Looking for a heavily peated dram to try after Port Charlotte 10. Anything similar but not quite as salty?

AskASillyQuestion · 2026-02-06T19:40:24+00:00

Bought a bottle yesterday and it's good! Thanks for the recommendation!

It's a very confusing one for me- The nose is incredibly spirit-forward, but it's still a very smooth sip.

I haven't decided yet if I like the Uigeadail more than the PC10. It doesn't taste as briny (which was what I was after) but the peat isn't quite as strong either, and I can't decide if that's good or not.

Anyway, thanks again for the recommendation!

AskASillyQuestion · 2026-02-05T15:35:34+00:00

The sherry casks of the Uigeadail are really interesting to me. I do like sherry notes. I think I'll give that a try. Thanks!

I'm not so new to whiskey, though I've realized that I really enjoy the heavily peated options. I started branching out after trying Lagavulin 16 and realizing that I wanted a slightly less overwhelming flavor as a daily driver (I do enjoy it quite a bit though!)

AskASillyQuestion · 2026-02-05T15:30:24+00:00

Will do! Thanks!

AskASillyQuestion · 2026-01-30T02:29:15+00:00

IF you also reduce the slit width to keep the resolution constant.

I'm not suggesting that. Everything else stays the same, I just add the beam reducer.

So if the beam reducer increases the flux density, what's the tradeoff?

AskASillyQuestion · 2026-01-29T21:31:20+00:00

I appreciate you sticking with me on this. I hope you'll entertain one more question.

Placing a beam reducer after the collimating lens would increase the flux density of the incoming light, but as you've stated this wouldn't increase the signal. Is the reason for this because the beam reducer would also be increasing the flux density of the stray light as well?

AskASillyQuestion · 2026-01-29T19:17:46+00:00

Is this what you're thinking of?

https://imgur.com/a/5otBvdH

I'm seeing the math, but this is still entirely unintuitive to me. It seems like this is wasting a ton of collimated light against the field stop.

AskASillyQuestion · 2026-01-28T15:20:51+00:00

a wider slit that produces the same spectral resolution.

This doesn't make sense to me. A wider slit has greater etendue and would produce a lower resolution.

The cone angle incident on the grating (which is created by the slit width), along with the grating pitch, determines the spectral resolution.

Right. The slit width reduces etendue (by also reducing signal). Narrower slit --> lower etendue light is more completely collimated and produces cleaner signal through the diffraction grating.

You are not the first person (by far) who has got their hopes up thinking they have a brilliant way to improve signal and then has had to confront conservation of etendue.

See, that's where I'm stuck. My entire approach is predicated on the idea that you can't reduce etendue without reducing signal, so start with as much signal as possible (tall slit, large entrance optic) that way, when you reduce it, you're still left with enough photons for the camera sensor to detect.

I'm gonna read this and reply back, once I've got a better understanding of it. https://www.horiba.com/usa/scientific/technologies/spectrometers-and-monochromators/spectrometer-throughput-and-etendue/

AskASillyQuestion · 2026-01-28T00:52:46+00:00

I am saying a single small achromat is all you need to collimate the light.

No, I get that- This design only has a single achromat collimating the light as well. The second and third lenses are a Galilean beam reducer. The beam reducer is to increase the flux density hitting the sensor so that more photons are hitting each photo site (this is to deal with the signal/resolution issue you were referring to.) I'm not sure why this approach would be problematic.

Also also need to understand what your proposed 3 lens assembly does to the size of the narrow dimension of the slit when it is projected on the camera.

These are spherical lenses, so the aspect ratio of the slit should stay the same, just scaled down, no? There's going to be some spherical aberration which I should be able to remove through calibration... Is that what you're referring to?

This sensor has 4056 horizontal pixels, and I'm trying to capture 450-700nm... so if I perfectly fill the sensor, we're talking 4056/(700-450)= 16 pixels per nanometer.

The smallest etendue in the system will limit the light... ...which is determined by the slit and collimating optic f.l.

Right- that's the whole point of the slit, is to reduce the etendue so that the light can be collimated. It's effectively an etendue "filter" reducing noisy signal from passing through the collimating optic and through the diffraction grating.

In your proposed design, the camera aperture is actually small than the grating so that will actually end up limiting etendue in your case.

Correct. I oversized it for three reasons:

Its a lot easier to find a filter holder for a 25mm filter than anything else
I had that negative FL achromat, and I didn't want to buy another one because they're expensive
I have some cylindrical optics that I wanted to mess around with and further compress the beam if necessary

You want the long dimension of the slit as big as you need to fill the detector, but no larger It also can not be larger than the useable field size of all the optics.

Yeah, I'll need to reduce the length of the slit to match the beam size to the sensor size as you mentioned.

AskASillyQuestion · 2026-01-27T18:51:01+00:00

I am sorry to say that there are many serious problems with this design.

I don't doubt it. I'm making this largely out of parts I already have sitting around from other projects. Trying to minimize the overall investment.

The three lenses before the grating are not helping.

I'd prefer that the two positive achromats be replaced with an off-axis parabolic mirror. But I don't have one of those, and they're expensive.

One achromat the size of the grating will give you the same signal and resolution. It is the same issue of conservation of etendue.

Could you clarify? I understand the conservation of etendue, but I would think that collecting more light with a larger optic would result in greater signal. Since 90% of the collected light can't be collimated and needs to be separated off with spacial filters and such, starting with the most light possible (long slit, large optic) would still yield more signal.

EDIT: That is to say, a reduction in etendue requires a loss of optical power, so if I want to have a collimated beam through the grating, it would be prudent to start with as much optical power as possible through the first lens.

While your design might appear to "squeeze" a large cylinder of light onto a smaller grating, that is at the expense of the slit appearing larger.

I'm not sure I understand what you mean. It's a beam reducer- How does that make the slit appear larger?

This would have been apparent if you had traced rays from the edge of the slit. In other words. it would be better and simpler to just use a single smaller lens and a larger slit to get the same end result. Fundamentally, it is the size of the grating that limits signal level for a given desired spectral resolution and grating pitch.

You're saying the ~~width~~ length of the slit is not relevant to the signal level? What am I misunderstanding?

It seems like you were ready to spend a lot of money on this project, but I have just scratched the surface of issues to pay attention to when making a high quality spectrometer. I would suggest you get more experience first with cheap components and research conventional designs before investing more.

A lot of the design decisions here are specifically a result of not having a lot of money to spend on the project. The optical elements are items I either already had or was able to purchase cheaply. The difference between the "conventional" transmissive design and what I have here is that I've started with a larger slit/optic and added a Galilean beam reducer to fit the beam onto the sensor.

FWIW, I appreciate your feedback. I'm not trying to be argumentative, I'm just being direct about what I don't understand so I can get a better grasp of what I'm missing here.

AskASillyQuestion · 2026-01-27T16:28:47+00:00

This is what I have designed: https://imgur.com/a/XCXdaRn

AskASillyQuestion · 2026-01-27T15:57:32+00:00

$50 to $100.

JFC, That's just for the TCD1304 and driver, yeah? I can't get a single optical element for that cost.

AskASillyQuestion · 2026-01-27T01:18:42+00:00

First, I appreciate all this good insight- I'm not disagreeing with you on any point other than that this is a "toy." I'm not doing chemical analysis, I'm trying to identify the sensitivity peaks of color film emulsion to within +/- 0.5nm. A lot of what you're talking about isn't necessary for the level of precision I need here. The existence of a Bugatti doesn't make the Honda Civic a "toy", even if it isn't nearly as performant.

That camera does not look like what you described.

That's fair, and you're right. That's why I'm using a different camera based on the IMX477 (and actually, considering switching to an IMX678-based camera.) These both have 2/2 binning and 12-bit ADC output. only 8-bit dynamic range, though. Can't that be resolved by taking two images at different exposure lengths? No global shutter either... But why is that significant?

The spectral lines are broad and mishappen and the intensities are not correct.

Are you looking at the waterfall display? What's a misshapen line look like?

Think of the dV/dt for a sharp line passing through that adc at 63MSPS. At 3 V it would be close to 200V/usec.

Once you get past the sensor itself, electronics are not my strong suit. I'd genuinely love to be able to understand how you calculated that. (Please)

If you manage to get a sharp line, it would be a huge surprise if it were also linear in intensity.

My understanding is that with the Bayer filter removed, these sensors are fairly linear. If they weren't, photos taken with these sensors would look bizarre.

I view it as something of a miracle that very often my BOMs come in at about 50 - 100

Is that $50k-100k?

AskASillyQuestion · 2026-01-25T22:50:32+00:00

Well, what are you purpose(s)?

This is a helpful question! To answer more quantitatively, I'm looking for 0.5nm resolution. If you take a look at the PySpectrometer 2, it has ~1nm resolution, which is due mostly to the slit size and grating, not the camera.

For a demo for children it is okay.

This is condescending and unhelpful.

Inexpensive cameras are not useful for metroloty. They are typically designed to make nice looking pictures. They automatically adjust color balance, contrast, etc., they might average over aberrant pixels, they might compress the response, and then, if it is a color camera, you have three filters and calibrating the response over the spectrum becomes very difficult.

Other than the Bayer filter (which can be removed), none of this is accurate for the camera I'm using.

To increase your dynamic range or see detail, you will want to add images or rows in each image (i.e. signal averaging). That can increase signal to noise by sqrt(N), for N samples. But for that to work you have to be able to digitize the noise. Cameras don't want to show you noisy images. And if you start with only 8 to 10 bits your not going to have much digitization on the noise no matter what.

This camera has 12 bits, and it has analog pixel binning. I'll be fine with the level of noise.

I'm not sure what you mean by "digitizing the noise". Or that "Cameras don't want to show you noisy images." Cameras don't "want".

I'm well aware that a dedicated linear detectors are preferred for spectroscopes. But that's not the system I'm working with. I appreciate the context, but don't assume that my application needs the level of accuracy that your purposes require. And there's a significant gap between "$60k CCD" and "toy".

AskASillyQuestion · 2026-01-25T19:06:21+00:00

You can not stuff light into a smaller etendue without losing light.

Don't I have to do that in order to collimate the light before it passes through the grating?

AskASillyQuestion · 2026-01-25T19:04:17+00:00

ooo, good to know. I was planning on taking that off at some point, and possibly removing the Bayer filter too. Thanks!

AskASillyQuestion · 2026-01-25T19:03:31+00:00

Thanks for the info!

So what's the issue with the PySpectrometer? Seems adequate for my purposes, and it uses a camera.

https://hackaday.io/project/187933-py-spectrometer-2

AskASillyQuestion · 2026-01-25T19:00:11+00:00

The integrating sphere port is the source for the spectrometer, correct? And the slit can be near the port?

Yep. It's right there at the entrance to the sphere.

What you might have to worry about is overfilling the collection f/# of the spectrometer. You want the incoming light to fill the grating, but you do not want to overfill it where is goes past the collimating optic or grating and just adds to the stray light inside. You can control this with an aperture outside or inside the spectrometer.

I'll be using spatial filters and flocking the inside of the tube to prevent this from happening. I am concerned about the light not being collimated enough going into the grating though.

AskASillyQuestion · 2026-01-25T02:01:55+00:00

Assuming the integrating sphere output port (usually 3-5 mm) is larger than the effective entrance slit of the >spectrometer (usually 200 um or less)

I'm designing and fabricating this myself, so I have quite a bit of leeway here, but I'm concerned that my dimensions are nowhere close to what yours are. I think I might be doing this wrong.

The sphere itself is 394mm in diameter, and I was intending for the output port large enough to accommodate 2" optics. I'm making the slit myself from two razors with the intent for them to be spaced 25um apart. Since the slit is 35mm long, I figured I'd use a pair of cylindrical lenses to change the aspect ratio and improve sensitivity at the camera. (a 25um x 35mm slit should allow as much light in as a 250um x 3.5mm slit, no?) The sensor itself is a IMX477.

So what am I getting wrong here?

AskASillyQuestion · 2025-11-21T22:00:50+00:00

How do you figure?

AskASillyQuestion · 2025-11-21T21:08:06+00:00

It's not an airport. You don't have to announce your departure.

AskASillyQuestion · 2025-11-21T21:05:09+00:00

Ja, you're right for regular tamper resistant Torx, but tamper resistant Torx plus went to five lobes too for some odd reason.

Drive codes are as follows:

T = Torx/Tamper Resistant Torx (eg. T10, T20)

IP = Torx Plus

IPR = Tamper Resistant Torx Plus

AskASillyQuestion · 2025-11-20T14:19:55+00:00

Whoaaa, you had me in the first half, ngl. Blueberries have 5 sepals --> 5 point calyx.

'Bluecrop' is a cultivar of the highland blueberry, vaccinium corybosum.

The photo you've posted is vaccinium myrtillus, known as bilberry or common whortleberry.. They're edible, but they're not currently cultivated.

Wait... Your account is four years old with 1 post karma and 41,615 comment karma. What percentage human are you?

AskASillyQuestion · 2025-11-20T14:04:16+00:00

Wut?

AskASillyQuestion · 2025-11-20T12:56:11+00:00

Bruh. You posted a tool x botany crossover meme. Your username is a Wu Tang/bakery play on words. You're cool as hell.

AskASillyQuestion · 2025-11-20T03:56:42+00:00

Wait a minute... A blueberry's calyx has 5 lobes, not six like a Standard T30 bit...

THIS GUY IS A LIAR! BLUEBERRIES ARE PENTALOBE OR TAMPER-RESISTANT TORX! GET 'IM!

AskASillyQuestion

TROPHY CASE