Cylinder dimensions for Armstrong Siddeley Deerhound?

Discussion in 'Engines' started by Algernon, Oct 4, 2015.

  1. Algernon

    Algernon New Member

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    #1 Algernon, Oct 4, 2015
    Last edited: Oct 4, 2015
    Hallo.

    Cylinder dimensions for Deerhound I in “British Piston Aero-Engines and Their Aircraft” (by Alec Lumsden):

    Bore: 5.26 in. (135 mm)
    Stroke: 4.95 in. (127 mm)
    Swept Volume: 2259.75 cu. in. (38.19 litre)

    But...

    5.26 in. is (circa) 133.6 mm
    4.95 in. is 125.73 mm
    Vol. (circa) 2259 cu.in. OR 37.0 litre

    OR

    135 mm is 5.315 in.
    127 mm is 5.0 in.
    Vol. 2330 cu. in. OR 38.175 litre

    Deerhound cylinder dimension in metric units? Or British (imperial) units? (Previous Armstrong Siddeley engines is in imperial units.)

    Thank You.
     
  2. GregP

    GregP Well-Known Member

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    #2 GregP, Oct 4, 2015
    Last edited: Oct 5, 2015
    Hi Algernon. I'm not sure what you are saying. Are you wondering why the displacement doesn't multiply out for radials by taking the displacement for the bore and stroke and then multiplyiny by the number of cylidners?

    If so, the answer is that radials are usually designed around a master rod. The other rods connect to the master rod and they do not have quite the same stroke as the master rod since they are connected in a circle around the exterior of the master rod.

    Here is a pic:

    xr600-radial-master-link-rods.jpg

    You can see the master rod rotates about the stated stroke, around the center of the master rod's bottom end, but the other slave rods rotate through the pins located around the master rod and are not on the same center as the master. The difference isn't great, but it is present. It is impossible to calculate the actual theoretical displacement without the exact dimensions of the master rod so you can calculate the stroke for each slave rod.

    But the standard formulas for non-master-rodded engines will get you quite close for comparison purposes. If you were to do a standard calculation (volume of a cylinder 8 number of cylinder based on master) and then use it for HP/cubic inch or HP.cubic meter (centimer, millimeter, etc), you'd be very close.
     
  3. camman

    camman New Member

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    He's wondering whether the cylinder dimensions are supposed to be in inches or in millimeters. I don't know the answer to his question, but I certainly understand the problem that he's facing.
     
  4. GregP

    GregP Well-Known Member

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    #4 GregP, Oct 4, 2015
    Last edited: Oct 4, 2015
    A dimension is a distance. Inches or millimeters don't matter since there is an exact conversion. By definition, there are 25.4 mm in 1 in., so conversion is simple.

    The British started looking at metrification in 1818 and agfain in 1824. They set up a metrification commission in 1838 and decided in 1841 that metric coinage was required first. Another Royal Commision was et up in 1869. There was much arguing back and forth in the 1860's - 1870's.

    In 1896, Parliament passed the Weights and Measures (Metric System) Act, legalizing metric units for all purposes but not making them compulsory. In 1902, an Empire conference decided that metrication should be compulsory across the British Empire. In 1904, scientist Lord Kelvin led a campaign for metrication and collected 8 million signatures of British subjects. On the opposition side, 1904 saw the establishment of the British Weights and Measures Association for "the purpose of defending and, where practicable, improving the present system of weights and measures". At this time 45% of British exports were to metricated countries. Parliament voted to set up a Select Committee on the matter.

    This Select Committee reported in 1907 and a bill was drafted proposing compulsory metrication by 1910, including decimalisation of coinage. The matter was dropped in the face of wars and depression, and would not be again raised until the White Paper of 1951, the result of the Hodgson Committee Report of 1949 which unanimously recommended compulsory metrication and currency decimalisation within ten years.

    So suring WWII, the measurements were probably whatever system Armstrog Siddeley was using in-house. If there are any original drawing around, and if you cna find them, you have your answer.

    However, I might be able to help. See the pic below.

    Armstrog_Siddeley_Data.jpg

    This is a data table from an Armstrong-Siddeley document. Note the units are in imperial measurements. This document came from the archives of Enginehistory.org and would seem to indicate that Armstrong-Siddleley was not using metric units in-house in the 1939 timeframe.

    Note the old-style practice of not specifying tolerances. The bore is 5.4" and not 5.4" ± .001". There is probably some sheet of the document that states, "Tolerances are ±0.X inches unless otherwise stated," but I didn't manage to locate the tolerance limits in that document, so the tolerances could be called out in some standard British procurement document that I am not aware of.

    I seriously doubt they'd make a major change like going from Imperial to metric during wartime, making me pretty sure the Deerhound was specified in Imperial units.

    I couldn't care less which system of units is used myself, but I really dislike machines and mechanical constructs that employ two or more system of units. I'd rather work on something that is all SAE or all metric or all British Standard Fine or Coarse and not mess around figuring out whether a bolt is SAE or metric based on which set of wrenches fits the best.
     
  5. gumbyk

    gumbyk Well-Known Member

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    The conversions have been incorrectly done by the look of it. Now, were the original figures done in metric or imperial?

    P.S. my annoyance is items that are 'nominal' i.e. 50mm pipe - isn't 50mm, its 2"
     
  6. GregP

    GregP Well-Known Member

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    The Armstrong-Siddeley documents from the time are in Imperial units. I thought that answered the question in it's entirety.

    All he needs do is find an Armstrong-Siddeley document that describes the Deerhound. He's looking for the original Imperial units specs. Pretty much the end of the query unless he wants someone else to spend their time to locate the docs for him, and that is not indicated in the original post.
     
  7. Milosh

    Milosh Well-Known Member

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    You sure about that Greg?
     
  8. GregP

    GregP Well-Known Member

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    #8 GregP, Oct 5, 2015
    Last edited: Oct 5, 2015
    Absolutely, Milosh.

    If the entire master rod moved in a circle at the same angle to a reference, all the pistons would move in a circle of the same diameter. In fact, the top end of the master rod stays in a linear path along the bore while the bottom end rotates on the crankshaft. The rest of the pins do NOT describe a circle ... only the center of the crankshaft hole on the master rod does. The difference isn't much, but it is definitely there.

    That's why no radial engine will come out to exactly the displacement of the master cylinder multiplied by the number of cylinders, as an inline will. The rest of the cylinders don't quite exactly have the same stroke. Ask any engineer. A good reference to show the engineer if he or she doesn't know radial engines is the pic I posted above.

    Here is an animation:


    View: https://www.youtube.com/watch?v=liGPjzMJNLU

    As you can see, all the slave rods rotate about a point different from the piston's wrist pin center point. They are displaced by a small distance fron the crank journal center to any particular slave pin. The master rod's rotation goes from the piston's wrist pin center to the center of the crankshaft journal ... which is a different distance from the piston wrist pin than the slaves rod crank pins are.

    Each of the slave pins on the master rod describes a slightly flattened oval. It isn't a big difference, but the dispalcement will not add up if you use the master rod stroke. It will be off by a small amount.

    If the angle between the master rod and the slave rods never changed, ... they would all rotate in a circle. But the piston keeps the top end inline with the bore and the rotation meas the slaves rods all form a continuously-changing angle with the master rod vertical centerline. The result is the slave rods all move in slightly flattened ovals and do NOT have quite the same stroke as the master ... close, but not exactly the same. Each pair of rods moving away from the master vertical centerline have identical strokes, but the stroke of the first pair will not quite be the same as the next pair, etc. The diffrence will be in the second or third decimal place.

    To verify, you can model it in CAD. I have in the past. As I said, the differences are small enough to not matter much to the HP/cubic inch guys, who usually quote only to 1 decimal place (as in 6.4 pounds per HP) and most radials have their displacement rounded to the nearst 10 cubic inches anyway for marketing purposes. But if you want to get down to exact numbers, the difference will show up.

    Probably only an engineer or mathematician would notice ... or care. The actual difference is only 2- 3 cubic inches in 2800, so most people don't bother with the rather complex math. I did maybe 20 years ago just for a mental exercise, but the file has disappeared over the years and I can no longer read the old hard drives anyway.

    Think about it though. If the slave rods stayed at a constant angle over their up and down travel , the rod lower ends would describe circles. But they don't stay that way. Instead the upper ends travels in straight lines and the lower ends travels in almost-circles.

    Said another way, the master rod is not quite perpendicular to the slave rod when any slave rod passes top dead center (TDC) or bottom dead center (BDC). The difference in distance would be the sine of the master rod angle away from 90° multiplied by the slave rod length when it passes TDC or BDC.

    Again, not much, but it's there if you want to be precise. Most people probably couldn't care less and it wouldn't ever change the 10-cubic inch roundoff in any case I can think of. It's just a long exercise in math for no particularly good reason.

    Attached is a paper by someone else I located several years back that confirms my own calculations some 25 years ago. Any radial engine designer would already have done this. I only did it because I am an engineer, but NOT a mechanical engineer who designs engines, and was curious. After having done it, I am no longer curious about it.
     

    Attached Files:

  9. WJPearce

    WJPearce Active Member

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    #9 WJPearce, Oct 5, 2015
    Last edited: Oct 6, 2015
    In "Parkside: Armstrong Siddeley to Rolls-Royce 1939 - 1994" (Rolls-Royce Heritage Trust Historical Series No 39) by Roy Lawton, the bore and stroke are given as follows:

    Deerhound - 135 mm bore, 127 mm stroke
    Deerhound II - 140 mm bore, 127 mm stroke

    That info is found on page 36 and 225. Page 225 has a chart where other engines are listed with imperial units. The Deerhound is in metric. I think the metric numbers are correct and AS changed to metric for the Deerhound, Boarhound, and Wolfhound.

    According to my handy-dandy calculator, the bore and stroke info breaks out as follows:
    Deerhound - 135 mm (5.315 in) bore, 127 mm (5.000 in) stroke, 38.175 L (2,329.599 cu in) total displacement (not accounting for individual cylinder stroke variance).
    Deerhound II - 140 mm (5.512 in) bore, 127 mm (5.000 in) stroke, 41.055 L (2,505.347 cu in) total displacement (not accounting for individual cylinder stroke variance).
     
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  10. GregP

    GregP Well-Known Member

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    Anything published in 1994 or later in the UK would naturally list things in metric units since they converted officially well atfer WWII and the cold war.

    But during WWII their documents that I have seen were all in Imperial units.
     
  11. WJPearce

    WJPearce Active Member

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    I found some additional information.

    In 1932, Armstrong Siddeley began to design a new series of engines that were very different than their previous engines. The old engines were named after felines (cats), so they decided to name the new engines after canines (dogs). The new series used metric. I do not know why Armstrong Siddeley switched and it seems no one else does either.

    "Armstrong Siddeley—the Parkside story 1896–1939" by Ray cook lists the Deerhound as 135 mm bore and 127 mm stroke, the Deerhound III as 140 mm bore and 127 mm stroke, and does not list the Deerhound II. This book also states that it is believed that all dog-named engines were metric.

    "Sectioned Drawings of Piston Aero Engines" by Lyndon Jones lists the Deerhound II as 140 mm bore and 127 mm stroke, but does not list the Deerhound I or III. This book also states that all dog engines were metric.

    All I am saying in this post is that I think the metric bore and stroke numbers are correct. Yes, all sources are from the RRHT, but their company did buy Armstrong Siddeley.
     
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  12. camman

    camman New Member

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    Well, that settles it for me. Off to edit wikipedia I go!
     
  13. GregP

    GregP Well-Known Member

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    That's funny.

    The only Armstrong-Siddeley drawings I have seen are in Imperial.

    But I have bever asked the RRHT for a drawing myself that they actally supplied, so perhaps you are correct. I tried to get Allison drawings and they declined. Good thing I already HAD some factory Allison drawings. The ones I asked for are of later dash numbers.
     
  14. Algernon

    Algernon New Member

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    #14 Algernon, Oct 8, 2015
    Last edited: Oct 10, 2015
    WJPearce: Thank you, very much! Your posts not only answer for cylinder dimensions, but with more useful informations for me. Thank You!
    I not known stroke variations (on connecting rods), but most probably not too high. As i see russian „remont manual“ for M-82FN engine (or Shvetsov ASh-82FN, Швецов АШ-82ФН), its piston strokes varied (bore × stroke is usually as 155.5 mm × 155 mm):
    cylinder with master rod (master con-rods in cylinder Nos. 5 12), stroke 155 milimetres;
    articulated connecting rods (one row, with master con-rod in cylinder No. 5):
    4 6 — 155.05 mm
    3 7 — 155.824 mm
    1 2 — 155.474 mm
    (dimensions without tolerances)
    Swept volume for M-82 engine calculated by simple formula from dimensions 155.5 mm bore × 155 mm stroke is 41.210 litres;
    volume calculated by piston strokes variations is 41.313 litres. Or 2514.833 cu. in. versus 2521.082 cu. in.
    (I this “computed” on calculator on my comp, with “Pi” is 3.1415926535897 . . . etc.) Difference between this is neglibile — below 2.5 ‰ of swept volume.
    Mea Culpa! Books from RRHT is great sources, but in my library is only few — and all only for RR engines (most for pistons, only Dart between them).
    For me not too clear for british firms (but only few, two, or three?) at this times “switched” to metric, but Armstrong Siddeley not one, also de Havilland (piston engines as Gipsy Queen / Six and Gipsy King / Twelve):
    1938 | 1783 | Flight Archive
    With bore 118 mm and stroke 140 mm. (Or post-war Gipsy Queen Srs. 70, with bore × stroke is 120 × 150 mm.)

    And Alvis, with his pre-war licences French Gnome et Rhône engines (Alvis Alcides, Pelides, Mæonides). Or companies as Wolseley (built Hispano-Suiza V8 engines as Python, Viper and Adder), R.A.F. (adapted or redesigned french Renault engines), Sunbeam and Siddeley-Deasy at Great War? (This companies buy for licences — not all his engines is licences —, as Renault or Hispano-Suiza engines, french tools, machines, measuring instruments, etc., or converted this engines to british standards?)
     
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