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Glad it helped.

Thank you - glad you found this helpful.

Yes this theodolite can be used with the stadia lines on a survey staff. Typically the distance from the instrument to the rod can be determined quickly by subtracting the bottom stadia reading from the top reading and multiplying the difference by 100. However please pay attention, the stadia distances have a low level of accuracy, a one mm error in staff reading gives a distance error of 0.1 metre

I had cramped a lot into 60 seconds. Remember - take your time, practise using your own calculator and this short is there to help.

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Glad you found this useful.

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Thank you.

Fantastic! It still amazes me how many views this series of videos get.

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Thank you.

Thank you for your feedback. Hope it was useful for you. If you want my teaching notes on this topic and several more, they are available on my website www.applied-maths.co.uk

Thank you - hope you have passed that test.

Thank you

A syntax error means that the command used contains an error. Look for misplaced functions, parentheses, commas, or arguments. Re type your calculation paying attention to the functions used to ensure none are missing or mistyped or misplaced.

Thank you. Glad it was helpful!

The Ah2 is part of the parallel axis theorem and is not needed in this application. This is because the axis of the two shapes coincide together at the same point which is at the overall centre of the shape. The parallel axis theorem is only applied when you have this offset.

@@applied-maths Ohh now I understand. When you can please do a question that includes the Ah2 because it is more complex. Thank you so much.

There's an example here th-cam.com/video/jipAxL-lwws/w-d-xo.html My student notes are also available here www.applied-maths.co.uk/home/structures#h.hz86l7d22dzv

Young modulus works out as Stress/Strain and this example you can calculate it as 0.166 N/mm2. This would indicate that the bar is composed of a mild steel material. The bar is 500mm long and 8mm in diameter.

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Thank you

Topcon Digital Theodolite DT-106

Topcon Digital Theodolite DT-106

Calculated, the strain is 100/70,000 = 0.0014238 or 1.4238 x 10^-3. A small rounding error on my part.

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The original length is 500mm and the bar was stretched to 503mm, therefore the change in length is 503 - 500 = 3mm.

The original length is given tin the problem as 500mm and the bar was stretched to 503mm, therefore the change in length is 503 - 500 = 3mm.

Thank you - glad you found it helpful.

Your welcome

Thank you. Hope it was useful.

I appreciate that!

Apologies for any errors - I was one of my first videos to upload to TH-cam. Check out my other video for booking the angles from a 4 point closed traverse using measured values. th-cam.com/video/igFLqsEuTHU/w-d-xo.htmlsi=8Ot7bHyg17HX1EwD then there’s part 2 for calculating the mean and then corrected angles. th-cam.com/video/yjiLb_gQiUQ/w-d-xo.htmlsi=4sMVddJJQjJABK88 - there’s a playlist with all the series of videos needed to complete the coordinate calculations as well. Thank you

218deg is the correct angle. It is used at the end to calculate the average angle.

If you watch my other video, it shows you which angles to record and how to book them correctly and also the calculations necessary at the very end. th-cam.com/video/QnbU5jT6PKY/w-d-xo.htmlsi=VIJuwhayVSTyzzym

@@applied-maths Thanks! So if we are measuring the internal angles of a triangle, I would set my initial angle as 0 from vertex A to B, then turn to the next vertex C, and record the angle, then change face and record the angle at vertex C again? Im confused about turning it 180 degrees to change face, how is doing that giving an accurate angle? Is it using the line AB as the reference still? Does it matter if i rotate it left or right when turning it 180 degrees? And when turning it to another point, does it matter if i turn it around left or right? Wouldn’t turning it one way measure the internal angle, and the other way measure the external angle?

The face left and face right measurements are to do with the accuracy / errors of the instrument. This explains it - The “Left” “Right” readings will tend to cancel out errors in the alignment of the telescope with the verniers or in more modern instruments the encoder disks. What this means is after a proper shop adjustment, the centerline of the telescope and the “Zero” mark on the verniers/encoders will be in as near perfect alignment as practical. There are ALWAYS systematic errors we must deal with. In this case, if you sight a line turning an angle and read 23 degrees 42 minutes 30 seconds with only one face. You will never know if your instrument is properly aligned - if you are very careful and do both faces you might see 23 degrees 42 minutes 30 seconds on the right face and 23 minutes 41 minutes 50 seconds on the left face. This 40-second difference would show that the centerline of the telescope is about 20 seconds out of alignment with the verniers or encoders. But the average of the two faces should give a fairly true reading. I measure angles in the clockwise direction. It does not matter to turn your instrument clockwise or anti-clockwise when taking the next set of measurements.

@@digitalsurveyor thank you! So turning the UTS around the long way to the next point won’t actually record external angle?

No it shouldn’t. Just make sure that you zero your instrument looking at your first station and then turn to look at the next station and record your angle. Just beware then I used an older theodolite, it could measure an angle as an acute angle in both the clockwise and anti-clockwise directions or rotation - there was a specific mode - this may or may not apply to your equipment.

Thank you.

Sorry but no hack - just how to use the powers or order function on your calculator

Sorry but I can’t help you with that one.

@@applied-maths yes sir

You're welcome 😊

If you need to calculate the second moment of area about another axis not through the centre of the shape, then you would need to use the parallel axis theorem. This is calculated as the second moment of area of a given shape about its own centre + Ah^2, where h represents the offset between parallel axis. There is some info on this on this website. www.applied-maths.co.uk/home/structures#h.hz86l7d22dzv

If you don’t understand- try to review my lecture notes for this topic which are available here. www.applied-maths.co.uk/home/structures#h.qpdb0qwbaupr

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Thank you. Hope it helps.

Glad it was helpful!

Glad it was helpful!

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Thank you

Thank you!!

Thank you