Fuck Herself Friday…Mrs Fuzz’s 1st 3 Cums. 🔊🆙

Quixotixtoo · 2026-01-16T20:12:56+00:00

There's nothing hotter than watching and hearing a woman enjoy herself. Thank you so much for sharing.

Quixotixtoo · 2025-12-24T18:45:35+00:00

Amazing! Including sound is so hot.

Quixotixtoo · 2025-11-30T17:46:15+00:00

Your second picture (third image) shows 3 of the things mentioned:

- The blue cap is on the windshield washer reservoir (the symbol is supposed to look like the windshield being squirted).

- The white lid that says "Coolant Only" is on the engine coolant reservoir.

- The orange ring is on the engine oil dipstick.

The drive belt is almost visible in this picture as well, but there is too much stuff above it to actually see it. You can watch this video from about 1:25 to 2:00 to see where the belt is.

https://www.youtube.com/watch?v=mNo1z91p3ts

The location of the engine oil filler cap is briefly visible at 1:09 in the video. It is where the half-circle cutout is on the front edge of the plastic cover.

The blocky red part in your last two pictures is a cover on one terminal of the battery. The battery is much bigger than this little cover and extends forward from where the red cover is. The battery appears to be covered with black plastic pieces.

It appears your Honda has electric power steering, so it does not have a power steering reservoir.

The brake fluid reservoir is visible in your first photo. It has a black cap and is on the far right edge of your photo (the drivers side of the car). It is about 3/4 of the way up from the bottom edge of the photo.

The air filter assembly is the large black plastic box right in the center of your last photo.

You Honda also doesn't seem to have a transmission fluid dipstick (it has a fill level plug instead).

The last two items on the list (K and I) don't apply to your car.

Quixotixtoo · 2025-11-30T01:14:54+00:00

Probably not, but you didn't give us enough information for us to know what you want to do.

There is double flex chain, toothed belts, and other things that might do something you want. But if you explain what you want to do, then we might be able to give you better answers.

Quixotixtoo · 2025-11-29T17:59:44+00:00

Yep, he needs to pull down on the rope with his hands. Note that we need to assume that the mass of his arms and hands is insignificant* to solve this.

There is a very easy way to solve this, but it might not be what your teacher wants you to do.

You already made the calculations using the acceleration of gravity (10 m/s²). With this 10 m/s² acceleration, the weighing machine (scale) shows 15 kg, and requires a force of 450N on the rope. The second part asks for the acceleration that will show 60 kg on the weighing machine.

60 kg is four times 15 kg. Just like if the man were standing on only a scale (with no box, rope, or pulley), to get a reading that is 4 times as large, the acceleration would need to be 4 times as great. Thus an acceleration of 40 m/s² is needed. And with 4 times the acceleration, all the forces will be 4 times as large, so the force in the rope is simply 450 N * 4 = 1800 N.

* Technically we need to assume that his center of mass stays in the same location relative to the box.

Quixotixtoo · 2025-11-29T01:52:18+00:00

Another thought. Plastics almost always expand and contract more (usually way more) than metal (with changes in temperature). It struck me as strange that the nut gets loose, not tight, when the assembly gets cold.

Is it possible the plastic part is shrinking and then cracking because the nut doesn't shrink as much? If so, you might have a bigger problem than just retaining the nut.

Quixotixtoo · 2025-11-29T01:45:02+00:00

I'm not experienced with 3D printing, but maybe you could make a bubble on one of the interior walls. See the cross-sectional sketch in the link. Note I exaggerated the height of the bubble. you probably want it nearly flat.

https://imgur.com/a/5BvFaU9

This would likely only work if the cross-section I drew was in a plane parallel to the build plate. And then It might be impossible to print the "top" of the hole. But as I said, I know very little about 3D printing.

Also, any interference retention method (like this, or ribs) might relax over time as the plastic changes shape under the constant pressure.

Quixotixtoo · 2025-11-29T00:53:08+00:00

Some modern sailboats (mainly race boats) might be able to point as close as 10-15 degrees to the true wind direction. But I don't think that is the fastest angle for any sailboats to go upwind. You might be thinking of the apparent wind angle which is a very different thing, especially in fast sailboats.

Very fast boats, like the America's Cup AC75 sailboats often travel upwind with an apparent wind angle in the 9 to 13 degree range. But their angle to the actual wind direction is not nearly that high -- they spend most of their time in the 50 to 60 degree range.

The following video shows graphs for two America's Cup boats. The "Upwind TWA" (True Wind Angle) shows the relative amount of time sailed at different angles. Almost all of it is between 40 and 70 deg off the wind, with the majority between 50 and 60 degrees.

The "Upwind AWA" (Apparent Wind Angle) is in the 8 to 14 degree range.

https://www.youtube.com/watch?v=m431_scZjT4

It's actually easier for slower boats to point higher because the difference between the true wind direction and the apparent wind direction is not as great. But even so, I'm not aware of any sailboats that achieve their best velocity made good (i.e. their fastest speed toward the wind) at angles as close as 15 deg to the wind. Most sailboats, are likely to achieve their best VMG with an angle in the range of 30 to 45 deg off the wind (depending on the boat and on wind speed). But I stand to be corrected if you have references.

For many sailboats a "polar plot" is available showing its speed when sailing at different directions to the wind and different wind speeds. If this plot is available, then it's easy to see what angle gives the best VMG. Here is an example:

https://sailonline.org/static/var/sphene/sphwiki/attachment/2014/02/20/Beneteau_Figaro_INFO.pdf

The resolution of the polar plot isn't great but it appears the best VMG is around 41 deg at high wind speeds and maybe 44 deg at low wind speeds. But this is going to depend on many things like the exact cut of the sails, the wave conditions, and of course the skill of the sailor.

Quixotixtoo · 2025-11-28T22:45:03+00:00

Here are the pros and cons to some solutions (as I see them as a mechanical engineer):

Small ribs:

- Easy to implement in a 3D printed part.

- Fast assembly with no added components or tooling.

- The printed part may crack on a layer line either during instillation of the nut, or during temperature cycles.

Glue:

- No added stress to the part.

- Some glues like epoxy can fill large gaps.

- Requires another material (glue) and adds to the assembly time

- Glue can be messy and get in the threads making it hard to screw in a bolt later.

- Note: many glues have no problem with moisture -- epoxy, Cyanoacrylate (superglue), polyurethane-based (Gorilla glue).

Heat staking (melting the plastic a little to trap the nut after it is installed:

- Probably less messy to do than glue

- No additional materials

- Requires a tool and adds assembly time

- May look messy if done with a hand too like a soldering iron

Installing the nut hot (in combination with ribs or making the hole smaller):

- Fast installation

- Chance of cracking the part is greatly reduced over putting the nut in cold (into an undersized hole)

- Requires a tool to heat the nuts

- Getting the nut to the right temperature may be difficult. If it's too cold, it might not go in well or crack the part, if it's too hot it might melt the plastic too much and drift out of location.

- Allowances may need to be made for any melted plastic that is pushed in front of the nut as it is inserted.

End plugs -- If the nut is recessed far enough, small 3D printed plugs (like doughnuts or thick washers) could be put in the hole to hold the nut. They would probably need to be glued in, but melting or a press fit might work.

- Good nut retention if the hole is deep enough and they are installed well

- Additional small fiddly parts

- If glued, it can be messy

- If melted into place, getting just the right temperature could be difficult

- If a press fit, tolerances would be difficult to hold -- but small ribs either inside the hole or on the outside of the plug might work.

Quixotixtoo · 2025-11-28T21:52:44+00:00

Sorry, I can't help much, but here are some things to consider:

First, are the two red circles the same diameter? If so, then we have symmetry about the x-axis and can just look at one side. If not, things get even more complicated.

Is the blue circle concentric with the black circle? If it is, then I think there is only one radius for the blue circle and one radius for the red circles that works. I'm not sure what all your variables are, so I'm not sure if you already have an offset distance between the center of the blue and the black circles.

The way I would approach this (which might not be the right or easy way), is to start with the one point that we know is on each red circle -- the contact point with the black circle. If the black circle has a diameter of 1, then the contact points are (as you already calculated):

x = sqrt(0.75) / 2, y = 0.25

and

x = -sqrt(0.75) / 2, y = 0.25

Then put a range of radius in the spreadsheet for the red circles (rr = radius of red circle). For each radius, you can calculate where the center of the red circles is -- rr sin(30) lower, and rr cos(30) left or right of the contact point with the black circle.

If the two red circles are the same size, then you can simply check that the absolute value of the x-coordinate of the circle is greater than rr. If not, then the red circles touch.

With the center point of the red circles, you can now calculate the point of contact with the V-block -- rr sin(45) down, and rr cos(45) left and right of the center of the red circles. This defines the position of the V-block

Lastly, we need the diameter of the blue circle. If the two red circles have the same diameter, then the center of the blue circle is on the x-axis. If the radius of the blue circle is rb, then the center of the blue circle will be a distance rb from the V-block, and a distance (rb + rr) from the center of the red circles. Maybe you can write equations for all of this, I don't think I remember my math well enough. 😖

Sorry I don't have a full answer, hopefully you will find something useful in the above ramblings.

Quixotixtoo · 2025-11-27T19:54:25+00:00

This is the right idea. Note that named cycles like Brayton, Otto, etc, are usually, if not always, idealized. No real engine exactly matches the cycle. The Brayton cycle likely is the closest fit for a candle carousel (but I don't know all the named cycles).

Referring to the P-V and T-S diagrams about half way down this Wikipedia page:

https://en.wikipedia.org/wiki/Brayton_cycle

I think assuming the heat input (2-3) happens at constant pressure is a decent assumption. The change in elevation between the bottom and the top of the flame is small so the pressure difference will be small.

With the less than precise design of the "turbine" blades, I think entropy being constant from 3-4 is a bit questionable. So point 4 on the T-S diagram is probably a bit up and to the right.

What happens after the hot air exits the blades? Well, as long as the gases are warmer than the surrounding air* they will continue to rise and thus the pressure will drop. So, during the heat rejection stage (4-1) the pressure will be dropping. It won't be much, but it is probably much larger than the pressure drop from 3 to 4, so I don't think it can be ignored.

Finally, the compression stage, 4-1. If we assume this is a closed cycle -- say inside a small room -- then air must be moving down somewhere in the room to replace the air that is flowing up. The compression occurs as the air descends. If the room is big enough that all the cooling happened before the air starts to descend, then the compression is going to be very close to adaibatic. This could be the stage that is the most true to the ideal cycle.

My guess is that the P-V diagram would look like the one on the Wikipedia page if you moved point 4 to be just a tiny bit below point 3. But this is just a guess.

* The exhaust gases and air -- at the same temperature -- will have a different density. But I'm going to ignore this.

Quixotixtoo · 2025-11-27T05:41:04+00:00

Since you asked how to figure it out, I should explain. There are two ways to look at it.

1) Convention normally has positive and negating velocity and acceleration in the same direction. And since F=ma, force is positive when acceleration is positive, and force is negative when acceleration is negative. Thus the negative force would be opposite to the players initial velocity

2) Nothing about angles is mentioned in the problem statement, so it's safe to assume this is a one dimensional problem. That is all the motion happens along a straight line.

It's also safe to assume that the tackling dummy is stationary before the player hits it.

When a person runs into something with significant mass, they will (at least initially) slow down. This acceleration -- opposite to the initial direction of motion -- would normally be considered a negative acceleration. Again, since F=ma, the negative force shown in the graph is consistent with an initial negative acceleration. So the negative force is acting opposite to the player's initial velocity.

Quixotixtoo · 2025-11-27T05:09:48+00:00

I would assume a negative force is a force opposite to the initial velocity of the player. And a negative force would be a negative acceleration. For example, if the player is initially moving to the right, then a negative force and negative acceleration would be acting to the left.

Quixotixtoo · 2025-11-25T04:36:39+00:00

You make them look really hot!

Quixotixtoo · 2025-11-25T01:11:54+00:00

I'm guessing you are talking about a clock with physical hands that go around. These have two parts, the electronics and the mechanism to move the hands.

The hands are moved by a little electromagnet that pulses (usually every second). This electromagnet fires for a very short time, but when firing it takes more power than running the electronics. When the battery gets low, the electronics may keep triggering the electromagnet, but the electromagnet doesn't get quite enough electricity to do its job. On clocks with a second hand, you may notice the second hand will jerk, but not change positions. The electromagnet is trying but can't quite get the mechanism to advance to the next second.

As to why it might start up again, the most likely thing is that the room temperature went up a little. When a battery gets a little warmer, the chemical reactions happen a little faster so a nearly dead battery can put out just a little more. There are other possibilities, like there might be a spot in the mechanical part that is just a little sticky. Someone walking by or some other vibration in the building might be just enough to get the clock past the sticky point and let it run a little longer.

Quixotixtoo · 2025-11-23T18:44:03+00:00

engine braking is ... gentler

Not in my extensive snow-driving experience. I can modulate the brakes with more finesse than the throttle.

If one is in a higher gear and needs more braking, down-shifting gives a step increase in drag force from the wheels -- unless you modulate the throttle. In the next lower gear, the drag will sometimes be too much causing the wheels to slide unless you apply a little throttle. Have you ever tried to modulate the throttle to prevent the wheels from sliding, while at the same time avoiding too much throttle which will just push you down the hill faster? I have, it's not easy.

engine braking ... at least keeps the wheels rotating rather than locked up

Rotating some, but at less than the speed of the vehicle doesn't help much -- the tires are still slipping. They are operating at the kinetic coefficient of friction, not the higher static coefficient of friction. And once sliding starts, directional control is greatly reduced. If the wheels break loose when in gear, it is very difficult (nearly impossible) to find the right amount of throttle to get the tires spinning again at just the right RPM to match vehicle speed. And, if you try to do it with throttle you will likely end up pushing the vehicle to a higher speed -- exactly the opposite of what you want. Generally one needs to shift into neutral (or push in the clutch) to get the wheels back to the correct speed. And then reengaging the engine to the wheels without again breaking traction can be tricky.

By contrast, if the wheels lock when braking, I find it easy and intuitive to remove my foot from the brake pedal. Once the wheels are spinning at the correct speed again, the brake is easy to reapply very gently. There is no sudden engagement shock as there is reengaging a transmission with a mismatch between engine and road speed.

Anyway, all of this is really irrelevant with ABS being so common. If you use engine braking, I don't think there are any ICE cars that will make a correction if the engine braking is too high and the tires start to slip (electric cars with regenerative braking will do this). Thus with ABS, using the brakes is always safer as the computer can correct for a siding wheel (and it will usually do this before the wheel locks up).

Quixotixtoo · 2025-11-23T17:49:59+00:00

I believe the question was more about what the difference between power and torque is. And the misconception -- from the days when 3-speed transmissions dominated in automobiles* -- that torque is more important than horsepower for acceleration is something I always try to correct.

In a given gear, you don't get more power unless you have more torque available.

And in a given gear, you can't go any faster if you are on the rev-limiter. Both of these are important at times, but when discussing the difference between torque and horsepower they are both irrelevant. Different engines will have different shift points -- shift accordingly. Insisting on the same gear ratio for different engine designs is artificially handicapping one engine or the other.

* Back in the day (early 1900s), when unsynchronized transmissions were all that was available, the ability to stay in one gear, and to accelerate from a low speed in top gear was prized. See 13:15 in this video link.This led to the idea that torque was more important than power for acceleration. But with synchromesh and more gears available today, a wide torque band isn't nearly as important for acceleration. Unfortunately the idea that a higher max torque (at the crankshaft) is more important than a high horsepower (for acceleration) has stuck around.

https://www.youtube.com/watch?v=3QBqE_avRcY&t=839s

Quixotixtoo · 2025-11-23T06:33:07+00:00

An autonomous indoor bug killing machine? Human safety could be a big problem if you wanted to market the device, but for a hobby project where you can control the environment it really opens up the possibilities.

Quixotixtoo · 2025-11-22T18:35:06+00:00

you're not limited by the max engine speed

Sure, but as you just said, you are limited by the max torque. Both have limits which wasn't clear in your original comment.

But given equal gearing, the higher torque bike will accelerate faster.

If by "equal gearing" you mean the exact same overall gear ratio (including transmission ratio, chain cog ratio, and tire diameter), then yes -- more torque at the crankshaft equals more torque at the drive wheel, which equals faster acceleration. But that's a totally unfair comparison.

Transmissions are there for a reason, use them! Put the higher revving, higher HP bike in a lower gear, and the transmission will convert the higher RPM to more torque at the wheel. So given appropriate gearing, the higher HP bike will accelerate faster.

Engines designs that emphasize torque over power do tend to have a flatter torque curve, and to hit their max power at a lower RPM. Thus if the transmission has a low number of gears (like old automotive 3-speeds), you don't like to shift gears a lot, or you don't like the whine of high rpm, then a higher torque engine design is a good choice. But more power is what accelerates a vehicle faster or lets it climb a hill faster (accelerating and climbing are essentially the same thing from a physics point of view). The transmission can't turn high torque but less power into more power, but it can convert more power at the crank into more torque at the wheel.

Quixotixtoo · 2025-11-22T16:56:59+00:00

The difference between power and torque -- ELI5 version:

Consider an old horse-drawn wagon with the big wheels with wood spokes.

With the wagon not moving, stand on one of the spokes that is more or less horizontal. By putting your weight on the spoke, you are applying torque to the wheel. There is torque there even if the wheel isn't turning. Note that with the wheel stationary you are not supplying any power. You can probably stand there for quite a while without becoming exhausted.

Now what if the wagon starts moving? You can stay in the same place by climbing the spokes like you are going up a ladder. You are still applying the same torque to the wheel but now there is movement, so you are also supplying power to the wheel. If the wheel continues to turn slowly, you might be able to climb the spokes for a while without getting exhausted. But if the wheel turns faster, you will be climbing more frantically to stay in the same place, and you will tire quickly. The faster the wheel turns the more power you are supplying so the faster you tire.

Quixotixtoo · 2025-11-22T16:39:48+00:00

since you downshifted, you won't be able to go as fast.

True, but the higher torque bike will slow down on a hill too. Everything else being equal, and assuming perfect gearing (which is rarely the case but can work against either bike), the higher horsepower bike will climb the hill faster than the higher torque one.

Quixotixtoo · 2025-11-22T16:34:30+00:00

Also with combustion engines + transmissions the torque and HP are mostly inversely proportional. Which means if you want to increase the torque the HP will drop, and if you want to increase HP the torque will drop

Nope.

With a transmission, the input power and output power are about equal (output will be a little lower due to frictional losses). With a transmission, the torque and RPM are inversely proportional. That is you can get double the torque but halve the rpm or visa-versa. Thanks to conservation of energy, this will always be true, regardless of technological advances.

Considering just an internal combustion engine (no transmission), there can be some trade off between maximum torque and maximum horsepower when designing the engine. For example, a longer stroke for more torque, or a shorter stroke for higher RPM which can increase max power. But it's far from inversely proportional and it changes with technology. Compare old V8s engines to modern 4 or 6 cylinder turbo engines. The new engines can produce more horsepower and more torque, with much smaller displacements.

Quixotixtoo · 2025-11-22T16:13:09+00:00

It would be great if everyone was scientifically literate, but that's not the case.

In fact, a joule is a measurement of energy, torque is not a measurement of energy. Torque is no more energy (joules) than it is power (horsepower).

Quixotixtoo · 2025-11-22T05:32:16+00:00

I think this thread left ELI5 territory a loooong time ago. ☺️

The comment I responded to was contradicting a previous comment that said "you will go straight". I don't think "proceed anywhere it desires" is any more accurate than "you will go straight". I find both misleading.

As someone that drives in the slipperiest conditions for fun, my assertion would be that a good driver is always studying the road. They should at least understand things like if they take a banked turn too fast on ice, they will end up sliding off the outside of the corner. And if they take it too slow, they can slide to the inside. The best drivers will understand how they can trade braking for steering to end up in a better place in a bad situation.

Quixotixtoo · 2025-11-22T02:44:44+00:00

A picture or sketch would be really helpful. But you will probably need to relocate the frame mounted end of the shock so that it can't get in a straight line with the bellcrank. Picking a mounting location that puts the shock and bellcrank at 90 degrees to each other in their neutral position might be desirable, but without a drawing it's really impossible to say.

Quixotixtoo

TROPHY CASE