HELP WITH CLUB WEIGHT VS. BALL SPEED DISCUSSION

pricenickypricenicky Members Posts: 122 ✭✭✭

If 2 drivers are exactly same weight ( 280 grams) / same length /loft / head / shaft and swung at the same speed and the only difference is 1 has a higher SW by 3 points ; which will produce a faster ball speed off the face? Your answer please.

Comments

  • ValtielValtiel Konica-Minolta Bizhub Members Posts: 2,333 ✭✭✭✭✭✭

    That amount of difference would be completely negligible. Besides, how would two drivers be exactly the same and have different swing weights? In order for that to technically be true, one driver would have to have 5 extra grams of weight on the head and 15 grams less in the grip, and at that point you have already introduced significant enough variables that it is no longer an apples to apples test. If what you want to know is whether or not a driver that is the same but has 5g more head weight would produce higher ball speeds when swung exactly the same, then the answer is technically yes, but only on a robot. Even the most minuscule differences in strike location or club head speed would be enough to completely override the fractions of 1mph that 5g could technically give you.

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  • mogc60mogc60 Members Posts: 975 ✭✭✭✭✭
    edited Aug 13, 2019 2:32am #3

    The one you hit the most solid in the best impact position. 3 swingweight is a lot and can seriously change path and impact angle. Me personally at D5 I can’t hit it consistently and at C9 I cant but at D2 or D3 with a driver I’m pretty darn consistent direction and distance

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  • musclefrontmusclefront Members Posts: 247 ✭✭✭

    Can’t all be the same but different swingweight

  • storm319storm319 Members Posts: 3,838 ✭✭✭✭✭✭✭
    edited Aug 13, 2019 3:40am #5

    @Valtiel hit the nail on the head.

    It is confusing as to why a supposed experienced club fitter/builder would be asking a question like this...
    (Post #95):
    https://forums.golfwrx.com/discussion/comment/19062373#Comment_19062373

  • Howard JonesHoward Jones Members Posts: 9,164 ✭✭✭✭✭✭✭
    edited Aug 13, 2019 5:28am #6

    @Valtiel said:
    That amount of difference would be completely negligible. Besides, how would two drivers be exactly the same and have different swing weights? In order for that to technically be true, one driver would have to have 5 extra grams of weight on the head and 15 grams less in the grip, and at that point you have already introduced significant enough variables that it is no longer an apples to apples test. If what you want to know is whether or not a driver that is the same but has 5g more head weight would produce higher ball speeds when swung exactly the same, then the answer is technically yes, but only on a robot. Even the most minuscule differences in strike location or club head speed would be enough to completely override the fractions of 1mph that 5g could technically give you.

    I agree with all of it, but i once made a chart for Head weight vs PTR value on drivers (cant find it now), where i tried to isolate the numbers to see if we gained anything by adding 15 grams of hot melt. (Head weight adjustment from 197 to 212 grams after going shorter)

    i cant find the chart anymore, only the text i wrote to it, and ive used a SPIN LOFT of 7.0* and then mathematical PTR changed from 197 grams at 1.473 up to 1.493 at 212 grams or a PTR gain of 0.020

    at 100 mph club speed x 1.473 = 147.3 mph ball speed
    same same...x 1.493 = 149.3 mph ball speed

    We gained 2.0 mph ball speed x 2 yards average = about 4 yards more carry by adding 15 grams, so at 3 SWP added .....(4 yards / 15 grams = 0.26 yards pr gram x 5 grams (1.65 pr SWP on a driver) = 1.3 yards more carry from adding 3 SWP....)

    So we can say that to gain 1 yard we need 4 grams average if Iron Byron is the player.

    Post edited by Howard Jones on
  • buntabunta Members Posts: 660 ✭✭✭✭✭
    edited Aug 13, 2019 5:11am #7

    @storm319 said:
    @Valtiel hit the nail on the head.

    It is confusing as to why a supposed experienced club fitter/builder would be asking a question like this...
    (Post #95):
    https://forums.golfwrx.com/discussion/comment/19062373#Comment_19062373

    lol d*mn you're just gonna put him on blast like that ?

    TS3, s55, SM7, Juno
  • buntabunta Members Posts: 660 ✭✭✭✭✭
    edited Aug 13, 2019 5:10am #8

    ..

    TS3, s55, SM7, Juno
  • Howard JonesHoward Jones Members Posts: 9,164 ✭✭✭✭✭✭✭

    @storm319 said:
    @Valtiel hit the nail on the head.

    It is confusing as to why a supposed experienced club fitter/builder would be asking a question like this...
    (Post #95):
    https://forums.golfwrx.com/discussion/comment/19062373#Comment_19062373

    Club makers and Club makers is not the same, and we are now into a area where less than 1 out of 1000 is able to answer, this DONT belong to any classes for this, its only Nerds like me who checks every rabbit hole there is...just to make sure, if we dont, we never find out, and thats what most club makers do, they dont go further than the books, and they stops expanding their knowledge way before we are at this level.

    If you look over my post above, you will also understand why its left out of the books, it does not matter, and real life adding 5 grams changes "everything" (for the player) , we are not like iron byron, so its all "on the paper only" and differences so small, we will NOT be able to isolate them in a practical test and say, YES we gained 1.3 yards...that want happen for THIS reason if we see that number, so real life we could loose or gain distance, we DONT use head weight adjustments for this reason, but to improve impact, and then we might see quite different numbers if the players PTR was low, but now moves to the top.

  • pricenickypricenicky Members Posts: 122 ✭✭✭

    YES I am a very experienced club maker and I WANTED to ask this question without any bias. 1 of the clubs( g400 without weight) had a 25 gram grip/ 203 gram head and was d5; the other( TS1 ) had a 52 gram grip and a 187 gram head and was d2.The answer is= FORCE = M X V SQUARED. The heavier head does generate more force at impact @ the same speed = higher ball speed.Thanks for all the replies

  • FadeFade Members Posts: 1,207 ✭✭✭✭✭✭

    Force is not M x V Squared.

  • ThinkingPlusThinkingPlus South TexasClubWRX Posts: 1,722 ClubWRX

    @pricenicky said:
    YES I am a very experienced club maker and I WANTED to ask this question without any bias. 1 of the clubs( g400 without weight) had a 25 gram grip/ 203 gram head and was d5; the other( TS1 ) had a 52 gram grip and a 187 gram head and was d2.The answer is= FORCE = M X V SQUARED. The heavier head does generate more force at impact @ the same speed = higher ball speed.Thanks for all the replies

    No. Just no. Force = mass * acceleration, momentum = mass * velocity, Energy = 1/2 * mass * velocity ^2. These are the simple linear motion formulas. There are rotational equivalents that are really more appropriate for the golf swing, but I don't have them committed to memory as well.

    Driver: Callaway GBB Epic 9° w/Project X HZRDUS T800 65 gm 6.0 flex
    3W: Callaway Rogue w/Project X Evenflow 5.5 Graphite R-flex
    Hybrids: Callaway Apex 3h, 4h w/MR Kuro Kage 80HY S-flex
    Irons: Maltby TS-1 5i-GW w/KBS Tour R-flex
    Sand Wedge: Titleist Vokey SM7 54/08 M Grind w/KBS Tour R-Flex
    Lob Wedge: Titleist Vokey SM6 58/04 L Grind w/TT Wedge Flex
    Putter: Scotty Cameron Futura X w/Super Stroke Claw 1.0
    Ball: Titleist AVX (wind) or ProV1X (tournaments) in yellow
  • pricenickypricenicky Members Posts: 122 ✭✭✭

    YES I am a very experienced club maker and I WANTED to ask this question without any bias. 1 of the clubs( g400 without weight) had a 25 gram grip/ 203 gram head and was d5; the other( TS1 ) had a 30 gram grip and a 187 gram head and was d2.The answer is= FORCE = M X V SQUARED. The heavier head does generate more force at impact @ the same speed = higher ball speed.Thanks for all the replies

  • JeffreySpicoliJeffreySpicoli Members Posts: 1,956 ✭✭✭✭✭✭

    It’s a little crazy in here!

  • ValtielValtiel Konica-Minolta Bizhub Members Posts: 2,333 ✭✭✭✭✭✭

    @JeffreySpicoli said:
    It’s a little crazy in here!

    YES! I am a very experience club mak.....ah nevermind. :D

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    Nike Tour Issue SQ2 17* Diamana Blueboard 103x || SQ2 15* Diamana Blueboard 93x
    PING Anser 20* Aldila Rogue Black 110MSI 105h Tour-X || Taylormade V-Steel 21* Project X Rifle Satin 6.5
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    Mizuno MP-H4 4i 24* Project X PXi 7.0
    Mizuno MP-59 4i-PW 24*- 48* Brunswick Precision Rifle FCM 7.0
    Vokey Mild Raw 8620 54* Brunswick Precision Rifle FCM 7.0
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    Vokey Special 62* Black Oxide V-Grind Brunswick Precision FCM 7.3 SSx2
    Scotty Cameron Santa Fe Bullseye shaft

    WITB Thread
  • Howard JonesHoward Jones Members Posts: 9,164 ✭✭✭✭✭✭✭

    @pricenicky said:
    YES I am a very experienced club maker and I WANTED to ask this question without any bias. 1 of the clubs( g400 without weight) had a 25 gram grip/ 203 gram head and was d5; the other( TS1 ) had a 52 gram grip and a 187 gram head and was d2.The answer is= FORCE = M X V SQUARED. The heavier head does generate more force at impact @ the same speed = higher ball speed.Thanks for all the replies

    Your club heads is not the same heads, neither is the shafts, and like ive explained, "on the paper", 15 grams added might give 4 yards "all else equal", but nothing is equal here, so your compare is not valid, you are not able to isolate head weight in your compare. because you dont use the same head and shaft or grips, that means you have not found the answer to why one of them is longer hitting then the other if that was the question.

    The head itself could be the reason, or impact spot on the face (actual CT or COR at impact spot vs point of measured club speed), only tracing impact and by using a LM you get to see why, but that cant be done when we introduce different heads to the picture.

  • Stuart_GStuart_G New HampshireMembers Posts: 23,844 ✭✭✭✭✭✭✭

    @ThinkingPlus said:
    No. Just no. Force = mass * acceleration, momentum = mass * velocity, Energy = 1/2 * mass * velocity ^2. These are the simple linear motion formulas. There are rotational equivalents that are really more appropriate for the golf swing, but I don't have them committed to memory as well.

    I just fall back on Tutelman's simple approximation:
    Vball = V_clubhead * ((1 + e)/(1 + m/M)) * cos(loft)
    e = COR, m = ball mass, M = club head mass, loft is dynamic loft not static

  • Howard JonesHoward Jones Members Posts: 9,164 ✭✭✭✭✭✭✭

    @Stuart_G said:

    @ThinkingPlus said:
    No. Just no. Force = mass * acceleration, momentum = mass * velocity, Energy = 1/2 * mass * velocity ^2. These are the simple linear motion formulas. There are rotational equivalents that are really more appropriate for the golf swing, but I don't have them committed to memory as well.

    I just fall back on Tutelman's simple approximation:
    Vball = V_clubhead * ((1 + e)/(1 + m/M)) * cos(loft)
    e = COR, m = ball mass, M = club head mass, loft is dynamic loft not static

    If you ask Trackman, they would say Dynamic loft is useless here, we need to use SPIN LOFT, but the math is the same.

  • Stuart_GStuart_G New HampshireMembers Posts: 23,844 ✭✭✭✭✭✭✭
    edited Aug 14, 2019 11:15am #19

    @Howard Jones said:

    @Stuart_G said:

    @ThinkingPlus said:
    No. Just no. Force = mass * acceleration, momentum = mass * velocity, Energy = 1/2 * mass * velocity ^2. These are the simple linear motion formulas. There are rotational equivalents that are really more appropriate for the golf swing, but I don't have them committed to memory as well.

    I just fall back on Tutelman's simple approximation:
    Vball = V_clubhead * ((1 + e)/(1 + m/M)) * cos(loft)
    e = COR, m = ball mass, M = club head mass, loft is dynamic loft not static

    If you ask Trackman, they would say Dynamic loft is useless here, we need to use SPIN LOFT, but the math is the same.

    True. It's loft relative to the path of the head (assuming face-to-path is zero). I wasn't thinking of "dynamic loft" from the TM definition - but probably should have realized most would assume I was.

  • ThinkingPlusThinkingPlus South TexasClubWRX Posts: 1,722 ClubWRX

    @Stuart_G said:

    @ThinkingPlus said:
    No. Just no. Force = mass * acceleration, momentum = mass * velocity, Energy = 1/2 * mass * velocity ^2. These are the simple linear motion formulas. There are rotational equivalents that are really more appropriate for the golf swing, but I don't have them committed to memory as well.

    I just fall back on Tutelman's simple approximation:
    Vball = V_clubhead * ((1 + e)/(1 + m/M)) * cos(loft)
    e = COR, m = ball mass, M = club head mass, loft is dynamic loft not static

    Yep. Was eating dinner out so didn't want to do search and destroy via phone. Thanks.

    Driver: Callaway GBB Epic 9° w/Project X HZRDUS T800 65 gm 6.0 flex
    3W: Callaway Rogue w/Project X Evenflow 5.5 Graphite R-flex
    Hybrids: Callaway Apex 3h, 4h w/MR Kuro Kage 80HY S-flex
    Irons: Maltby TS-1 5i-GW w/KBS Tour R-flex
    Sand Wedge: Titleist Vokey SM7 54/08 M Grind w/KBS Tour R-Flex
    Lob Wedge: Titleist Vokey SM6 58/04 L Grind w/TT Wedge Flex
    Putter: Scotty Cameron Futura X w/Super Stroke Claw 1.0
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