B Need an on-line calculator for a simple calculation

[guyburns]

This is a simple question, but after 30 minutes on Google, no sensible results.

I want to solve this equation: 1/(2.04Sin(29.36 + 0)) on a regular basis. My idea was to have that term as a Google search, saved as a bookmark. The idea has worked well for years for another formula, but without the Sine term.

So, I search and up comes Maths Solver, but it's in radians. It can be changed, but I couldn't get it to remember the formula. Several other attempts at other calculator sites, led me to no joy, and so I'm here.

Any suggestions for an on-line calculator that allows me to:

• Bookmark the search term so that I'm taken to a calculator ready to go, with the formula already in place.

• I change the 0 to the actual number I want, I press return, and there's my answer.

Thanks for any suggestions.

 

[Filip Larsen]

Perhaps Wolfram Alpha will suit you. There you can can do it like: https://www.wolframalpha.com/input?i=1/(2.04Sin(29.36+deg+++0+deg))

 

[apostolosdt]

“Ordinary” calculators executing such tasks should be programmable. Try any HP emulator, like HP-15C.


Finding one online might be tricky; the above suggestion is better.

 

[guyburns]

Thanks Filip. Your link works well, can be bookmarked, and saves the formula so it can be changed.

While I was trying the Maths solver from Google, a box popped up asking if I had any suggestions for improvement. My goodness, did I have some suggestions! Wasted more than 30 minutes of my time trying to get that dopey thing to work.

 

[jack action]

For Google, the angle must be in radians.

https://www.google.com/search?q=1/(2.04Sin(29.36 + 0))

If you want the angle in degrees:

https://www.google.com/search?q=1/(2.04Sin(29.36 deg + 0 deg))

OR:

https://www.google.com/search?q=1/(2.04Sin((29.36 + 0) deg))

 

[fresh_42]

This is a simple question, but after 30 minutes on Google, no sensible results.

I want to solve this equation: 1/(2.04Sin(29.36 + 0)) on a regular basis. My idea was to have that term as a Google search, saved as a bookmark. The idea has worked well for years for another formula, but without the Sine term.

So, I search and up comes Maths Solver, but it's in radians. It can be changed, but I couldn't get it to remember the formula. Several other attempts at other calculator sites, led me to no joy, and so I'm here.

Any suggestions for an on-line calculator that allows me to:

• Bookmark the search term so that I'm taken to a calculator ready to go, with the formula already in place.

• I change the 0 to the actual number I want, I press return, and there's my answer.

Thanks for any suggestions.

Here is a long list for any occasion:

https://www.physicsforums.com/threa...h-physics-earth-and-other-curiosities.970262/

 

[guyburns]

Thanks Jack Action. So Google can be made to work; just put in "deg" in the appropriate place. Google gives the result much quicker, but the Wolfram site is more flexible. I think I'll bookmark both.

Next problem
Given four values of V, I want to solve this equation for R…

V = 2.0397 • SIN (29.358º + Rº)

I tried doing it in the Wolfram site, but the answer always came back as 0.47 and I know that's wrong – or I'm doing something wrong.

How do I solve that equation for these values of V: 1.111, 1.388, 1.852, 2.5?

They are actually the inverse of more sensible figures: 0.9, 0.72, 0.54, 0.4 which I use in Premiere to scale an incoming image to full screen. If you want to know more, keep on reading.

Background
This equation is very useful to me. I should have gone into it years ago. I will be using it to set the Scale in Premiere. I produce slide shows based on slides and prints. Often the horizon or poles are not straight, and I insist on non-destructive processing of my scans. So how do I straighten in Photoshop non destructively, without rotating the image? I create a selection rectangle, rotate it to match the crooked horizon, save that selection rectangle, and export the selection to suit the Blu-rays I create. Then in Premiere, I rotate in the opposite direction. Voila – straight horizon!

The problem is, PS cannot crop to a sloping selection rectangle. It crops to the edges of the rectangle, so you end up with a larger image than you want. What size do I make the export, so that when I rotate the opposite way in Premiere, the vertical dimension is 1080P and the horizon is straight? That's why I need to know the vertical dimension of a rotated rectangle in PS. It can't be 1080P, it has to be larger, so that when I straighten in Premiere, the vertical dimension is exactly 1080P (or larger).

For various reasons I never export exactly at 1080P. For scans that need rotation, I have several standard dimensions: 1080, 1200, 1500, 2000. In the past I just guessed which one to use. Horizon at 3.7º in Photoshop? Should I use 1200 or 1500, so that when straightened in Premiere, the vertical pixels are more than 1080. Better to scale-down than scale-up because of quality issues.

If I can solve the equation, I will know before I export, what standard to use. The equation for 1920 x 1080 video has been given above. The equation has been normalised vertically to 1. If you put R=0 into the equation (no rotation), the answer should be 1.

I want to solve that equation for R, given the V for my various break points. For example, if I export at 1200P, I have 1200/1080 = 1.11 times as many pixels as I need vertically. If I sacrifice some because of rotation, that won't matter – as long as I have enough. What is the maximum rotation for that amount of scaling? I don't know. In the past I've just guessed. From experience I know it's about 4º. If I guess wrong, I have to do the whole lot again – the export from PS and the scaling in Premiere.

 

[jack action]

How do I solve that equation for these values of V: 1.111, 1.388, 1.852, 2.5?

https://www.wolframalpha.com/input?...397)+-+29.358+for+V+=1.111,+1.388,+1.852,+2.5

Or better, we set $V = \frac{P}{1080}$:

https://www.wolframalpha.com/input?....0397)+-+29.358+for+P+=1200,+1500,+2000,+2700

Note: the answer for $V = 2.5$ is a complex number because you cannot get a real value when $V > 2.0397$.

Here's the graphic solution for P = 1080 * 2.0397 * SIN (29.358º + Rº):

https://www.wolframalpha.com/input?...(29.358+++R)*pi/180)+for+-29.358+<+R+<+60.642

 

[MidgetDwarf]

Just download a calculator application.

I believe hp prime is free download on desktop, and a small fee on android/ipad ios.

 

[robphy]

Next problem
Given four values of V, I want to solve this equation for R…

V = 2.0397 • SIN (29.358º + Rº)

I tried doing it in the Wolfram site, but the answer always came back as 0.47 and I know that's wrong – or I'm doing something wrong.

How do I solve that equation for these values of V: 1.111, 1.388, 1.852, 2.5?

2.0397sin( (29.358+R) degrees)=1.111
https://www.wolframalpha.com/input?i=2.0397sin(+(29.358+R)+degrees)=1.111

R≈0.4735537190082645 (760.2654221687300 n + 7.69805360052734), n element Z
R≈0.4735537190082645 (760.2654221687300 n + 248.4346574838376), n element Z

Note that 0.4735537190082645*7.69805360052734 =3.64544191165

This gets the solution you probably want:
2.0397sin( (29.358+R) degrees)=1.111; 0<R<90
https://www.wolframalpha.com/input?i=2.0397sin(+(29.358+R)+degrees)=1.111;+0<R<90

R≈3.645441911654683

Possibly more useful (easier to modify):
https://www.wolframalpha.com/input?i=1.111=2.0397sin(+(29.358+R)+degrees);+0<R<90
or
https://www.wolframalpha.com/input?i=0<R<90;2.0397sin(+(29.358+R)+degrees)=V;V=1.111

---

Here's a desmos version that does it numerically (agreement to 3 decimal places)
2.0397\sin\left(29.358+R\right)\sim V
V=1.111
https://www.desmos.com/calculator/yylohjuqsd

R=3.64534141807

 

[guyburns]

Thanks for all the feedback. This is more complicated than I thought. I buggered up in a big way wanting a solution for 2.5. V can't get that big (but if it could, exporting at 2700P could handle it).

And there's something else I hadn't thought of. If a photo had the horizon at more than 45º, I wouldn't set up a selection rectangle to suit. I'd rotate the image 90º. The maximum angle I'd ever encounter is 45º.

I need to go back to the drawing board and give this some more thought. But I have the answers I was looking for.

• If the horizon is horizontal, I export at 1080P
• Up to 3.6º… 1200P (tallies with my guess of 4º)
• Between 3.6º - 13.5º… 1500P
• Between 13.5º - 35.9º… 2000P
• Between 35.9º - 45º… 2700P
• At 45º there is a discontinuity. Above that angle I rotate the image 90º first.

 

[pbuk]

Why don't you just use the spirit level tool in photoshop: this is what it was designed for?

 

[guyburns]

I'm just about to try the "spirit level" technique. Why I didn't try it first up is because… well, you did ask…

Workflow
Ten years ago, at the start of my AV projects, I experimented with rotating the image in PS, but it didn't fit easily in my workflow. I've exported thousands of images to my Premiere projects, and the workflow is detailed. Depending on the situation, I have to change the colour space, the resolution (six standards), the format (jpg, PSD, PNG), crop the image, and save it to the correct folder. Whoa!

30 Actions
So I set up about 30 Actions in PS. When the image is ready to export, I bring the selection rectangle on screen, select the relevant Action, click Play, and the export is done. No mistakes, no thinking.

However, if an image has to be rotated, that would have to be done before the Action. I'd have to…

• Bring the selection into play
• Work out the angle of rotation (or look it up, but that adds time)
• Rotate the image in the opposite direction
• Call upon the relevant Action.
• And make absolutely sure I don't save the rotated image. I insist on a non-destructive workflow.

Selection rectangle disappears
The problem was how to keep the selection rectangle on screen during the rotation. The damn thing disappears. I'd have to find it and load it again after the rotation.

So I decided that all exports would be done with the selection rectangle as is, using the Actions, and any rotations would be done in Premiere. Most images require Position and Scale adjustments in Premiere anyway (for pans and zooms), so a rotation every now and then is really not a burden.

A second look
Recently I decided to take a second look at this export process. First with the maths, because I was interested in that aspect, and then with the "disappearing" problem. Turns out there is a simple workaround, (but not commonly known) for the disappearing selection rectangle – two "Q"s (see here).

New Procedure
I am now about to test a new procedure to see if it's efficient, or whether I just stick with rotations in Premiere.

1. Load the selection rectangle.
2. If horizontal, play the Action.
3. If not horizontal, find rotation angle, "Q", rotate, "Q", Action.

It's the time involved for #3, and the possibility for errors, that I'm not convinced is an improvement – except in one way. In PS you can set parameters to four figures; Premiere only three. So there would be an imperceptible improvement in image quality if I do the rotation in PS. I wouldn't have to rotate and scale in Premiere.