topic Re: how to loft with tangent arc? in Inventor Forum

how to loft with tangent arc?

jureumin — Sun, 19 Nov 2023 11:14:11 GMT

i’m new to this software and i have this project. i don’t know how to loft with only given radius, any idea how to do this? thank you!

Re: how to loft with tangent arc?

89198826955 — Sun, 19 Nov 2023 14:12:20 GMT

here is a snapshot with the build browser

Re: how to loft with tangent arc?

jureumin — Sun, 19 Nov 2023 14:30:24 GMT

thank you, will try this ^^ may i know the order to do this? did you first make the polygons and loft it? how about the radius of the arc? i’m having a hard time doing it with the available dimensions

Re: how to loft with tangent arc?

89198826955 — Sun, 19 Nov 2023 14:53:52 GMT

no, I did it through the surfaces and sculpture

Re: how to loft with tangent arc?

89198826955 — Sun, 19 Nov 2023 15:12:24 GMT

here are the sketches

Re: how to loft with tangent arc?

SBix26 — Mon, 20 Nov 2023 00:25:32 GMT

Loft is not the right solution if a simpler method will do the job. A radiused face is a simple cylinder, whereas a loft is usually far more complex and less well defined.

The attached file is my (Inventor 2024) solution to these models:

I notice that the octagonal part is not completely defined by the drawing, so I guessed at what might be appropriate for the top width across flats.

Sam B

Inventor Pro 2024.2 | Windows 10 Home 22H2

Re: how to loft with tangent arc?

jureumin — Mon, 20 Nov 2023 02:06:37 GMT

thank you so much!

Re: how to loft with tangent arc?

jureumin — Mon, 20 Nov 2023 02:06:57 GMT

thank you!

Re: how to loft with tangent arc?

johnsonshiue — Tue, 21 Nov 2023 22:01:59 GMT

Hi! This can be done using Guide Rail Sweep also (one profile, one straight path and one guide rail to scale the profile).

Many thanks!

Re: how to loft with tangent arc?

89198826955 — Tue, 21 Nov 2023 22:20:53 GMT

By Murray Gell-Mann (The Quark and the Jaguar) I received a call from a colleague about a student. He felt he had to give him a zero on a physics question, whereas the student demanded a 20. The teacher and the student agreed to choose an impartial arbiter and I was chosen. I read the exam question: Show how it is possible to determine the height of a building using a barometer. The student replied: You take the barometer from the top of the building, attach a rope to it, drag it down to the ground, then pull it up and measure the length of the rope. The length of the rope gives the height of the building. The student was right since he had answered the question correctly and completely. On the other hand, I could not give him his points: in this case, he would have graduated in physics when he had not shown me any knowledge of physics. I offered to give the student another chance. saying giving him six minutes to answer the question with the caveat that for the answer he had to use his knowledge of physics. After five minutes, he still hadn't written anything. I asked him if he wanted to give up but he replied that he had many answers for this problem and was looking for the best one. I apologized for interrupting him and asked him to continue. In the minute that followed, he hastened to answer me: — We place the barometer at the height of the roof. It is dropped by measuring its fall time with a stopwatch. Then using the formula x = 1/2*g*t*t, we find the height of the building. At that moment, I asked my colleague if he wanted to give up. He replied in the affirmative and gave the student almost 20. As I left his office, I called the student back as he said he had several solutions to this problem. 'Well,' he said, 'there are several ways to calculate the height of a building with a barometer. For example, it is placed outside when it is sunny. We measure the height of the barometer, the length of its shadow and the length of the shadow of the building. Then, with a simple calculation of proportion, we find the height of the building. - Good, I replied, and the others. — There is a fairly basic method that you will appreciate. We go up the floors with a barometer and at the same time we mark the length of the barometer on the wall. By counting the number of lines, we have the height of the building in barometer length. It is a very direct method. Of course, if you want a more sophisticated method, you can hang the barometer from a rope, swing it like a pendulum, and determine the value of g at street level and at roof level. From the difference in g the height of the building can be calculated. In the same way, we attach it to a large rope and while being on the roof, we let it descend to about street level. It is made to swing like a pendulum and the height of the building is calculated from the period of the oscillations. Finally, he concludes: — There are still other ways to solve this problem. Probably the best is to go to the basement, knock on the concierge's door and say, I've got a great barometer for you if you tell me how tall the building is. I then asked the student if he knew the answer I was looking for. He admitted he was, but he was fed up with college and professors trying to teach him how to think. »

Re: how to loft with tangent arc?

89198826955 — Tue, 21 Nov 2023 22:34:56 GMT

your teacher will like a lot of solutions to this problem 😀