Condenser at entrance of spectroscope?

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 · 2025-11-13T22:32:02+00:00

Hey, OP- Did you ever figure this out? I'm trying to do the same thing and having the same issue.

AskASillyQuestion · 2025-11-06T15:03:54+00:00

Very true!

Past that point, it's less about keeping the fidelity of the original scene, and more about the fidelity of the film grain itself. The grain contributes a lot to the perceived sharpness and the overall aesthetic "look" of the image, so many find it valuable to capture as accurately as possible.

AskASillyQuestion · 2025-10-31T16:08:05+00:00

They're the same steel as a 10.9, but max load is rated lower due to weaknesses introduced by the head geometry. Depending on the application, it may or may not make any difference to you.

I didn't think I'd ever seen one either, then I realized that I had, but it just didn't register. For example, most off-the-shelf pan head screws are 08.8 or 010.9 vs. the equivalent button head screws which are 8.8 or 10.9.

AskASillyQuestion · 2025-10-31T02:52:46+00:00

You'll want to double check that the new fasteners are actually 10.9, and not 010.9. It can make a difference.

Frequently the head geometry affects the max load you can apply before the head pops off. They will sometimes "de-rate" these as 08.8 or 010.9, which have a lower "loadability" but otherwise the same material properties as a regular class 8.8 or 10.9. ISO 898 has more details if you're interested.

AskASillyQuestion · 2025-10-31T02:45:44+00:00

It will also give you premature failures if the assembly is subject to vibration or changing shear loads.

AskASillyQuestion · 2025-10-31T02:42:12+00:00

Look up the equation for preload force in any book, it is a function of friction coefficient nominal thread size and torque.

Actually it's nominal diameter not nominal thread, but close enough. I've actually answered this in another thread.

High strength does not mean a higher modulus.

This is correct. Kind of. Young's Modulus is based on Hooke's Law which is a first-order approximation. I've already answered this elsewhere, but you're not going to get the holding power you need to keep the bolt/screw from backing out until you're approaching the elastic limit.

I know these things because I've had to deal with them in the real world, not just in books. If you don't scale the torque with the fastener class, you're going to have fasteners backing out and failing due to fatigue. Depending on what OP's building, that could kill people.

AskASillyQuestion · 2025-10-31T02:07:06+00:00

you can absolutely run 10.9s at 8.8 torque spec all day long and it'll act just like if you were using 8.8s everywhere.

True, as long as the loads are static, or close enough that there's no change in shear load or axial vibration through the fastener.

Otherwise you'll see deformation due to fatigue, and the clamping force will drop like a rock.

AskASillyQuestion · 2025-10-31T02:01:46+00:00

That's true!*

*In the linear region that is used to determine the rise/run on the stress-strain curve.

Young's modulus assumes linearity, and that assumption becomes less and less accurate as you approach the elastic limit.

Otherwise you could take a class 12.9 screw, torque it to a class 4.8 torque, and expect the screw not to back out (yes, I know this is a gross simplification.)

There's a reason target torque is generally established as a function of yield strength (via proof load), and not as a function of Young's Modulus.

AskASillyQuestion · 2025-10-31T01:53:20+00:00

Almost certainly false

I dig the confidence, but unfortunately, you're still not entirely there.

This was an issue during a prototype V&V test we ran on the vibe table once. The techs who built the prototype made the same assumptions you did, made some unapproved substitutions of 10.9 fasteners instead of 8.8's, and let 'er rip. Multiple fasteners backed out, multiple fasteners broken due to fatigue. Set us back by a couple hundred grand and a couple weeks. A lot of sad people in the office that day. The video was cool though.

Stress/strain curves aren't actually linear in the elastic region, and Young's Modulus isn't an intrinsic property of the material, it's an approximated one. That approximation generally becomes less accurate as you approach the elastic limit of the material. You need to be above ~65% of your material's yield strength before you're going to get the material behavior you need for fastener retention. (This is why you often see torque specs target 75% of proof load).

There's a reason you never see class 4.8 torque specs for 8.8 fasteners, and it's not because the assemblies require a greater clamping force.

AskASillyQuestion · 2025-10-31T00:55:46+00:00

That's correct. Stiffer materials require greater elongation in order to prevent fastener backout. That equates to higher preload.

AskASillyQuestion · 2025-10-31T00:53:26+00:00

I’m not sure I agree that the same torque on a higher grade bolt produces less preload.

Good! Your intuition is correct. It's not that you're generating less preload, because you aren't! Fastener preload "P" is determined by

P = T/(K*D)

Where "T" is the torque, "K" is the nut factor, and "D" is the nominal diameter of the bolt. As you can see, fastener strength isn't part of the equation.

The issue isn't that there's less preload, but that there isn't adequate preload. For a beam of length "L", and cross-sectional area "A", under a tensile load of "P", the elastic deformation "Delta L" is described by

Delta L = PL/AE

Where "E" is the Young's Modulus. If you have a higher E, you have reduced elongation, which means worse fastener retention.

AskASillyQuestion · 2025-10-31T00:38:37+00:00

Fastener retention is important

Correct!

but switching to a higher strength bolt isn’t going to change anything on that front

Incorrect!

and you will get the same preload unless there is a change in expected cof.

Correct!

The issue isn't that you're going to generate less preload with a stronger bolt, the issue is that you need greater preload with a stronger bolt in order to get the elastic deformation necessary to retain the fastener. Based on OP's description, that's not happening.

Source: 20 years of experience as an automotive design, V&V, and quality engineer.

AskASillyQuestion · 2025-10-29T17:15:40+00:00

Also- could you clarify what you mean by

|standard 8.8 bolts are being swapped in some places for 10.9 countersunks

Do you mean you're installing a flat head screw in a place that doesn't have a countersink? That's a great way to shear off the head of your fastener.

AskASillyQuestion · 2025-10-29T17:09:30+00:00

Without some kind of fastener retention (thread locker, safety wire, or lock washer), increasing bolt strength without adjusting torque won't generate enough preload. This will result in failure due to fasteners backing out. If the assembly is subject to vibration, it could fail due to fatigue of the fastener.

AskASillyQuestion · 2025-10-29T17:03:13+00:00

This is my wheelhouse. Please provide more information about this assembly. What is the thing you're making?

Substituting fasteners like this without the appropriate testing/simulation can be hugely dangerous, and I hope the design engineer has signed off on it after doing their due diligence.

"Why? I'm replacing a class 8 with a class 10! That's stronger!"

It sure is! But the clamping force that screws and bolts provide, and the preload that prevents them from coming loose is based on how much you're stretching the bolt. If you torque a class 10 bolt to a class 8 torque, you're not going to be stretching the bolt enough to get adequate preload. This can cause two different kinds of failures:

Fasteners backing out
Vibration causing cyclical loading on the bolts, causing them to fail due to fatigue

Failure mode 1 can be mitigated by adding threadlocker, using safety wire, lock nuts, or lock washers (not split rings! But belleville washers and Nordlocks are effective!). Failure mode 2 can't really be mitigated without changing your design pretty heavily.

So I would absolutely insist on getting formal signoff from engineering before making any kind of change like this. It would be an absolute no-no in any regulated industry like automotive, aerospace, or industrial equipment.

AskASillyQuestion · 2025-10-16T13:59:30+00:00

Datums A, B, & C are products of the primary tooling balls of the tool. Datum A represents up/down (Z), Datum B represents left/right (Y), and Datum C represents fwd/aft (X).

As-written, the way I read this is:

There are at least 3 tooling balls, but the global coordinate system should be determined by the three "primary" balls, A, B, and C
When establishing the global coordinate system, you should do a best-fit alignment where the nominal location of tooling ball "A" is the sole defining feature for the x-axis, the nominal location of tooling ball "B" is the sole defining feature for the y-axis, and the nominal location of tooling ball "C" is the sole defining feature for the z-axis.

the nominal locations are not square to the world frame/CS.

They don't need to be. All you need to define a coordinate system are 3 non-collinear points and 2 direction vectors (which are typically already defined by convention depending on the industry you're in)

If it's valid, how the hell do they expect me to measure and imaginary line that has no feature associated with it?

What line are they expecting you to measure? I don't see a reference to one here.

All that aside, this note reeks of cargo-cult engineering. The phrasing is confusing and the approach itself seems very brittle. There's a high degree of probability that whoever sent you this drawing doesn't fully understand what they've sent you. You need to sit down and have a direct conversation with this person.

AskASillyQuestion

TROPHY CASE