radians. Whatever the length of the arm (whether it is fully extended or not), the length of the radian in cm is equal to that of the radius. A fully extended arm, say 100 cms long, moved 1 radian each will travel precisely 100 cms. If the arm is contract to just 10 cms long, it will travel by just 10 cms.
– Patricio
Dec 1 at 4:26




OK, I guess that's the beauty of radians. Whatever the length of the arm (whether it is fully extended or not), the length of the radian in cm is equal to that of the radius. A fully extended arm, say 100 cms long, moved 1 radian each will travel precisely 100 cms. If the arm is contract to just 10 cms long, it will travel by just 10 cms.
– Patricio
Dec 1 at 4:26










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You need one other piece of data. You know you are $2$ cm from the object but you have not told us you know what angle it is at. Without that it could be anywhere on a $2$ cm circle around your current point. It sounds like you are approaching it, so it can't be behind you but that still leaves a large range. You know $D=2$ but you have no data to evaluate $theta$ in what you have given.






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    You need one other piece of data. You know you are $2$ cm from the object but you have not told us you know what angle it is at. Without that it could be anywhere on a $2$ cm circle around your current point. It sounds like you are approaching it, so it can't be behind you but that still leaves a large range. You know $D=2$ but you have no data to evaluate $theta$ in what you have given.






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      You need one other piece of data. You know you are $2$ cm from the object but you have not told us you know what angle it is at. Without that it could be anywhere on a $2$ cm circle around your current point. It sounds like you are approaching it, so it can't be behind you but that still leaves a large range. You know $D=2$ but you have no data to evaluate $theta$ in what you have given.






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        You need one other piece of data. You know you are $2$ cm from the object but you have not told us you know what angle it is at. Without that it could be anywhere on a $2$ cm circle around your current point. It sounds like you are approaching it, so it can't be behind you but that still leaves a large range. You know $D=2$ but you have no data to evaluate $theta$ in what you have given.






        share|cite|improve this answer












        You need one other piece of data. You know you are $2$ cm from the object but you have not told us you know what angle it is at. Without that it could be anywhere on a $2$ cm circle around your current point. It sounds like you are approaching it, so it can't be behind you but that still leaves a large range. You know $D=2$ but you have no data to evaluate $theta$ in what you have given.







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        answered Dec 1 at 6:12









        Ross Millikan

        291k23196370




        291k23196370






























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