Piston stroke question

  • Thread starter Thread starter Joe Ferreira
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Joe Ferreira

Joe Ferreira

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South Africa
Cape Town
Assume an engine has a piston stroke of 100mm and is for the purpose of this question positioned at bottom dead centre. The crankshaft now rotates exactly 90 degrees in its rotation towards top dead centre. At this point how far has the piston travelled up the cylinder? This is not a trick question! Post your answers please and your reasoning.
 
Can't give a accurate answer without knowing the length of the connecting rod. Need to know the angle the conrod makes with the centerline of the cylinder bore when the crank is rotated ninety degrees, because the distance traveled by the piston is a function of the cosine of that angle.
 
The length of the conrod is actually of no consequence to the question. You have the stroke length which is 100mm. You have a crank shaft that rotates 90 degrees. The crankshaft only has to turn another 90 degrees to move the piston up to top dead centre. So, I ask what (percentage if you wish) of the stroke is accomplished during the first 90 degree revolution of the crankshaft from bottom dead centre.
 
I believe the length of the connecting rod is necessary. Stroke 100mm would result in a turning radius r=50mm. If the rod is connected in the center of the piston head, it will always create an angle in any other configuration from 0 and 180 and thus a triangle with only one leg known in length. All other information is missing.

given the rod length d it would be:
h180-h0=100mm
h0=d-r, h180=d+r, -> r=50mm
h90=sqrt(r^2-d^2)

pos=h180-h90

not solvable with a under-defined triangle

for any given crank angle use law of cosines

Edit: Googleing after my try confirms my assumption that rod length is necessary
Edit2: lol sign error
 
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I'm trying to understand why the answer to this isn't just 50mm?

edit:
Wait... wait... almost... getting... head... round...
.. no gone again.

edit 2:
Okay, right I understand now. Yeah, rod length is required.

edit 3: Tree'd by Imari!

At 90° the end of the big end of the con-rod has gained 50mm (the radius of the crankshaft sweep) in vertical displacement (height) thanks to the crankshaft rotation, however since its also moved out by 50mm, the small end of the con-rod has only achieved a height of √(Con-rod length² - 50mm²). So for example, if the con-rod was 200mm, then it's vertical height when the crank is 90° is 193.65mm (it's lost 6.35mm because it's at an angle!), therefore the 50mm it gained - the 6.35mm it lost, means it actually on moved up by 43.65mm.
 
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Rod length of 200mm = piston is 43.6mm from the bottom.
Rod length of 1000mm = piston is 48.7mm from the bottom.

Rod length matters.
 
BobK & third Reign you are actually 100% correct you will need the length of the conrod. I did not expect an answer to the nearest 1000th of a mm. You can use a conrod 200mm centre to centre for your calculation. The reason I said the conrod length was of no consequence because if you take a conrod at 300mm centres the difference between the two conrods only differs in effect by approx 0.14%. I only brought up this question to dispell the general perception that the stroke is 50/50. Your calculation may well be more accurate than what I have from physically measuring an engine stroke. I'll still be glad to hear your result.
 
With a 200mm conrod the piston will be 43.65mm off the bottom.

x = r cos a + sqrt(l^2 - r^2 sin^2 a)
Because a = 90 degrees, cos a = 0 and sin^2 a = 1
x = sqrt(l^2 - r^2)
l=200mm=0,2m - conrod lenght, r=50mm=0,05m - crankshaft radius
x = sqrt(0,04 - 0,0025)
We get that x = 0,19365m(=193,65mm), it's the piston's position from the crankshaft center, so we take out 150mm, and get 43,65mm.
 
If an approximate answer is satisfactory then I'll go with "less than 50", which still illustrates the point I think you're trying to make.
 
Yes BobK you're quite right that was really my point, I did this exercise many years ago and over the years I have broached this subject with many mechanics and a few big boys in the motor industry, engine re-builders etc. Not one of them actually believed that this is so and still don't. I posted this question because I thought that it is an interesting aspect of an engine and a bit of trivial knowledge is better than none. You have only to ask a few of your friends in the automotive industry and I'll bet that they will not agree with you. Good thinking all of you that responded and I must admit the calculations were a bit over the top for me, I left school in 1957 and forgot most of what I learnt.
Cheers guys!
 
Jet Badger
50% or 50mm. Easy, assuming your engine uses scotch yoke.
Click to expand...

I somehow missed this post earlier. Jet Badger is quite correct.

Incidentally, why was this thread moved to the Cars in General subforum? It concerns a linkage that is found in many places besides automobiles. It's not even specific to internal combustion engines; it also applies to reciprocating steam engines. And air compressors.
 

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