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.