If there is a noticeable amount of friction in the bearings, going for the larger wheels should be beneficial. They would allow a small number of revolutions per distance.
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I took this to mean that the surface is sufficiently smooth to have zero practical impact. Larger wheels are less affected by unsmooth surfaces. For example I’d rather have a tractor tyre than a scooter tyre if going over pot holes!
In extremis, if the surface was completely smooth (this is sometimes assumed for mathematical questions) then the wheels would skid rather than turn. Hence smaller wheels (assuming lighter) would reduce mass and therefore benefit from any given initial push.
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I’m assuming that “… very smooth …” means “devoid of bumps” as opposed to “… like a sheet of (frictionless) ice …”
But what else is Matthew suggesting about the downhill course that would be really important in relation to the choice of large or small wheels ?
Co-efficient of friction
Gradient
Camber (Lateral curvature)
Straight line (to navigate)
Curves left and/or right (to navigate)
Quite separately, what aspects of the wagon would be really important .
C of G (eg can the driver lay below the axle level)
Weight (or mass)
Wheelbase (lateral/longitudinal)
Steering arrangement
Choices …
Enigma - person or thing, mysterious or difficult to grasp
Based on our collective responses so far, Matthew’s wagon seems to fit the bill called “Enigma” but hopefully, the enigma below will be provide a short spell of mental relaxation !
It starts with five hundred and ends with the same
The middle’s a straightforward five
The first of our letters and first of all numbers
Are sitting each side of that five
When seen as a whole, it reminds me of slings
And especially biblical kings
Alpha?
Is it DAVID
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Straight or curved.
Given the road surface is very smooth, if course is curved, a larger tyre surface area in contact with the road would be beneficial, to create more grip to steer.
Therefore the larger wheel would be better.
If the coarse is straight, no grip to steer is required. Less tyre surface area would be beneficial as there will be less rolling friction.
Therefore the smaller wheel would be better.
That’s my guess.
Nice one.
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It most certainly is - Well done !
Firstly, apologies for setting a teaser and then dissappearing on holiday!
So, I think Fred needs to know if the course is straight or curved, and the degree of cornering that would be required. The length of the course could also be important.
I think Ravvie was on the right line of thought for a straight long course.
Family photo album
Mrs D took ten photos today for the family album. They depicted me, each of our three children and each of our six grandchildren.
She would like to send one or more of these photos to various friends and relatives but can’t make up her mind regarding how many photos or which combination of photos to send to various people.
I suggested she simply send all ten to each recipient, but she said that was too easy. Alternatively, I suggested just sending the one of me to each recipient – but that was met with a glaring stare !!
“Choices, choices” I heard her say, “too many choices” she added. “Well,” I interjected, in as helpful a tone as possible, “you actually have hundreds of ways in which to choose combinations of those photos to send, including just one or all”.
Just how many choices did she have ?
Obviously, there is a quick way and a more time-consuming way to derive the number of choices.
I have yet to suggest she takes a photo of herself, thus making eleven photos from which to form combinations … !!!
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But what is the theory why he would choose large or small wheel if the coarse is straight.
And what is the theory why he would choose the opposite if the coarse was curved.
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Mrs R solved it methodically, though it didn’t take too long.
I went for the direct route, but I have an unfair advantage as I could use a technique that I am very familiar with. It has appeared from time to time on this thread.
Good questions Fatcat, good questions.
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Gravity Racer. MatthewT’s teaser !!!
Pure serendipity. I have just watched a TV programme about Guy Martin’s record breaking run of a free-wheeling go-kart down a French mountain road and break a world record at just over 85 mph.
During the design & development stage, the engineering team experimented with 20" v 26" wheels. Based on the experimental results, they opted for the 20" wheel set.
The downhill course was steep, relatively straight over 2km, and with a fairly smooth black-top surface. The Gravity Racer had to be steered and to keep the wheels in good contact with the road surface, they kept adding extra weight (bottles of water). They also had a decent set of brakes to slow the kart down at the end of the run.
To cut a long story short, 20"wheels 
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I think I would describe my initial method as “trial & error”
I looked at having just two photos, then three photos, then four. By this time, I could see a pattern emerging and an associated formula, which even covered the situation of just one photo.
Mrs R’s solution:
1 way of sending 10 photos
10 ways of sending 9 photos (choose each 1 of 10 not to send)
Choosing any 8 from 10 photos is a combination C(8,10) = 10!/(8!x2!) = 10x9/(2x1) = 45
7 photos: C(7,10) = 10!/(7!x3!) = 10x9x8/(3x2x1) = 120
C(6,10) = 10!/(6!x4!) = 10x9x8x7/(4x3x2x1) = 210
C(5,10) = 10!/(5!x5!) = 10x9x8x7x6/(5x4x3x2x1) = 252
C(4,10) is same as C(6,10) by symmetry
C(3,10), C(2,10) and C(1,10) follow similarly
Teaser requires at least 1 photo to be sent, so don’t need C(0,10).
Mrs R didn’t finish the arithmetic, but her method is correct. Answer is 1023 possibilities.
There is a shortcut that Mrs R forgot about that could reduce the arithmetic even further, possibly within the realm of mental arithmetic (at a pinch!).
There is an even more direct approach where the answer drops out instantly, though this method isn’t likely to be obvious. Hint: adding Mrs D’s selfie would increase the possibilities to 2047.
I will leave it another day before posting my solution(s).
Inching towards our solutions, those two numbers, 1023 and 2047 are correct (of course) and in my route, stem from 1024 - 1 = 1023 and 2048 - 1 = 2047
I think many people could manage the mental arithmetic up to about 15 photos. They could go a lot, lot further with pencil and a sheet of paper !
I have a feeling this could be the same or similar to your “… where the answer drops out instantly…”
Mrs R correctly noted that “combinations”, specifically mentioned in the teaser, suggests making use of Pascal’s Triangle which has appeared a few times before on this thread. She ruled it out, exclaiming: “I’m not writing out 10 rows of Pascal’s Triangle!”
Direct Method
I reminded her that she could go directly to the tenth row and complete the first half:
1
1 x 10/1 = 10
10 x 9/2 = 45
45 x 8/3 = 120
120 x 7/4 = 210
210 x 6/5 = 252
Instant Method
As Pascal’s Triangle can be produced by adding pairs of numbers from the row above, it means the row totals double each time. Hence the total of row 10 is 2^10 = 1024. The answer is reduced by 1 since the teaser specifies at least one photo must be sent.
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Nicely done Ravvie.
Yes, after a few rows, starting with 1, 2, 3 … photos, I came up with …
If n = number of photos, then the number of combinations ‘c’ will be c = (2^n) - 1
Which is quite easy to manipulate.
And as you say, the row totals double each time, so that with pencil and paper it’s quite easy to generate the figures for large values of ‘n’