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May 19

Feynman Lectures. Exercises PDF. Exercise 4-4 JPG1 JPG2

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4-4. The truss in the figure is made of light aluminium struts freely pivoted at each end. At C is a roller which rolls on a smooth plate. When a workman heats up member AB with a welding torch, it is observed to increase in length by an amount x, and the load W is thereby moved vertically an amount y.
a) Is the motion of W upward or downward?
b) What is the force in the member AB (including the sense, i.e., tension or compression).

— R. B. Leighton , Feynman Lectures on Physics. Exercises

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4.4. A) Imagine that the rod of the truss is deleted. Then under the action of the load W the hinge B will go down to the right, and the hinge A - go up to the left. It is therefore clear that the rod AB was stretched. However, removing the rod, we can also keep the truss in balance by hanging two equal loads on fixed blocks as shown in the figure. If the length of AB increases, the loads rise and their potential energy increases. To ensure that the potential energy of the entire system remains unchanged, the load W must drop. Therefore, when the rod AB is heated, the weight W is lowered.
B) The weight of the loads we are considering is equal to the tensile force of the rod T (the truss is in equilibrium). If you increase the length of AB by an amount x, the loads will rise upward, the sum of the changes in the heights of the loads also equal to x, and the weight W will drop by y. Therefore, Tx = Wy, that is, T = (y / x) W. Figure

— MEPhI , Solutions (Google Translate)

I don't understand how the strut can have potential energy, when it is not compressed or tensioned. When it is heated up, there are no elastic deformations. Heat expansion can lift any weight while pressure in the body is less than breaking stress.

Username160611000000 (talk) 05:21, 19 May 2017 (UTC)[reply]

That answer may be mathematically correct but as explanations go it seems obtuse to me. AB increases in length so the flag moves to the left so the right hand part of the structure is rotated anticlockwise, thereby allowing the point C to move rightwards, while lowering the entire assembly. The strut has (elastic) PE, 1/2.k*dx^2. k.dx=T Greglocock (talk) 02:06, 20 May 2017 (UTC)[reply]

It seems that x from heat expansion and dx from the formulae E=0.5k(dx)² & T = k(dx) are different things. The member AB is already tensioned in both cases (before and after heating). Using the method of joints (proportions and angles are taken from the figure) I've found T_AB = 0.4568W before heating and 0.7586W after heating (I took arbitrary y/x=0.6058) and T_AB directed inside the member AB (tension). So if we know the area and lenght of the member AB, we can calculate stiffness and so dx.
Username160611000000 (talk) 08:34, 20 May 2017 (UTC)[reply]

They are indeed different. The elastic potential energy is 1/2*T^2/k (same equation as in my previous post rearranged to eliminate dx). That does not change when the bar is heated. The work done by heating the bar is T*x, where x is the change in length due to the heating. Greglocock (talk) 01:59, 21 May 2017 (UTC)[reply]

And by the principle of virtual work T.x=W.y obviously, since no other forces move in their own directions. Quite why the official answer includes hard to imagine weights and pulleys and so on is a good question.Greglocock (talk) 06:36, 20 May 2017 (UTC)[reply]

Little loads and pulleys are absolutely clear for me and they are in line with Lecture 4. In the Lecture 4 Feynman associates all movements with lifting loads (I think it's because we officially don't know neither forces nor work, nor center of mass).
Username160611000000 (talk) 08:34, 20 May 2017 (UTC)[reply]

To test the idea that ignorance might help me to explain things simply, I looked at this for a bit sheerly as a puzzle without making any attempt to study trusses... this is what I came up with; the catch is I'm not sure about it!

1) The force of the weight is straight down; the force of the roller must be straight up. Therefore, the remaining pivot cannot exert force left or right either and ought to be replaceable by a roller, and is also straight up.

2) Naming the unnamed points from left to right starting with D, the pivot (D) is attached to E and F. The pivot D is constrained to rotate in a circle around F by segment DF.

3) DF can apply no torque to D regarding this rotation. The upward pressure of the ground (see 1) applies a torque that is clockwise. Therefore, DE applies a counterclockwise torque, so it is under compression, in order that D is not moving relative to F.

4) DE can apply no torque to D where its rotation about E is concerned. The ground again applies a clockwise torque. Therefore, the strut DF must apply an anticlockwise torque, so it is under tension.

3-4a) We can calculate the forces in principle - the torques from D to B and C to B must be opposite, so if we can measure those distances we know the upward force on D. Once we have that, we know the torque around DF and DE involves the sines of the angles each strut or upward force make with the strut to the point around which rotation is considered, and we just divide by those sines to get forces. What I don't understand is the problem seems to expect an answer for force (ultimately at AB, which can be reached by repeating such calculations) but the relevant angles and measurements seem absent, unless there's something I'm missing. Still, we can definitely infer directions from the diagram.

5) Point F would rotate about E if the tension from segment DF were not compensated by a tension from segment BF.

6) The tensions from DF and BF would move point F if that were not opposed by a force (not torque this time) from EF. So EF is under compression.

7) If EF and ED are both under compression, E would rotate about B clockwise if it were not for a compression from AE.

8) Likewise A would rotate about B if it were not for a compression from AG to counter AE. And G would rotate if it were not under compression at GI.

9) Now if A is under compressions from the two outer points (AE and AG) then there must be tension from B to make up for that; this is part of the main work lifting the weight. So AB is under tension - which is one answer.

10) Lengthening AB incrementally should allow B to drop slightly, but it also allows B to move right. Moving right means moving up as the point rotates around H. The problem I have here is that if all the struts are completely rigid, triangles DEF and BEF cannot change, nor the three triangles from B to C ... so B cannot move. But if the struts have limited rigidity, then figuring out whether the relaxation downward for B and rightward about F outweighs the push up from H does not seem trivial. I mean, I *think* it goes down but how do I tell?

Wnt (talk) 17:16, 20 May 2017 (UTC)[reply]

Carpet bombing

3 related questions: (1) In a typical World War 2 air raid using heavy bombers (such as the Avro Lancaster), what was the typical creep-back distance from the raid leader's aiming point? (2) For a Lanc flying at the normal airspeed and altitude, how far did it normally travel between the instant when the bombardier hit the bomb release and the instant when the last bomb was dropped (i.e. what was the typical creep-forward distance)? (3) If the aforementioned Lanc was loaded with napalm bombs (purely a hypothetical question -- napalm was rarely used in Europe, and was never dropped from Lancs to the best of my knowledge) of the typical size, and dropped from the normal altitude at the normal airspeed (as in part 2 of this question), how far forward would the napalm spread from the point of impact under its own momentum? 2601:646:8E01:7E0B:B167:959E:4F51:222F (talk) 07:45, 19 May 2017 (UTC)[reply]

The aiming stuff would have to be looked up from the bomb aimer's manuals - @Maury Markowitz: has done a lot of editing in this field. The US manuals are easily available as reprints, UK are rather harder to find.

Heavy bombers did drop oil incendiaries in WWII, particularly the Luftwaffe. However, unlike gellified napalm containers, these were bombs that dropped intact and then opened with a small bursting charge on impact. So they had almost no spread before impact. Also bombs from heavies at altitude have almost no forward momentum on landing, having lost it due to air resistance and being redirected downwards by their tail drag. This is another reason why napalm and cluster weapons are dropped from low level. Andy Dingley (talk) 12:02, 19 May 2017 (UTC)[reply]

So there are two different numbers here. One is creep-back, which refers to the bomb aimer's tendency to drop early so they could turn around and leave ASAP. Generally, the first bombers in the raid would drop on the target, but as soon as the smoke of the bombs obscured it the following bombers would drop on the leading edge of the smoke area. Now you have a larger smoke area extending back along the direction of approach, and then the next guys would drop on that leading edge of that area and so on. I seem to recall the typical creep for USAAF being as much as a mile, but I can't find a ref now.

The RAF also saw this problem on night missions, where the following bombers would drop on the leading edge of any fires and they had creep in the miles. They solved it by having a "master bomber" flying around, often in a mosquito, and periodically dropping new sets of target indicator to keep the bombing centered. They would also call in corrections over the radio to bombers still on the approach, things like "bomb 500 yards to the north of the markers". Their Mark XIV bomb sight made this somewhat easier than the US Norden bombsight, which didn't like last-second corrections.

The other issue is "range", which is the distance the bombs move forward due to momentum after being dropped. For typical B-17 drops range was about 1.5 miles. You can find tables for typical bombs and altitudes here. Now since the aircraft is still powered, and the bombs slow due to drag, by the time they hit the ground the aircraft is in front of them. This "trail" is dependant on the aircraft speed. The typical trail was on the order of 500 yards or so.

As to the spread along the bomb line, this was typically only a few hundred yards, but could be adjusted based on the "interlovemeter" (RAF terminology). The bomb aimer would pre-set this, what is today known as "ripple", and then set the bombsight so it would drop slightly early so the middle of the ripple was dropped on the aim point. They talk about how to do this in the MK. XIV manual.

The final question I can's answer, but watching footage of napalm dropped from low altitude jets in Vietnam I suspect the spread would be pretty low. For high altitude drops from aircraft with much lower speeds, the bombs were falling pretty much vertical (about 70 degrees) so I think any spread would be due entirely to splash as opposed the residual motion of the bombs. Maury Markowitz (talk) 15:15, 19 May 2017 (UTC)[reply]

I can't add much to User:Maury Markowitz's splendid answer, but you may be interested in Graph to show the accuracy of night bombing of German cities. Note that 1941 Butt Report defined a successful attack as one where the bomb landed "within 5 miles" of the target area, and in 1941, "only about 5℅ of bombers setting out bombed within five miles of their target" (to quote our article). Alansplodge (talk) 16:34, 19 May 2017 (UTC)[reply]

Thanks, all! A couple more questions: (1) Typically, how much time would pass between when the target markers were dropped and when the actual bombs would start falling? (2) Given the extremely low accuracy of night bombing during WW2, if a massive bombing raid was launched on Gerolstein (get out your maps!), is it likely that Schleiden and/or Steinborn would also be hit just from the stray bombers dropping their bombs all over the place? (Assume that this is late in the war, so the bombers fly on a straight course to their targets.) 2601:646:8E01:7E0B:B167:959E:4F51:222F (talk) 00:16, 20 May 2017 (UTC)[reply]

For (1), our Pathfinder (RAF) article states in the Equipment section: "Candle Aircraft, TI, [and] Bomb, Type H: the candle was the basic indicator. About 2 feet long by about 2 inches in diameter, it sequentially ejected flare pellets that burned for 15 seconds each. The type H was filled with alternately coloured pellets (red/yellow or red/green or yellow/green), and illuminated for about 5 1/2 minutes in total". Alansplodge (talk) 01:46, 20 May 2017 (UTC)[reply]

Thanks! So, if you saw these things floating down, that means you have less than 5 minutes to take cover or otherwise try to escape? 2601:646:8E01:7E0B:B167:959E:4F51:222F (talk) 03:54, 20 May 2017 (UTC)[reply]

The Germans had a sophisticated air raid warning system, so you'd have known about it long before then. BTW, the linked article is from 1943, when the existence of radar was still secret, but I'm certain that "listening devices" means radar. Alansplodge (talk) 18:24, 20 May 2017 (UTC)[reply]

Following on from what User:Maury Markowitz was saying. The RAF High Command realized that the bombardiers would naturally want to release their bombs as soon as they 'thought' they were over the target. Which was often too soon. So, successive waves of bombers where given a off-set to calibrate the bomb sights to slightly delay the release. The Americans did not employ this technique. Hence the term American carpet bombing because their bombing pattern appeared like a carpet being rolled out and away from the intended target area. Just like the British, their air crews too, wanted to drop their bombs as quickly as possible and head back for home but they where not afforded this calibration off set and dropped many tens of thousands of tons of bombs on Germany with little strategic benefit.Aspro (talk) 16:27, 21 May 2017 (UTC)[reply]

German Path Finders liked to drop their flairs (and anti-personal parachute bombs) about ten minutes before the arrival of the bombers. Form hight, the bombers could see these markers a long way off. The didn’t fly 'straight' to the target. They also had to skirt around areas, that day time reconnaissance had suggested held a concentration of AA batteries. So, they would fly to one-side and turn into the wind so as to reduce their ground speed, making for more accurate bombing. Air raid warning went off so often that people didn’t bother to go to a shelter until they heard the bombers approaching and saw they were turning their way for a bombing run. Air-raid shelters were so common place back then, that is only took 30 seconds to walk to the nearest one and less than ten seconds if one noticed that the approaching bombers had the bomb doors open (they mostly often just flew over head on the way to another target). Aspro (talk) 16:55, 21 May 2017 (UTC)[reply]

Physics Question about a clock meant to address time dilation

Will this clock solve the time dilation problem

The position of four mirrors A, B, C and D are such that AB=BC=CD=DA as shown in the picture of a light clock. The scale of the clock can be chosen and so its orientation by the observer. Similarly, things can be assumed if missing.

The observer in the clock at point A is Oc

A pulse of light takes 1 second or less depends upon the scale of the clock to complete a path of ABCDA if fired at point A for all stationary observers.

Let the clock starts moving at high speed in space relative to another stationary observer “Os”.

Would the clocks of Oc and Os be synchronized relative to each other if Oc is moving relative to Os? — Preceding unsigned comment added by 2001:56A:7399:1200:B9AC:FB08:382B:CEE1 (talk) 18:01, 19 May 2017 (UTC)[reply]

The answer is at the given link. It concludes thus: "...were we to invoke the Pythagorean Theorem for the three sides of our observed right triangle and rearrange terms, we'd find dT = dt/sqrt(1 - (v/c)^2) which is the usual time dilation equation. Sorry it's not an anti dilation clock at all." Blooteuth (talk) 20:42, 19 May 2017 (UTC)[reply]

Moving Light Clock and Pythagoras

Let there is a stationary observer “A” relative to which a light clock is moving at high speed in space. The observer in the light clock is “B”. A pulse of light bounces in between two mirrors of light clock.

Now ct’, ct and vt’ are the magnitude of hypotenuse, perpendicular and base of Time Dilation Triangle (TDT) respectively. So

Is the magnitude ct of perpendicular of TDT taken relative to #1- “A” or #2- “B”

If it is #1 then the path made by pulse of light relative to "A" is hypotenuse

If it #2 then the path made by a pulse of light relative to "B" is vertical but in time t' (stationary frame time t') not moving time t. #1 is not possible, as "A" didn’t notice any vertical path of the pulse of light

Although the path traced in case #2 is vertical but then it doesn’t make a perfect triangle with the hypotenuse and base of TDT due to the measurement of c in t' (stationary frame time t')2001:56A:7399:1200:3D60:F122:A71A:F537 (talk) 16:19, 22 May 2017 (UTC)eek[reply]

Honey

Honey doesn't spoil. What about a granulated sugar and water mixture that has the similar chemical composition as honey, would that be spoil-free as well? Scala Cats (talk) 19:58, 19 May 2017 (UTC)[reply]

Here's an explanation of why honey has a long shelf life:[1] A key point is that there is very little water in honey, and that fact prevents spoilage. ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:26, 19 May 2017 (UTC)[reply]

So that's a yes? Scala Cats (talk) 22:49, 19 May 2017 (UTC)[reply]

More like a no. (See Semantic's excellent comments below.) If you could exactly replicate the chemistry of honey, it could be more like a yes. ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:19, 19 May 2017 (UTC)[reply]

Yes, lots of sugar with a bit of water will resist spoiling. Sugar has been used as a preservative for millennia, see Food_preservation#Sugaring. The good article Bugs linked explains how honey's low pH also helps to prevent bacterial from colonizing, and a dry/thick simple syrup like you propose would have a higher pH, so it might need to be a bit drier to resist spoilage to the same degree. SemanticMantis (talk) 20:37, 19 May 2017 (UTC)[reply]

I should add that our article on honey also has lots to say on why it stays good a long time, including lots of physics and chemistry. It says that honey is around 13%-18% water, and fermentation doesn't occur up until about 25%. This [2] magazine article claims 2:1 simple syrup should stay good for 6 months, but to be honey-like, your cane-sugar paste would have to be over 5:1 sugar to water. A key concept here is water activity, which is what actually determines if bacteria can grow, not water content. The idea is water can be sort of bound up and inaccessible to bacteria. Here [3] is a good table showing what water activities various foods have and what sort of stuff will grow on them.

That's the chemistry and biology of why you can expect cane paste to last in edible condition for a very long time, but nothing lasts forever, and a lot of this depends on storage. In case you don't know, dry granulated sugar also has a very long shelf life, so while the science is interesting I can't think of why we'd want to moisten our sugar before storing :) SemanticMantis (talk) 23:11, 19 May 2017 (UTC)[reply]

The problem with granulated sugar is that it's not all that soluble in water. You generally need to add it to a hot water-based liquid, like coffee or tea, and then stir, the get it to dissolve, and even then expect some residue. Liquid sweeteners are more likely to work in cold beverages, like iced tea. (Sweetening the whole batch of tea before you add the ice works, too, but this assumes that everyone wants the same level of sweetness.) Then there's the issue of adding sugar to solid foods, such as toast. Honey on toast works, as does cane syrup, but granulated sugar would need melted butter to hold it in place, and even then remains gritty and undissolved. Some people don't mind this grit, but others do. StuRat (talk) 00:27, 20 May 2017 (UTC)[reply]

My mother used (very sparingly) the same tin of Tate & Lyle's Golden Syrup ("out of the strong came forth sweetness") for the best part of a decade, if I recall correctly. Alansplodge (talk) 01:37, 20 May 2017 (UTC)[reply]

You can get liquid gula melaka [4] and other forms of palm sugar [5] [6] [7]. I think when sold commercially these are always made by not dehydrating the sugar syrup enough to form a solid mass, but I could be wrong. You can also get liquid sugars of various types including white sugar, although these are mostly sold for industrial and commercial food manufacturing purposes. Again these are generally made by not crystalising the sugar syrup rather than by rehydrating it after the fact [8] [9]. Of course these are the commercial products (although is perhaps a good reminder that pretty much any crystal or solid sugar is likely to have been liquid earlier in processing), if you need some syrup at home and either can't purchase, find it more expensive, don't need such a large quantity; you can just rehydrate/moisten the crystallised sugars. Nil Einne (talk) 02:36, 20 May 2017 (UTC)[reply]

I see some references to "fossilized honey" [10][11] - I haven't raided Sci-Hub for the latter, but it doesn't sound all that palatable from the description. I'll believe the 5000-year-old honey when I see it. ;) Wnt (talk) 17:23, 20 May 2017 (UTC)[reply]

It goes well with 5000-year-old bread. ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:41, 20 May 2017 (UTC)[reply]

May 19

Feynman Lectures. Exercises PDF. Exercise 4-4 JPG1JPG2

Carpet bombing

Physics Question about a clock meant to address time dilation

Honey

Feynman Lectures. Exercises PDF. Exercise 4-4 JPG1 JPG2